B-Rob's Advanced Math 3rd hr. 09 - 10

Dustin's Final Reflection

2010-05-26T09:30:00.002-05:00

This year has been a tough year, but a good year. Although my grades weren't great, I learned alot. The major things we covered were trig and calc. I loved trig when it was just triangles, then this year I saw a different side of trig. It was difficult, but I understood it in the end. It will help me in Calculus and it will help me in college, so it was worth the hard work.

We also learned a lot of calculus towards the end of the year. That was the one thing that I just perfectly understood at the end of the year. This year was great and I loved my class. I could've done much better and I will remember to do that next year after seeing how my grades turned out.

Devin's Reflection

2010-05-24T21:31:00.000-05:00

How to Find the Inverse of a Function:

Replace f(x) with y
Reverse the roles of x and y
Solve for y in terms of x
Replace y with f-1(x)
check: should equal to x

Example:

f(x) = √x + 4

(x)^2 = (√y + 4)^2

x^2 = y + 4

y = x^2 - 4

f-1(x) = (x^2 - 4)

f(f-1(x)) = f(x^2 - 4) = √(x^2 - 4) + 4 = x

f-1(f(x)) = f-1(√x + 4) = (√x + 4)^2 - 4 = x + 4 - 4 = x

Logarithm Properties:

logb MN = logb M + logb N
logb M/N = logb M - logb N
logb M^K = K logb M
logb b^k = k
b^logb^k = k

Changing Bases: (Done when you can't solve a log)

Rewrite it as an exponential
Take the log of both sides
Move the variable to the front
then solve

Example:

log5 10 = x

5^x = 10

log 5^x = log 10

x log 5 = 1

x = 1/log 5

Devin's Reflection

2010-05-24T21:29:00.000-05:00

Completing the Square:

You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².

For example:

x² + 6x - 2 = 0

x² + 6x = 2

x² + 6x + 9 = -2 + 9

(x + 3)² = 7

x + 3 = √7

x = -3 ± √7

(-3 + √7,0) (-3 -√7,0)

Rational Root therom:

Example: f(x)= 2x^3 + 3x^2 - 8 + 3

Step 1: find all possible roots..
p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2

*p is the leading constant term & q is the leading coefficient
possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2

Step 2: plug roots in calc & the zeros will be: 1, 1/2, -3

Step 3: synthetic division: (x - 1) (2x^2 + 5x + 3)

Step 4: slove further (factor): (x - 1) (2x^2 + 5x + 3)= (x - 1) (2x - 1) (x + 3)

Answer: x = 1, 1/2, -3

Domain & Range of functions:

Polynomials-domain of all polynomials is (−∞, ∞).

Fractions-you set the bottom to zero, solve for x, and then set up intervals

Square Roots-domain: set the inside = to zero, then set a # line, try values on either side of each #, and get ride of the negatives-range:graph

Absolute Value-domain: (- ∞ , + ∞)-range: [0 , + ∞)

Devin's Reflection

2010-05-24T21:28:00.000-05:00

SOLVING TRIG EQUATIONS

for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2

y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift

Identities:

check identities
algebra
identities

Reciprocal Relationships

cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ

Relationships with Negatives

sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ

Pythagorean Relationsihps

sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ

Cofunction Relationships

sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)

Devin's Reflection

2010-05-24T21:26:00.001-05:00

Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)

1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°

Unit Circle

sin theta = y/r

cos theta = x/r

tan theta = y/x

csc theta = r/x

sec theta = x/y

cot theta = x/y

r=sqrtx^2+y^2

90 pi/2
180 pi
270 3pi/2
360 2pi

sin pos in one and two and neg in three and four

cos pos in one and four and neg in two and three

tan pos in one and three and neg in two and four

cot pos in one and neg in two three and four

sec pos in one and neg in two three and four

csc pos in one and two and neg in three and four

AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)

RIGHT TRIANGLES

hypotenuse opposite angle
a=1/2bh
SOHCAHTOA

sin theta=opposite/hypotenuse
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent

LAW OF SINES
sinA/a sinB/b sinC/c

used when you know pairs in a non-right triangle
you are setting up proportions

Right Triangles

A=1/2bh
SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443

Area of Inscribed Shapes
A=nr²sinΘcosΘ

Devin's Reflection

2010-05-24T21:24:00.001-05:00

Triangle trigonometry:
sine = opp/hyp
cos = adj/hyp
tan = opp/adj
csc = hyp/opp
sec=hyp/adj
cot=adj/opp

Moving between quadrants:

I to IV = make it negative and add 360º
I to III = add 180º
I to II = make it negative and add 180º

II to IV = move 180º

Law of sines (sinA/a) = (sinB/b) = (sinC/c)
Law of cosines (opp leg)² = (other adj leg)² -2(adj leg) (adj Leg) cos (angle b/w)

sinΘ = y/r
cosΘ = x/r

tanΘ = y/x
cotΘ = x/y
cscΘ = r/y

secΘ = r/x

r=√(x² + y²)

Graphing Trig functions:

y=Asin(Bx-h)+C

A = amplitude or height

B determines period(p)

p = (2π/B)

h = ((phase shift-horizontal shift)/(opposite))

C= vertical shift

Trigonometric Identites:
csc = 1/sin x
sec = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x+cos^2 x = 1
1+tan^2 x = sec^2 x
1+cot^2 x = csc^2 x
sin x = cos(90*-x)
tan x = cot (90*-x)
sec x = csc (90*-x)
cos x = sin (90*-x)
cot x = tan (90*-x)
csc x = sec (90*-x)
tan x = sinx/cosx
cot x = cosx/sinx

Devin's Reflection

2010-05-24T21:22:00.001-05:00

Ch. 6 Conics

The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)

To find the intersection of a line and a circle:

1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.

If your x value is imaginary, then there is no point of intersection.

EX: find the center and radius.(x-3)^2+(y+7)^2=19

c:(h,k)

center: (3,-7)

radius: square root of 19

--Parabolas have no major axis and no asymptotes.

Axis of symmetry x=-b/2a

Finding the vertex
(-b/2a, f(-b/2a))

or

complete the square to get vertex form
y=(x+a)^2+b a&b are #'s

(-a,b) vertex

focus: 1/4p= coeff of x^2 then add p.

directrix is p units behind vertex. subtract p.

EX: 1/8y^2

v(0,0)

Focus: p=2(2,0)

directrix: x=-2

Devin's Reflection

2010-05-24T21:20:00.001-05:00

CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)

ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph

HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom

to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes

Devin's Reflection

2010-05-23T21:57:00.001-05:00

1.) area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)

Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.

A = (1/2)(2)(6)sin(65)

A = 5.438

2.) Law of Sines(used to non-right triangles):

Sin A/a = Sin B/b= Sin C/c

Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.

A = 1/2 (4) (5) Sin 30 degrees

A = 10 Sin 30 degrees which is aproximately = 5

3.) Law of Cosines (used when you can't use Law of Sines):

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)

Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.

7^2=6^2+5^2-2(5)(6)

cos a7^2-6^2-5^2= 2(5)(6)

cos acos a= 7^2-6^2-5^2 / -2(6)(5)

a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))

a= 78.463 degrees

4.) For any line : m = tan (alpha)
**m = slope , (alpha) = angle of inclination

5.) For a conic: tan 2 (alpha) = B/A-C
**if A=C then pie/4 (always)
**A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2

Examples:

1. Find the angle of inclination of x^2 - 2xy + 3y^2 = 1.

tan 2 (alpha) = B/A-C

A = 1 , B = -2 , C = 3

tan 2 (alpha) = -2/1 -3 = 1

tan 2 (alpha) = 1

2A = tan^-1 (1)

2 (alpha) = 45 , 225

alpha = 45/2 , 225/2

alpha = 22.5 , 112.5

2. x^2 + y^2 - 3xy + 4x - sqrt.

x = 1alpha = 1 (because A = 1 & C = 1 so A = C)

Devin's Reflection

2010-05-23T21:54:00.000-05:00

**Logs

Condense:

Ex) logm + log7 + 4logn

= log7mn^4

Ex) 5loga + logd + log6

= log6da^5

Ex) 4logt - logc

= t^4/c

Ex) logn - 3logh -logy

= n/yh^3

Expand:

Ex) log5gh^2

= log5 + 2logh +logg

Ex) m^3b^7/f

= 3logm + 7logb - logf

**The Unit Circle

90 degrees, (0,1), pi/2

180 degrees, (-1,0), pi

270 degrees, (0,-1), 3pi/2

360 degrees, (1,0), 2pi

**6 Trig Functions

sin = y/r

cos = x/r

tan = y/x

csc = r/y

sec = r/x

cot = x/y

**Degrees & Radians

Degrees to radians= Degree * pi/180

Radians to degrees= Radians * 180/pi

**To solve coterminal angles, either add or subtract 360 to the angle.

Devin's Reflection

2010-05-23T21:52:00.001-05:00

Simplifying Trig Function

1. Check identities

2. Algebra (factoring, combining like terms, and fraction)

3. Check identites

Some Proofs to help

-cotx= cosx/sinx

-tanx= sinx/cosx

-1+cot^2x=csc^2x

-1+tan^2x=sec^2x

sin^2+cos^2=1

The way to simplify is to find and rearrange the functions in a way the makes them resemble one of the proofs. After you have done that then you use the proofs to replace something in the equation. And then you keep repeating those steps until you can not do it anymore without making the equation bigger.

Ex. Prove sec^4x-tan^4x/sec^2x

(sec^2x-tan^2x)(sec^2x+tan^2x)/sec^2x

1=sec^2x-tan^2x

1(sec^2x+tan^2x)/sec^2x

sin^2x/cos^2x/1

=1+sin/2x

Devin's Final Reflection

2010-05-16T23:46:00.002-05:00

This year has been filled with much knowledge and alot (ALOT) of difficulty. But this year I learned alot of information that I will much need in the next educational level. I learned trig and even though it was difficult, it will be a major help in calc and college. Even though trig was hard, it was not the hardest thing for me to learn this year. The section that gave me the most difficuly, was the section dealing with sequences.

Taylor final reflection

2010-05-16T22:44:00.003-05:00

The major concepts of Advanced math were all the things dealing with trig
the focus and most stress of this year was learning and dealing with trig
all trig stems from the unit circle and the trig chart so ill include that

The Unit Circle

90 degrees, (0,1), pi/2

180 degrees, (-1,0), pi

270 degrees, (0,-1), 3pi/2

360 degrees, (1,0), 2pi

TRIGCHART
0° *sin 0= 0 *cos 0= 1 *tan 0= 0 *csc 0= undefined *sec 0= 1
*cot 0= undefined

30° * sin π/6= 1/2 *cos π/6= √3/2 *tan π/6= √3/3 *csc π/6= 2 *sec π /6= 2 √3/3
*cot π/6= √3
45° *sin π/4= √2/2 *cos π/4= √2/2 *tan π/4= 1 *csc π/4= √2 *sec π/4= √2
*cot π/4= 1
60° *sin π/3= √3/2 *cos π/3= 1/2 *tan π/3= √3 *csc π/3= 2 √3/3 *sec π/3= 2
*cot π/3= √3/2
90° *sin π/2= 1 * cos π/2= 0 *tan π/2= undefined *csc π/2= 1 *sec π/2= undefined *cot π/2= 0

i feel like this year really helped me with advancing my math score on the act
when i took the act last june i made a ninteen in math but this time when i took it in april i made a twenty five
i really did learn alot thus year

finally the one moment where something really clicked was when i missed the lesson on refrence anglwa
when mrs robinson broke down the lesson into three easy steps i got it easily

there are three steps to finding a refrence angle
STEPS
pre step- set up ____ trig function____
the rest will plug into this
#1- discover which quadrent the given angle is in
#2- determine if the given trig function ((which will go in the "trig function" spot)) is positive or negative in that quadrent ((this will go into the first blank))
#3- change given angle to some number between 0 and 90 degrees by subtracting 180 until the angle is between 0 and 90 degrees. ((the number that you get which is between 0 and 90 degrees will go in the second blank))

ALAINA'S FINAL REFLECTION

2010-05-16T21:45:00.002-05:00

This year we have covered a lot of material. B-rob has helped and has been a great teacher. The class wasn't "easy" but it wasn't "hard" either. I struggled on some concepts but i eventually understood them.
the major concepts were:
review on algebra 2
solving polynomials
inequalities
domain and range
exponents
conics
trigonometry
trig with triangles
solving trig equations
trig formulas
polar
sequences and series

What did I gain?
As this course is coming to an end, I feel as if I've gained a mutual knowledge of the subject. I learned how to graph conics and actually understood it and used trig to build a bridge.

Methods that helped me:
Well first off, B-rob is a great teacher. She's very visual and explanitory. She gives a lot of examples of each type of whatever it is we're learning. Also, I have to work some things on my own in order to understand them.

Final Blog!

2010-05-16T20:34:00.002-05:00

Okay, so I think that one of the most major concepts that we covered this year was knowing the Trig Identites:
csc = 1/sin x
sec = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x+cos^2 x = 1
1+tan^2 x = sec^2 x
1+cot^2 x = csc^2 x
sin x = cos(90*-x)
tan x = cot (90*-x)
sec x = csc (90*-x)
cos x = sin (90*-x)
cot x = tan (90*-x)
csc x = sec (90*-x)
tan x = sinx/cosx
cot x = cosx/sinx

To me this and the trig chart were two of the most important things we learned because you had to memorize all these identities, and without them you would not be able to do a lot of the problems in trig, and memorizing the chart helps you simplify an answer to the simplest form possible.

An activity that really helped me this year was when B-rob told me that the whole trig chart can be derived by just knowing the first few lines. sin cos and tan's answers can be flipped and that gives you the csc sec and cot's answers.

Devin's Make up

2010-05-12T22:44:00.001-05:00

Exponents:

1. b^x * b^y = b^x + y....example: 2^3 * 2^5 = 2^8

2. b^x/b^y = b^x - y....example: 5^7/5^4 = 5^3

3. (ab)^x = a^xb^x....example: (3 * 7)^3 = 3^3 * 7^3

4. (a/b)^x = a^x/b^x....example: (3/5)^3 = 3^3/5^3

5. (b^x)^y = b^xy....example: (2^2)^3 = 2^6

6. b^x/y = y^√b^x....examples: 5^3/4 = 3^√5^3

7. to solve for exponents:

write as the same base
set exponents equal
then solve for x

examples:

(a). 5^3x = 5^7x - 2

Examples:
log3^9=2
In order to solve, you need to put this into exponential form...
3^2=9
...and that is your answer.

log2^16=4
2^4=16

log5^125=x
5^x=125 (if it asks for exponential form)
x=3 (if it says to solve)

log x=2
10^2=x
x=100

logx^8=3
x^3=8
x=2

Devin's Make up

2010-05-12T22:42:00.001-05:00

When it is b^x*b^y= b^x+y. When it s b^x/b^y= b^x-y. When it is (ab)^x= a^xb^y. When it is (a/b)^x= a^x/b^x. When it is (b^x)^y= b^xy. When it is b^x/y= y{b^x}.

To solve for an exponent
a. write as the same base
b. set exponents equal
c. solve for x

Simplfy
(b^2/a) ^-2

-b^-4/a^-2

-1/b^41/a^2

= a^2/b^4

With double fraction, you have to multiply the outsides by each other, and the insides byeach other.

Ex. (a^-2+b^-2)^-1

-(1/a^2+1/b^2)^-1

-(b^2/b^2*1/a^2+1/b^2*a^2/a^2)^-1

-(b^2/a^2b^2+a^2/a^2b^2)^-1

-(b^2+a^2/a^2b^2)^-1

=a^2b^2/b^2+a^2

This week we also covered logirythms (I doubt I spelled that correctly).
logb x=a

- b^a=x

Ex. log2 8=x

-2^x=8

x= 3

Domain of logs and ln = (0,infinity)
Range of logs and ln = (-infinity, infinity)

Devin's Make up

2010-05-12T22:39:00.001-05:00

Logarithm Properties:

logb MN = logb M + logb N
logb M/N = logb M - logb N
logb M^K = K logb M
logb b^k = k (this one i don't get..maybe i copied it wrong)
b^logb^k = k

Here are some examples:

1. log 2 + log 3 + log 4 = log 24 (mulitply: 2 x 3 x 4)

2. log 8 + log 5 - log 4 = log 10 (mulitply: 8 x 5 then divide: 40/4)

3. 2 ln 6 - ln 3 = ln 12 (raise 6 to the 2nd power = 36 the divided by 3 = 12)

4. log M - 3 log N = log M/ N^3

5. ln 2 + ln 6 - 1/2 ln 9 = ln 12/3 = ln 4

6. Expand logb MN^2....logb M + 2 logb N

7. Condense log 45 - 2 log 3....log (45/9) = log 5

8. Rewrite in exponetial form: log36 6 = 1/2....36^1/2 = 6

9. Rewrite in logarithmic form: 2^2 = 4....log2 4 = 2

Changing Bases: (Done when you can't solve a log)

Rewrite it as an exponential
Take the log of both sides
Move the variable to the front
then solve

(use the same steps when solving for x as an exponent when you can't write them as the same base)

examples:

1. log5 10 = x

5^x = 10

log 5^x = log 10

x log 5 = 1

x = 1/log 5

2. 2^x = 7

log 2^x = log 7

x log 2 = log 7

x = log 7/log 2

Devin's Make up

2010-05-12T22:36:00.001-05:00

How to change bases:
1)rewrite problem in exponential form
2)take the log of both sides
3)move the variable to the front
4)solve
Eg. log5of10=x
5^x=10
log5x=log10
xlog5=1
x=1/log5

Graphing exponential functions (ab^x):
if b is greater than 1, the graph goes up whereas if it is less than one, the graph goes down
Eg. f(x)=5(3)^x the graph would go up because b is greater than one

The formulas are pretty ease but remembering them will be pretty hard and using them in the correct problem will also be kinda hard.

A(t)=Ao(1+r)^t
Ao = what you start with
r = rate
t = time

A(t)=Aob^t/k
Ao = what you start with
t = time
b = double, half, etc.
k = regular time to double, half, etc.
Eg. half-life of 5 days
b = ½
k = 5
A(t)=Ao(1/2)^t/5

P(t)=Poe^rt
Po = wha tyou start with
r = rate
t = time
you only use the problem with p when it says compounding continuously in the problem

then theres the limit thing and rule of 72 (72 / r% = how long it takes to double

Devin's Make up

2010-05-12T22:28:00.000-05:00

Exponential functions

All you have to do is basically plug things in and then ur set. There are 3 different formulas:
A(t)=Ao(l+r)^t
A(t)=Aob^t/k
Ao=what you start with
b=double, half, etc (use this if u see double, half, etc in the problem)
k=time reg. to double, half, etc
t=time
P(t)=Poe^rt
Po=what you start with
r=rate
t=time
(only use this when compounding continuously)

All you basically have to do is plug the numbers in and solve if it says so.

The steps for condensing logs are easy
First you have to remember the relations
Mn = m+n
m/n = m-n
m^k = k log M
sub b B^k = k
b^log sub b^k = K

so any problem will fit into one of these relations
Ex: expand log sub b MN^2
Log sub b M + 2 Log sub b N

Alicia's Final Reflection :)

2010-05-10T19:46:00.002-05:00

Okay so the senior's final exam is tuesday... chapters 1-6 and 13!!! I am going to review some stuff from a few of these chapters:

Exponents:

* b^x * b^y = b^x + y

* b^x/b^y = b^x - y

* (ab)^x = a^xb^x

* (a/b)^x = a^x/b^x

* (b^x)^y = b^xy

* b^x/y = y^√b^x

Changing Bases:

* Rewrite it as an exponential

* Take the log of both sides
* Move the variable to the front
* solve

Ch. 13

* Arithmetic- tn*t1+(n-1)d

n=term # t1=first term d=what you add

* Geometric- tn=t1*r^(n-1)

r= what you multiply by t1= first term

Examples:

1. Find the formula for the nth term of the arithmetic sequence: 3,5,7,...

tn = 3 + (n-1) (2)

tn = 3 +2n - 2

tn = 1 + 2

2. Find the formula for the nth term of the sequence: 3,4.5,6.75,..

* Don't forget to divide the 2nd term by the 1st term to find r

4.5/3 = 3/2 = r

tn = 3 (3/2)^(n-1)

taylor rodriguez reflection 10 may 2010

2010-05-10T08:11:00.003-05:00

soving and sketching parabolas

write this in your notes as you see it posted. it should help your graphing problem.

**#1
you need to see if the parabola will open up or down. think of it this was: if the first thing you see in the equation is a negative sign relate that to which way negative numbers go on a graph or think "if some thing is negative you get a thumbs down" like wise "if something is positive it gets a thumbs up"

so first thing you see at the front of the equation is a negative sign? thumbs down therefore the parabola opens down. If the first thing you see is a positive number? thumbs up therefore the parabola opens up.

(using analogies like this is good for memory. If you start thinking in terms of analogies you get faster at retaining information)

**#2
deciding the number of X intercepts is also an easy remembering problem to fix.
first you need to answer the problem

bsquared - 4(a)(c)

as you said you are very good at plugging in this formula because you have remembered it well.
look at your answer to that and

remember: positive answer is two x intercepts
negative answer is none
zero for an answer is one X intercept

its better to have two than none
so POSITIVE thing to have TWO
NEGATIVE thing to have NONE

(i dont have a trick to remember zero.. i think its just a process of elimination thing.. if i didnt get a positive answer or a negative anser that means its not two x intercepts nor is it no X intercepts,, well that means its one X intercept)

**#3
to find an x intercept you solve for X

it says that in your question

"find X intercept"
remember "find X"

(dont forget to put answer into point form. when solving for x you will always wind up having to square root. you know this meas the answer will be +/-. be sure to show this when convertine to point form. {I.E. (#,0) & (-#,0)} in many of the problems we had there was also a matter of carring a number to the other side. this is no big deal you just tack it on also. for example... if you ended with

X-2= +/- square root 6/2

you would add 2 to both sides and put in point form. therefore you'd have
(squareroot 6/2+2,0) & (- squareroot 6/2+2,0))

**#4
y- intercept is just taking the 0 in the y spot for the last answer and plugging it into the x spot in the equation. which then leaves you only the Y variable to solve for.

Remember: "Find Y intercept"
"find Y"

common sense will tell you the only way to do that is to plug something into the X spot.. and i told you what to plug in

**#5
Axis of semmatry is a simple conversion formula you'll have to memorize the same way you did for the quadratic formula. by writing it down everytime you solve for axis of semmatry until you see the formula in your sleep.

the Formula (in case it isnt written down) is

X= -b/2(a)
(the a and b plug ins of course come from the original equation)

your answer will be the point to put your DOTTED LINE on. because this formula solves for X you know it will pass through that point on the X line. You also know its a vertical line. so no worries.

**#6
the vertex is also just a matter of plugging in
remember this step follows the step ahead of it so it retains the answer for X

that means half of your vertex is solved
you already have your X point for the vertex answer

that answer is also plugged into the original equation which again leaves you to solve for y.

this means you now have your vertex point
because you solved for X in step 5
and your answer after plugging that in gave you the Y

FINALLY! now that you have turned everything into points its just a matter of locating them and marking them all on your graph.

After each point is marked connect the dots.

just as a quick check look back and see if your parabola is supposed to open up or down if your graph matches then

congratulations! everything seems to have gone
well.i know everything i've given you is alot to remember. just write down the hints and keep the sheet as a reference. it doesnt have to be word for word. just putting "step one- opens up or down? work {b-4ac} *positive thumbs up *negative thumbs down"

reading and rereading tricks to help you remember will pay off i promise

i need help on finding inverses like those is chapter four

Reflection on old stuff

2010-05-09T22:11:00.002-05:00

Conics:
The steps to find the intersection of a line and a circle are: solve the linear equation for y, next substitute in circle equation, after this you solve for x, and last you plug x value in to get y value **If your x value is imaginary, then there is no point of intersection.

Example:
x^2+y^2+12y+16x-5=0

First you rewrite the problem in order with x's in front and y's in back, or vice versa, and you get this:
x^2+16x__+y^2+12y__=5

Next you would fill in the blanks with the number that belongs, for this you divide the x and y by 2 and then square it. For this problem you would use 16x and 12y, and you would get 64 and 36. So the answer would be:
x^2+16x+64+y^2+12y+36=5

After this you add the new numbers to the other side of the problem and you would get this:
x^2+16x+64+y^2+12y+36=5+64+36
or
x^2+16x+64+y^2+12y+36=105

Then you factor out the x's and y's:
(x+8)^2+(y+6)^2=105

In the end you would get:
Center=(-8,-6) Radius=square root of 105

alaina's blog, 9 May 2010

2010-05-09T21:35:00.001-05:00

There are 2 main types of sequences:

1.) Arithmetic- tn*t1+(n-1)d

n=term # t1=first term d=what you add

2.) Geometric- tn=t1*r^(n-1)

r= what you multiply by

Example: find the formula for the nth term of the arithmetic sequence

3,5,7

tn= 3+(n-1)(2)
tn=3+2n-2
tn=1+2n

Example: find the formula for the nth term of the sequence

3, 4.5, 6.75

divide the second term by the first to get your r.

4.5/3= 3/2 r= 3/2

tn=3(3/2)^n-1

I could use some help on the problems that ask you specifically to find the 200th term for example. I could also use some help with the recursive definitions. THANKSS!!!

ALSO, if anyone can exlpain how to find the equation of a graph when given a graph, i'd like your help.

Dustin's Blog

2010-05-09T21:28:00.002-05:00

Ok, haven't done this in a while. I figured my grade is terrible and this is how I can raise it up. My blog won't be ridiculously long like Amy's, but I'll try to teach how to do something. One of my favorite things is...................................(can't think of anything).......................Sigma Notation.

Let's say the equation is 2x+5. Underneath the sigma is x=2 and above the sigma is 5. What this means is that you have to solve 2x+5 for numbers from 2-5.

So, first we plug in 2 and get 9.

Then, we plug in 3 and get 11.

Next, we plug in 4 and get 13.

Finally, we plug in 5 and get 15.

Our expanded answer is 9, 11, 13, 15.

Thats about it to expanding sigma notation.

Some things I don't understand are how to use some of the formulas on sequences and series. I also forgot how to graph conics. Thats about it for this blog i guess.

Amy's Reflection #38

2010-05-09T17:45:00.003-05:00

our last reflection!! okay anyway, y'all here are some stuff from chapters 1 - 6 & 13..im gonna print it out and use it to study..i hope y'all will do the same :) good luck

Completing the Square:

You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².

For example:

x² + 6x - 2 = 0

* anytime you are solving a quadratic you’re finding x-intercepts

* Move the constant term to the right side:

x² + 6x = 2

* Take half of the coefficient on the x-term (divide it by two, and keeping the sign), and then square it. Add the squared value to both sides of the equation:

x² + 6x + 9 = -2 + 9

* Convert the left-hand side to squared form. Simplify the right-hand side:

(x + 3)² = 7

* the # half of the coefficient goes in the parentheses.

* Square-root both sides:

x + 3 = √7

* Solve for "x =". Remember to put the "±" on the right side and that it gives you two solutions.

x = -3 ± √7

* The two points for this solution are:

(-3 + √7,0) , (-3 -√7,0)

Rational Root therom

Example: f(x)= 2x^3 + 3x^2 - 8 + 3

Step 1: find all possible roots..

p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2

*p is the leading constant term & q is the leading coefficient

possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2

Step 2: now you can plug all of the possible roots in your calculator to find the roots that work

* the zero will be: 1, 1/2, -3

Step 3: use synthetic division to factor all of the roots that work

you should get: (x - 1) (2x^2 + 5x + 3)

Domain & Range of functions:

Polynomials:

the domain of all polynomials is (−∞, ∞).

For example:

f(x) = x^2 - 3x^2 + 2x - 1

D: (−∞,∞ )

f(x) = x^2 + 3

D:(−∞,∞ )

Fraction:

* you set the bottom to zero
* solve for x
* then set up intervals

For example:

f(x) = 1/x-2

x-2=0

x=2

D: (-∞ , 2) (2, ∞ )

Absolute Value:

D:(- ∞ , + ∞)

R: [0 , + ∞)

For example:

f(x) = |x + 8| - 9

D: (- ∞ , + ∞)
R: (-9, ∞)

f(x) = |x -7| + 5

D: (- ∞ , + ∞)
R: (5, ∞)

Square Roots:

to find the domain:

* set the inside = to zero
* then set a # line
* try values on either side of each #
* get ride of the negatives
* set up intervals

to find the range:

* graph

For example:

√9 - x^2

(solve for x...)

9 - x^2 = 0

-x ^2 = -9

√x^2 = √9

x = ±3

(# line)

(#s on either side..)

f(-4) = √9 - (-4)^2 = √-7
f(0) = √9 - (-4)^2 = √9
f(x) = √4 - (-4)^2 = √-7

√9 = ±3 so...

D: [-3, 3]

(graph....)

R: [0, 3]

How to Find the Inverse of a Function:

* Replace f(x) with y
* Reverse the roles of x and y
* Solve for y in terms of x
* Replace y with f-1(x)

Example 1 - f(x) = 2x + 3

1. write the function as an equation: y = 2x + 3
2. solve for x: x = (y - 3)/2
3. now write f-1(y) as follows .
f -1(y) = (y - 3)/2 or f -1(x) = (x - 3)/2
4. Check:

* f(f -1(x))=2(f -1(x)) + 3
=2((x-3)/2)+3 =(x-3)+3 =x
* f -1(f(x))=f -1(2x+3)
=((2x+3)-3)/2 =2x/2 =x

Example 2 - f(x) = √x + 4

1. (x)^2 = (√y + 4)^2
2. x^2 = y + 4
3. y = x^2 - 4
4. f-1(x) = (x^2 - 4)

* f(f-1(x)) = f(x^2 - 4) = √(x^2 - 4) + 4 = x
* f-1(f(x)) = f-1(√x + 4) = (√x + 4)^2 - 4 = x + 4 - 4 = x

Exponents:

1. b^x * b^y = b^x + y....example: 2^3 * 2^5 = 2^8

2. b^x/b^y = b^x - y....example: 5^7/5^4 = 5^3

3. (ab)^x = a^xb^x....example: (3 * 7)^3 = 3^3 * 7^3

4. (a/b)^x = a^x/b^x....example: (3/5)^3 = 3^3/5^3

5. (b^x)^y = b^xy....example: (2^2)^3 = 2^6

6. b^x/y = y^√b^x....examples: 5^3/4 = 3^√5^3

7. to solve for exponents:

* write as the same base
* set exponents equal
* then solve for x

here are some examples:

(a). 5^3x = 5^7x - 2

In this first part we have the same base on both exponentials so there really isn’t much to do other than to set the two exponents equal to each other and solve for x.

3x = 7x - 2

2 = 4x

x = 1/2

So, if we were to plug x = 1/2 into the equation then we would get the same number on both sides of the equal sign.

(b). 4^t^2 = 4^6 - t

t^2 = 6 - t

t^2 - t - 6 = 0

(t - 2) (t + 3) = 0

t = -3, t = 2

In this case we get two solutions to the equation. That is perfectly acceptable so don’t worry about it when it happens.

(c). 3^z = 9^z + 5

Now, in this case we don’t have the same base so we can’t just set exponents equal. However, with a little manipulation of the right side we can get the same base on both exponents. To do this all we need to notice is that 9 = 3^2. Here’s what we get when we use this fact:

3^z = (3^2)^z + 5

Now, we still can’t just set exponents equal since the right side now has two exponents.

3^z = 3^2(z + 5)

We now have the same base and a single exponent on each base so we now set the exponents equal. Doing this gives us....

z = 2(z + 5)

z = 2z + 10

-10 = z

...a solution of z = -10.

Step 4: solve further

(this can be factored...)

= (x - 1) (2x^2 + 5x + 3)

= (x - 1) (2x - 1) (x + 3)

(set x = 0 )

x = 1, 1/2, -3

Logarithm Properties:

* logb MN = logb M + logb N
* logb M/N = logb M - logb N
* logb M^K = K logb M
* logb b^k = k (this one i don't get..maybe i copied it wrong)
* b^logb^k = k

Here are some examples:

1. log 2 + log 3 + log 4 = log 24 (mulitply: 2 x 3 x 4)

2. log 8 + log 5 - log 4 = log 10 (mulitply: 8 x 5 then divide: 40/4)

3. 2 ln 6 - ln 3 = ln 12 (raise 6 to the 2nd power = 36 the divided by 3 = 12)

4. log M - 3 log N = log M/ N^3

5. ln 2 + ln 6 - 1/2 ln 9 = ln 12/3 = ln 4

6. Expand logb MN^2....logb M + 2 logb N

7. Condense log 45 - 2 log 3....log (45/9) = log 5

8. Rewrite in exponetial form: log36 6 = 1/2....36^1/2 = 6

9. Rewrite in logarithmic form: 2^2 = 4....log2 4 = 2

Changing Bases: (Done when you can't solve a log)

* Rewrite it as an exponential
* Take the log of both sides
* Move the variable to the front
* then solve

(use the same steps when solving for x as an exponent when you can't write them as the same base)
examples:

1. log5 10 = x

5^x = 10

log 5^x = log 10

x log 5 = 1

x = 1/log 5

2. 2^x = 7

log 2^x = log 7

x log 2 = log 7

x = log 7/log 2

(remember b-rob might use random symbol so don't panic)

Conics

Ellipses

Steps:

1. find the center
2. determine the major axis
3. find the vertex (± √big denom)
4. find the other intercept ( ± √small denom)
5. find the focus (c^2 = a^2 + b^2)
6. determine the length of the major axis (2√big denom)
7. find the length of the minor axis (2√small denom)
8. finally graph

Example 1: Graph the following ellipse. Find its major intercepts, length of the major axis, minor intercepts, length of the minor axis, and foci.

x^2/4 + y^2/9 = 1

This ellipse is centered at (0, 0). Since the larger denominator is with the y variable, the major axis lies along the y-axis.

Since a^2 = 9 then a = 3 & Since b^2 = 4 then b = 2

Major intercepts: (0, 3), (0, –3)

Length of major axis: 2 √9 = 6

Minor intercepts: (2, 0), (–2, 0)

Length of minor axis: 2√4 = 4

c^2 = a^2 + b^2

= 9 - 4

= 5

= √5

Foci: (0, √5) , (0, -√5)

then you graph your points..

Parabolas:

how to find the axis of symmetry, vertex, focus, & directrx??

1.) to find the axis of symmetry: x = -b/2a

2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:

y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex

3.) to find the focus: 1/4p= the coefficient of x^2 and then add p

Note:

*If opens up, add to y value from vertex, if opens down, subtract

*If opens right, add to x value to vertex, if opens left, subtract)

4.) directrix: is p units behind the vertex

Note:

*If opens up, subtract; if opens down, add from y-value of vertex.
*If opens right, subtract x-value
*If opens left, add x-value

Example: x^2 + 1

~vertex:

x = -b/2a

x = 0/2(1) = 0

0^2 + 1 = 1

(0,1)

~Focus:

1/4p = 1

4p = 1

p = 1/4

(0, 1 + 1/4)

(0, 5/4)

~directrix:

y = 1 - 1/4

y = 3/4

CIRCLES
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)

If the equation is not in standard form, you must complete the square to put it in standard form.

If you are given a center and a point, you can use the distance formula to find the radius.

To find the intersection of a line and a circle:

1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.

***Reminder. If your x value is imaginary, then there is no point of intersection.

EX: find the center and radius.

(x-3)^2+(y+7)^2=19 c:(h,k)

center: (3,-7) radius: square root of 19

EX: find the eqn of the circle with the center (1,4) through (3,7)

in the problem you are given a center and a point so you would plug into the distance formula.

square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13. **13 has no root.

Your answer should be (x-1)^2+(y-4)^2=13

Hyperbola

(x^2-h/a^2)+(y^2-k/b^2)=1

* Your center is (h,k)
* your major axis has the larger denominator

13-1

1. sequence-list of numbers

2. two main types: 1). arithmetic-add or subtract 2).geometric-multiply

Formulas:

1. arithmetic-used to find a term: tn . t1 + (n-1)d

**n=term #, t1=first term, d=what you add, tn=term #

2. geometric: tn=t1 . r^(n-1)

**r=what you multiply by..

Examples:

1. Find the formula for the nth term of the arithmetic sequence: 3,5,7,...

tn = 3 + (n-1) (2)

tn = 3 +2n - 2

tn = 1 + 2

2. Find the formula for the nth term of the sequence: 3,4.5,6.75,..

**divide the 2nd term by the 1st term to find r

4.5/3 = 3/2 = r

tn = 3 . (3/2)^(n-1)

13-2

Formula for a sequence that involves the previous term: (an - 1)

Examples:

1. Find the recursive definition of: 81, 27, 9,3,...

an = an - 1/3

2. 1, 2, 6, 24, 120, 720, ....

n = 1: 1

n= 2: 2

n = 3: 6

an = n . an - 1

13 -3

Series-List of added or subtracted numbers

**Leave it as a list: do NOT add

Formulas:

1. Arithmetic: Sn = n(t1 + t2)/2

**Finds the sum of the first n terms

2. Geometric: Sn = t1 (1 -r^n)/1-r

Examples:

1. Find the sum of the first 25 terms of the series: 11 + 14 + 17 + 20 + ....

Sn = n (t1 + tn)/2

t25 = 11 + (24)(3)

Sn = 25 (11 + 83)/2

= 1175

2. Find the sum of the first 10 terms of the series: 2-6 + 18 - 54 +...

**This is a geometric sequence and that is because you have to add or subtract the same number for it to be an arithmetic sequence, got it??

r = -6/2 = -3

Sn = t1(1 - r^n)/1-r

= 2(1 -(-3)^10)/1 - (-3)

= 2(-59048)/2

= -29524

Make up Blogs

2010-05-04T17:55:00.002-05:00

Taylor reflection 2 may 2010

2010-05-03T11:13:00.001-05:00

TRIG EXAM PT 1 TOMORROW!!
study study study

good luck everyone

for mw the main thing is to remember the formulas
Sum and Difference formulas for Cosine and Sine:
cos (alpha + or - beta) = cos(alpha)cos(beta) - or + sin(alpha)sin(beta)
sin (alpha + or - beta) = sin(alpha)cos(beta) + or - cos(alpha)sin(beta)

Half-Angle and Double-Angle Formulas:
sin(2alpha) = 2sin(alpha)cos(alpha)
cos(2alpha) = cos^2(alpha)-sin^2(alpha)=1-2sin^2(alpha)=2cos^2
(alpha)-1
tan(2alpha) = 2tan(alpha)/1-tan^2(alpha)
sin(alpha/2)= +- sqrt(1-cos(alpha)/2)
cos(alpha/2)= +- sqrt(1+cos(alpha)/2)
tan(alpha/2)= +- sqrt(1-cos(alpha)/1+cos(alpha))=sin(alpha)/1+cos
(alpha)=1-cos(alpha)/sin(alpha)

Recriprocal Relationships:
csc=1/ sin(theta)
sec=1/cos(theta)
cot=1/tan(theta)

Relationships With Negatives:
sin-theta= -sin(theta)
cos-theta= -cos(theta)
tan-theta= -tan(theta)
csc-theta= -csc(theta)
sec-theta= -sec(theta)
cot-theta= -cot(theta)

Pythogorean Relationships:
sin^2(theta)+cos^2(theta)=1
1+tan^2(theta)= sec^2(theta)
1+cot^2(theta)= csc^2(theta)

Cofunction Relationships:
sin(theta)= cos (90degrees-theta)
cos(theta)= sin (90degrees-theta)
tan(theta)= cot (90degrees-theta)
cot(theta)= tan (90degrees-theta)
sec(theta)= csc (90degrees-theta)
csc(theta)= sec (90degrees-theta)

Alicia's Reflection #37

2010-05-02T22:37:00.003-05:00

Okay so I've been studying for so long I have decided to stop and go to bed. Hopefully I remember all the formulas for chapters 8 and 10. The notecards that I bought from the math club in the beginning of the year were definitely helpful. Well, here is the trig chart. Tomorrow is the non calculator portion so if you do not know the trig chart you will probably fail :(.....

0°
sin0=0
cos0=1
tan0=undefined
sec0=1
cot0=0

30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3

45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1

60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2

90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

The 6 trig functions are also very helpful to memorize:

sin y/r csc r/y

cos x/r sec r/x

tan y/x cot x/y

Also to move from quadrants:

I-II make - then add 180

I-III just add 180

I-IV make - then add 360

Sin is positive in I and II... Negitive in III and IV

Cos is positive in I and IV... Negative in II and III

Tan is positive in I and III... Negative in II and IV

Goodluck to everyone tomorrow!!!

Terrio's Reflection

2010-05-02T21:05:00.002-05:00

I'm done. I worked on these packets for like three straight weeks and studied a total of at least fifteen full hours this weekend for this huge Trig Exam and I'm still going to fail.

Area of Non-Right Triangle:

A=1/2(leg)(leg)sin(angle between)

Two sides of a triangle have lengths 7cm and 4 cm. The angle between the sides measures 73 degrees. Find the area of the triangle.

sides=7 and 4

angle between= 73 degrees

A=1/2(7)(4) sin(73)

A=13.388cm^2

Stephanie's Reflection

2010-05-02T20:49:00.000-05:00

Sum and Difference formulas for Cosine and Sine:
cos (alpha + or - beta) = cos(alpha)cos(beta) - or + sin(alpha)sin(beta)
sin (alpha + or - beta) = sin(alpha)cos(beta) + or - cos(alpha)sin(beta)

Half-Angle and Double-Angle Formulas:
sin(2alpha) = 2sin(alpha)cos(alpha)
cos(2alpha) = cos^2(alpha)-sin^2(alpha)=1-2sin^2(alpha)=2cos^2
(alpha)-1
tan(2alpha) = 2tan(alpha)/1-tan^2(alpha)
sin(alpha/2)= +- sqrt(1-cos(alpha)/2)
cos(alpha/2)= +- sqrt(1+cos(alpha)/2)
tan(alpha/2)= +- sqrt(1-cos(alpha)/1+cos(alpha))=sin(alpha)/1+cos
(alpha)=1-cos(alpha)/sin(alpha)

Recriprocal Relationships:
csc=1/ sin(theta)
sec=1/cos(theta)
cot=1/tan(theta)

Relationships With Negatives:
sin-theta= -sin(theta)
cos-theta= -cos(theta)
tan-theta= -tan(theta)
csc-theta= -csc(theta)
sec-theta= -sec(theta)
cot-theta= -cot(theta)

Pythogorean Relationships:
sin^2(theta)+cos^2(theta)=1
1+tan^2(theta)= sec^2(theta)
1+cot^2(theta)= csc^2(theta)

Cofunction Relationships:
sin(theta)= cos (90degrees-theta)
cos(theta)= sin (90degrees-theta)
tan(theta)= cot (90degrees-theta)
cot(theta)= tan (90degrees-theta)
sec(theta)= csc (90degrees-theta)
csc(theta)= sec (90degrees-theta)

Stephanie's Blog Topic Response

2010-05-02T20:43:00.002-05:00

In order to tell what kind of graph you have, you look at the equation.
If the equation is r = a + b sin θ OR r = a + b cos θ, then you have a limacon.
If the equation is r = a - b sin θ OR r = a - b cos θ, then you have a cardiod.
If the equation is r = a θ + b, then you have an archimedes spiral whereas if the equation is r = a b ^ θ, then you have a logarithmic spiral.
If the equation is r = a sin(nθ) OR r = a cos(nθ), then it is a rose where n = the number of petals.

Stephen's Reflection

2010-05-02T20:37:00.002-05:00

Ok so this week we have trig exam..one thing i forgot is shapes in graphs like circles and roses and spirals. I now know it so im going to give the formulas for each shape in a graph...

Limacon
r=a+b sin(theta)
r=a+b cos(theta)

Cardioid
a-b sin(theta)
r=a-b cos(theta)

Rose
r=a sin(n theta)
r=a cos (n theta)

*n=how many petals

Archimedes Spiral
r=a theta+b

Logarithmic Spiral
r=a b^theta

Ok so i forget stuff about logs like formulas and examples on how to work them so if i can get an example of each i would be happy

Amy's Reflection #37

2010-05-01T20:34:00.001-05:00

so we pretty much just reviewed for the test..here's some examples of the stuff we gonna have to know..

Example 1: Find the exact value of sin 80 cos 130+ cos 80 sin 130

Let alpha = 80 and beta = 130 then sin 80 cos 130 + cos 80sin 130 = sin alpha cos beta + cos alpha sin b

= sin (alpha + beta)
= sin (80+130)
= sin (210)
= sin (180 + 30)
= - sin 30
= -1/2

Example 2: Find the exact value of

cos^215 - sin^215 = cos(2alpha)
= cos(30)
= sqrt 3/2

Example 3: cos 15

alpha = 45, beta = 30

cos (alpha - beta) = cos 45 cos30 + sin45 sin 30

cos (45 - 30) = (√(2/2) (√(3/2) + (√(2/2) (1/2)

= √(√6 + √2)/all over 4

Example 4: (1 + cot^2) (1 - cos 2x)

(csc^2x) (1-cos2x)

(csc^2x) (1-(1 - 2sin^2x)

(1/sin^2x) (2sin^2x)

= 2

Example 4: Find the exact value of sin22.5

alpha=2(22.5)

alpha=45

**if you are given an angle with a decimal you use the half-angle formula. To find alpha, you multiply by two.

Example 5: Sum Formula for Cosine

cos 75 cos 15 + sin 75 sin 15

=cos (75-15) = cos 60 = 1/2

Example 6: Find the exact value of tan15 +tan30/1-tan15 (tan30)

= tan(15 + 30)

= tan(45)

= 1

Examples:

1. Express 2 cis 50degrees in rectangular form

2 cos 50 + 2 sin 50 i

2. Express -1-2i in polar form

radius = +- sqrt of ((-1)^2 + (-2)^2)) = +- sqrt of (5)

theta = tan^-1(-2/-1)

theta = tan^-1(1)

*tangent is positive in the first and third quadrants, 63.435 and 243.435
*63 is positive for cosine so it goes with the positive sqrt of 5
*243 is negative for cosine so it goes with the negative sqrt of 5

z= sqrt of 5 cis 63.435

z= sqrt of 5 cos 63.435 + sqrt of 5 sin 63.435 i

z= negative sqrt of 5 cis 243.435

z= negative sqrt of 5 cos 243.435 + negative sqrt of 5 sin 243.435 i

De Moivre's Theorem: z^n = r^n cis(n)(theta)

Examples:

1. z=2cis20degrees Find z^2

z^2=2^2cis2(20degrees)

z^2=4cis40degrees

2. 4cis15degrees Find z^4

z^4=4^4cis4(15degrees)

z^4=256cis60degrees

Limacon
r=a+b sin(theta)
r=a+b cos(theta)

Cardioid
a-b sin(theta)
r=a-b cos(theta)

Rose
r=a sin(n theta)
r=a cos (n theta)

*n=how many petals

Archimedes Spiral
r=a theta+b

Logarithmic Spiral
r=a b^theta

Examples:
1. r=theta+2
2. r=2+3cos(theta)
3. r=5
4. r=3sin(4 theta)
5. r=1/2(3^theta)
6. r=2sin(theta)

1. Archimedes spiral
2. limacon
3. circle with its center at the pole
4. rose with 4 petals
5. logarithmic spiral
6. circle that intersects with the pole

i hope this helped..don't forget to study for our trig!!!

Amy's Blog Topic Response

2010-04-29T12:04:00.001-05:00

**Explain how to solve a non-right triangle with Law of Sines and Law of Cosines

1.) Law of Sines(used to non-right triangles):

Sin A/a = Sin B/b= Sin C/c

Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.

A = 1/2 (4) (5) Sin 30 degrees

A = 10 Sin 30 degrees which is aproximately = 5

2.) Law of Cosines (used when you can't use Law of Sines):

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)

Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.

7^2=6^2+5^2-2(5)(6)

cos a7^2-6^2-5^2= 2(5)(6)

cos acos a= 7^2-6^2-5^2 / -2(6)(5)

a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))

a= 78.463 degrees

i hope that helps...

Alicia's Blog Topic Response

2010-04-28T17:07:00.002-05:00

Okay so I am going to explain the 2nd topic.

Law of Sines

(sinA)/a = (sinB)/b = (sinC)/c

*The law of sines is used when you know pairs in non-right triangles

*Basically, all your doing is setting up a proportion

Law of Cosines

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 -2(adjacent leg)(adjacent leg)cos(angle between)

*When doing law of cosines, you should always use an angle to orient yourself like SOHCAHTOA!

Topics

2010-04-27T09:20:00.002-05:00

Write a blog on one of the following

1. Explain in detail the steps to graphing a trig function

2. Explain how to solve a non-right triangle with Law of Sines and Law of Cosines

3. Explain how to know what type of polar graph you have given an equation

Stephen's Reflection

2010-04-26T20:56:00.002-05:00

Ok so this week we did more stuff in the study guides for trig exam and yea..anyway im going to show you the half and double angle formulas for yall...

Half-Angle and Double Angle Formulas:

sin(2alpha) = 2sin(alpha)cos(alpha)
cos(2alpha) = cos^2(alpha)-sin^2(alpha)=1-2sin^2(alpha)=2cos^2
(alpha)-1
tan(2alpha) = 2tan(alpha)/1-tan^2(alpha)
sin(alpha/2)= +- sqrt(1-cos(alpha)/2)
cos(alpha/2)= +- sqrt(1+cos(alpha)/2)
tan(alpha/2)= +- sqrt(1-cos(alpha)/1+cos(alpha))=sin(alpha)/1+cos
(alpha)=1-cos(alpha)/sin(alpha)

Ok so like i said in class, everything with graphs is gonnneee. I need help working on conics and graphing them so if anyone can help i would be grateful

Trig Stuff

2010-04-26T20:50:00.000-05:00

SOHCAHTOA:
Sin=opp/hyp
Cos=adj/hyp
Tan=opp/adj

Trig functions:
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y

Law of sines:
(sinA)/a = (sinB)/b = (sinC)/c

Law of cosines:
(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 -2(adjacent leg)(adjacent leg)cos(angle between)

Taylor reflection for 25 April 2010

2010-04-26T08:45:00.002-05:00

Trig Identities

recriprocal relationships:
csc=1/ sin(theta)
sec=1/cos(theta)
cot=1/tan(theta)

relationships with negatives:
sin-theta= -sin(theta)
cos-theta= -cos(theta)
tan-theta= -tan(theta)
csc-theta= -csc(theta)
sec-theta= -sec(theta)
cot-theta= -cot(theta)

pythogorean relationships:
sin^2(theta)+cos^2(theta)=1
1+tan^2(theta)= sec^2(theta)
1+cot^2(theta)= csc^2(theta)

cofunction relationships:
sin(theta)= cos (90degrees-theta)
cos(theta)= sin (90degrees-theta)
tan(theta)= cot (90degrees-theta)
cot(theta)= tan (90degrees-theta)
sec(theta)= csc (90degrees-theta)
csc(theta)= sec (90degrees-theta)

***i need someone to give me tricks asto when to use each set of relationships to solve an equation
i.e. i need someone to help me recognize when each set of trig identities is necessary to be used

Alicia's Reflection #36

2010-04-25T22:13:00.002-05:00

Alrighty so I hope everyone enjoyed their weekend with senior day and prom :) Back to school and some more review for the trig exam on May 3rd and 4th. I am going to review some material for our trig exam from chapter 9 which was triangles.

SOHCAHTOA:

S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)

C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)

T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)

*the hypotenuse is opposite the right angle.

*A= 1/2 bh*

*To find the area of a non right triangle use this formula:

*A= 1/2 (leg)(leg)SIN(angle b/w)

*When you have a non right triangle that has pairs, use the law of sines:

Sin A/a = Sin B/b= Sin C/c

*All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)

Stephanie's Reflection

2010-04-25T22:10:00.003-05:00

Reflection

2010-04-25T22:02:00.001-05:00

Some Trig Stuff...

The Unit Circle

90 degrees, (0,1), pi/2

180 degrees, (-1,0), pi

270 degrees, (0,-1), 3pi/2

360 degrees, (1,0), 2pi

6 Trig Functions

sin = y/r

cos = x/r

tan = y/x

csc = r/y

sec = r/x

cot = x/y

Degrees & Radians

Degrees to radians= Degree * pi/180

Radians to degrees= Radians * 180/pi

Amy's Reflection #36

2010-04-24T23:37:00.003-05:00

Since our trig exam is coming up here a review on chapter 10...

Sum and Difference formulas for Cosine and Sine:

cos (alpha + or - beta) = cos(alpha)cos(beta) - or + sin(alpha)sin(beta)
sin (alpha + or - beta) = sin(alpha)cos(beta) + or - cos(alpha)sin(beta)

Rewriting a Sum or Difference as a Product:

sin(x) + sin(y) = 2sin(x+y/2)cos(x-y/2)
sin(x) - sin(y) = 2cos(x+y/2)sin(x-y/2)
cos(x) + cos(y) = 2cos(x+y/2)cos(x-y/2)
cos(x) - cos(y) = -2sin(x+y/2)sin(x-y/2)
**we didn't use these formulas for anything..so i got no idea where to use them..

Half-Angle and Double Angle Formulas:

sin(2alpha) = 2sin(alpha)cos(alpha)
cos(2alpha) = cos^2(alpha)-sin^2(alpha)=1-2sin^2(alpha)=2cos^2
(alpha)-1
tan(2alpha) = 2tan(alpha)/1-tan^2(alpha)
sin(alpha/2)= +- sqrt(1-cos(alpha)/2)
cos(alpha/2)= +- sqrt(1+cos(alpha)/2)
tan(alpha/2)= +- sqrt(1-cos(alpha)/1+cos(alpha))=sin(alpha)/1+cos
(alpha)=1-cos(alpha)/sin(alpha)

now here are some examples:

1. tan α = 2 and tan β=1, find tan (α - β)

= tan α + tan β/1-tan α tan β

=2+1/1-(2)(6)

=3/-1

=-3

2. Find the exact value of: tan 15+tan 30/1-tan 15 tan 30

tan α = 2 and tan β=1

find tan (α - β)

= tan (15 + 30)

=tan (45)

=1

3. Find the exact value of sin 15degrees

*exact value means you use your trig chart
*think of two numbers from the trig chart can either add or subtract to give you 15
*since it's (45-30), you would look for the formula that uses sin

sin (a-B) = sin a cos B - cos a sin B

* plug #s into equation..

a=45 degrees B=30 degrees

sin (45-30) = sin 45 cos 30 - cos 45 sin 30

sin (a-B) = sin a cos B - cos a sin B

sin (45-30) = sin 45 cos 30 - cos 45 sin 30

sin 15 = (square root of 2 over 2)(square root of 3 over 2) - (square root of 2 over 2)(1/2)

sin 15 degrees = (square root of 6 over 4) - (square root of 2 over 4)

= square root of 6 - square root of 2 all over 4

i hope this will help refreshen someone's memory :)

Terrio-Reference Angles

2010-04-22T20:29:00.002-05:00

Reference Angles have to be between 0° and 90°. First, find which quadrant the angle is in. Next, determine whether the sign is positive or negative in that quadrant. Last, subtract 180° until the angle is between 0° and 90°.

Moving To Different Quadrants:
I to II/II to III/III to IV= make negative and add 180°
I to III = add 180°
I to IV = make negative and add 360°
II to IV = add 180°

Reference Angles

2010-04-21T23:49:00.018-05:00

A reference angle must be between oº and 90º, it is used when an angle is not found on the trig chart or unit circle.

To find the reference angle: First find what quadrant the angle is in, next determine the sign of the quadrant (positive or negative?) that the trig function is in, lastly subtract 180º until the angle is between 0º and 90º

Amy's Blog Topic

2010-04-21T23:38:00.002-05:00

Reference Angles:
(must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)

1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant

**To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°

1. The reference angle in the 1st quad is equal to the angle measure.

Example- θ = 60 = 60

2. To find the reference angle in the 2nd quad just subtract the angle from 180°.

Example- θ = 130° = 180° -130°= 50°

3. To find the reference angle in the 3rd quad subtract 180° from the angle given.

Example- θ = 240° = 240° - 180° = 60°

4. To find the reference angle in the 4th quad simply subtracting the angle from 360° will provide the reference angle in the 4th quad.

Example- θ = 315° = 360° -315° = 45°

other examples:

1. sin^-1(-√2/2) = 45°
45 + 180 = 225° = θ
I to IV
-45 + 360 = 315°
θ = 225°, 315°

2. Find the reference angle for sin210°
1. lies in quadrant 3
2. sin is negative in quadrant 3
3. 210-180=30
sin210°= -sin30
you would then look to your trig chart and find a reference to sin 30 (π/6)
sin 210°=1/2

Stephen's Reflection

2010-04-21T22:44:00.001-05:00

Ok so this blog is about law of sines and cosines. You use this to find angles and sides of non right triangles.

When you have a non right triangle that has pairs, use the law of sines:

Sin A/a = Sin B/b= Sin C/c

*All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)

I forget how to use steps for circles and ellipses and stuff like that

Alicia's Blog topic

2010-04-21T21:45:00.002-05:00

okay so im going to explain the 3rd topic, how to find a reference angle.

Finding a reference angle is used when the angle does not appear on the trig chart or unit circle.

First, set up your blanks before and after your trig function
Example: ____SIN_____

The first blank is where you put positive or negative depending on the trig functions quadrant
The second blank is where you put the reference angle you find

The point of the reference angle is to get your angle that you are given between 0 and 90 degrees.

When finding reference angles, you can either add or subtract 360 and 180 to your angle depending on how big or small the number is.

The final number that you get should be between 0 and 90 degrees. This is the number that goes in the second blank

To determine positive or negative, you have to know your trig functions quadrants.
Example: SIN is positive is quadrant I & II. Negative in III & IV

Hope this helped!!!
Goodluck on the trig exam.
If anyone wants to make a study group I would like to be in it :)

Stephanie's Reflection

2010-04-21T15:09:00.002-05:00

you use a unit circle with the trig chart
it is also used when you want to change what quad you are in
and you can use it to figure out where sin, cos, etc. are positive and negative

to change what quad you are in, you can change the number to negative and add 360 or change the number to negative and add 180 (i forget the other one)

to figure out where sin, cos, etc. are positive and negative, you simply see what it equals (eg: sin = y/r means that wherever y is positive, sin is positive and wherever y is negative, sin is negative)

taylor blog topic from brob #1

2010-04-20T21:33:00.003-05:00

i chose to answer
3.How to find a reference angle

there are three steps to finding a refrence angle
STEPS
pre step- set up ____ trig function____
the rest will plug into this
#1- discover which quadrent the given angle is in
#2- determine if the given trig function ((which will go in the "trig function" spot)) is positive or negative in that quadrent ((this will go into the first blank))
#3- change given angle to some number between 0 and 90 degrees by subtracting 180 until the angle is between 0 and 90 degrees. ((the number that you get which is between 0 and 90 degrees will go in the second blank))

Thats it! good luck
study study study

Blog Topic

2010-04-19T18:27:00.002-05:00

The Second Blog topic for this week is to explain one of the following:

1. How to use the unit circle

2. How to solve for an angle

3.How to find a reference angle

Taylor reflection for 18 April

2010-04-19T09:08:00.002-05:00

NOTES FOR TRIG REVIEW CHAPTER 9 ((Triangles))

SOHCAHTOA:

S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)

C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)

T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)

*the hypotenuse is opposite the right angle.

*A= 1/2 bh*

*To find the area of a non right triangle use this formula:

*A= 1/2 (leg)(leg)SIN(angle b/w)

*When you have a non right triangle that has pairs, use the law of sines:

Sin A/a = Sin B/b= Sin C/c

*All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)

i need someone to give me the formula for area of an inscribed shape please =)

Amy's Reflection #35

2010-04-18T23:40:00.002-05:00

okay here is some stuff we learned this week..

Vector Operations with Coordinates
Vector Addition: v + u = (a, b) + (c, d) = (a + c, b + d)
Vector Subtraction: v - u = (a, b)-(c, d) = (a - c, b - d)
Scalar Multiplication: kv = k(a, b) = (ka, kb)

The Dot Product
If v1 = (x1, y1) and v2 = (x2, y2) then the dot product, denoted by v1•v2 is defined by: v1•v2 = x1x2 + y1y2.

Properties of the Dot Product
1. u • v = v • u
2. u • u = |u|²
3. k(u • v) = (ku) • v
4. u • (v + w) = u • v + u • w

**if two vectors are orthogonal, the dot product=0

**if two vectors are parallel, then x2/x1=y2/y1

Mid point formula: (x1+x2/2, y1+y2/2, z1+z2/2)

Example:

1. u= (3,-6) v= (4,2) w= (-12, -6)

show that u&v are perpendicular and that v&w are parallel.

u . v= 3(4)+(-6)(2)=0

-12/4 = -6/2
-3 = -3

2. Find the midpoint of (2,2,2) and (2,4,6)

= 2+2/2, 2+4/2, 2+6/2

= (2, 3, 4)

okay i also need help in remembering some trig stuff so any review on anything would be appreciated..

Vector Formulæ

2010-04-18T23:39:00.000-05:00

New Vocabulary: Orthogonal = Perpendicular

Vector Operations with Coordinates

Vector Addition: v + u = (a, b) + (c, d) = (a + c, b + d)

Vector Subtraction: v - u = (a, b)-(c, d) = (a - c, b - d)

Scalar Multiplication: kv = k(a, b) = (ka, kb)

The Dot Product

If v1 = (x1, y1) and v2 = (x2, y2) then the dot product, denoted by v1•v2 is defined by

v1•v2 = x1x2 + y1y2.

Properties of the Dot Product

1. u • v = v • u

2. u • u = |u|²

3. k(u • v) = (ku) • v

4. u • (v + w) = u • v + u • w

Stephen's Reflection

2010-04-18T21:44:00.002-05:00

Ok so this week we learned about vectors and some other stuff i dont really remember..ANNYYYWAYY..one thing i know is whether vectors are orthogonal or parallel. there are a few formluas to figure out if they are orthogonal..4 to be exact..

1. u . v= v . u
2. u . u= /u/^2
3. k(u . v)= (ku) . v
4. u . (v+w)= u . v+ u . w

It is orthogonal if it equals 0.

Ok so i forget alot of trig stuff like law of sines and law of cosines so i need help with that

Alicia's Reflection #35

2010-04-18T20:40:00.002-05:00

Alrighty so this week we learned Chapter 12 Vectors and we had a take home test this weekend thats due tomorrow... Dont Forget!!! I am going to review what we learned in chapter 12.

Another word that means perdencidular is orthogonal.

To figure out if two vectors are orthogonal, find the dot product.

v1 . v2=x1 x2 + y1 y2

**the dot does not imply multiplication.

Properties

1. u . v= v . u
2. u . u= /u/^2
3. k(u . v)= (ku) . v
4. u . (v+w)= u . v+ u . w

u=(x1 , y2)
y=(x2 , y2)
z=(x3 , y3)

Prove the 4 Properties

if two vectors are orthogonal, the dot product=0

if two vectors are parallel, then x2/x1=y2/y1

Example:

u= (3,-6) v= (4,2) w= (-12, -6)
show that u&v are perpendicular and that v&w are parallel.

u . v= 3(4)+(-6)(2)=0

-12/4=-6/2
-3=-3

I could use some help with determinants like on our take home test where there is 4 columns and 4 rows.

Reflection

2010-04-18T20:11:00.002-05:00

This week we learned a lot of new stuff. We learned about vectors, did some midpoint stuff, distance formula stuff, and all kinds of different math STUFF haha. This week wasn't too bad, I thought it was all pretty easy for the most part. Here is an example of one of the things we did this week.

MIDPOINT OF 3-D VECTORS:

Find the midpoint of (2,2,2) and (2,4,6)

-plug it into the formula m = (x1+x2/2, y1+y2/2, z1+z2/2)

-so 2+2/2, 2+4/2, 2+6/2

then you simplify it and you get (2, 3, 4)

Stephanie's Reflection

2010-04-18T14:05:00.000-05:00

Trig Chart:

0°
sin0=0
cos0=1
tan0=undefined
sec0=1
cot0=0

30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3

45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1

60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2

90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

Taylor reflection 11 April

2010-04-12T08:38:00.002-05:00

This week we will start reviewing for the trig exam
so i will start posting trig notes for each chapter

Angles are measured in DEGREES (possibly with either minutes' or minutes' andseconds")
and RADIANS

To find minutes
multiply what is behind the decimal by 60
and take whole number as minutes'

To find seconds
multiply what is behind the decimal of minutes by 60
then whats behind that decimal by 3600 to get seconds'

To get radians

degrees x pi/180degrees

((***Remember to always ues exact answers))
((*** Remember to never plug pi into calculator))

Unit Circle:

90 degs. = (0,1) pi/2

180 degs. = (-1,0) pi2

70 degs. = (0,-1) 3pi/2

360 degs. = (1,0) 2pi

sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y

i need someone to provide me with the formulas neccessary to have memorized for chapter eight and then i need someone to help me figure out any tricks they may know for solving chapter eight

a quick tip overview of chapter eight may be easiest.

Stephen's reflection

2010-04-11T21:15:00.001-05:00

Ok so yea i dont want school tomorrow and i forgot a bunch of stuff sooo yea..anyway i will talk about some trig stuff like SOHCAHTOA, law of sines and law of cosines..these are the formulas you use and the rest is reallllyyyy easy soooo yea.

SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°

Ok so what i dont really remember is i can use help on limits and sigma notation..i forgot all that stuff

Stephen's Make up blog over easter

2010-04-11T21:12:00.002-05:00

Ok soooo i completely shut out all school work and forgot to do this and my mom hasseled me to do it today soooooo yea...this is a few things from chapter 13..just a formula or two.

1. sequence-list of numbers

2. two main types: 1). arithmetic-add or subtract 2).geometric-multiply

Formulas:

1. arithmetic-used to find a term: tn . t1 + (n-1)d

**n=term #, t1=first term, d=what you add, tn=term #

2. geometric: tn=t1 . r^(n-1)

Ok theres alot i forgot..but if someone can give me the formulas from our last section i will be happy :)

2010-04-11T21:09:00.002-05:00

Well spring break is over...this sucks. Back to math, here's some old stuff that we did back in the gap.

SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°

Something that I need some help with, and it seems like a lot of people are having trouble with this too, is Integral Coefficients....

alaina's blog, 11 april 2010

2010-04-11T20:52:00.000-05:00

i'm going over old stuff, and for the most part, i remember everything. so i'm explaining how to complete the square.

when given an equation (ax^2+bx+c=0) where "c" is the constant, first add or subtract the constant over (ax^2+bx+_=-c). Second, divide the linear term (b) by two and then square it (b/2)^2. Your would then add your squared, lets call it "d", term to both sides of the equal sign (ax^2+bx+d=-c+d). You would then factor to get (ax+b/2)^2=-c+d. After factoring, you would take the square root of both sides (ax+b/2)=+/- √-c+d. Then, you would add or subtract (b/2) to both sides. And finally, you would divide by "a" to get "x" by itself. You would then write in coordinate form.

What I don't understand from the exam review is how to get an equation from INTEGRAL COEFFICIENTS!!!!! help.

alaina's makeup blog for 28 March 2010

2010-04-11T20:48:00.002-05:00

Okay, so this week I have been completely bogged down with studying for exams and keeping up with notes and homework in everything else. And, I almost forgot to do my reflection. So here goes.

First, I understand conics a lot better than I did last year, or durring the summer.
Circle
(x-h)^2+(y-k)^2=r^2
I always thought circles were easiest.

Your center is (h,k)
radius=r; take square root of r^2
to graph, plot your center and count your radius all the way around.
Ellipse

(x^2+h/a^2)-(y^2+k/b^2)=1

Your center is (h,k)
Your major axis is the non-negative denominator
minor is negative
Hyperbola

(x^2-h/a^2)+(y^2-k/b^2)=1

Your center is (h,k)
your major axis has the larger denominator

Parabola

y-y⊂1&sub/=m(x-x&sub/1)

**don't understand**

sigma notation i could use a little help on, also INTEGRAL COOEFFICIENTS!!!

Stephanie's Reflection

2010-04-11T15:23:00.000-05:00

3) (2+4i)²
(2+4i)(2+4i)
4+8i+8i+16i²
4+16i-16i²
-12+16i²

.25 = i = complex # = √(-1)
.5 = i² = -1
.75 = i³ = -i
whole # = i^4 = i
(divide exponent by 4)
a+bi form

4) 5-2i
4+3i
5-2i 4-3i
4+3i * 4-3i
20-15i-8i+6i²
16-10i+12i-9i²
20-23i-6
16+9
14-23i
25
=14/25+23/25i

5) 2+6i 4+2i
4-2i * 4+2i
8+4i+24+12i²
16+4i²
8+28i+12
16+4
20+28i
20
=1+7/5i

Amy's Relfection #34 (Easter Blog)

2010-04-10T01:38:00.001-05:00

okay now here is a review on some trig questions...
Example 1: Find the exact value of sin 80 cos 130+ cos 80 sin 130

Let alpha = 80 and beta = 130 then sin 80 cos 130 + cos 80sin 130 = sin alpha cos beta + cos alpha sin b

= sin (alpha + beta)
= sin (80+130)
= sin (210)
= sin (180 + 30)
= - sin 30
= -1/2

Example 2: Find the exact value of

cos^215 - sin^215 = cos(2alpha)
= cos(30)
= sqrt 3/2

Example 3: cos 15

alpha = 45, beta = 30

cos (alpha - beta) = cos 45 cos30 + sin45 sin 30

cos (45 - 30) = (√(2/2) (√(3/2) + (√(2/2) (1/2)

= √(√6 + √2)/all over 4

Example 4: (1 + cot^2) (1 - cos 2x)

(csc^2x) (1-cos2x)

(csc^2x) (1-(1 - 2sin^2x)

(1/sin^2x) (2sin^2x)

= 2

Example 4: Find the exact value of sin22.5

alpha=2(22.5)

alpha=45

**if you are given an angle with a decimal you use the half-angle formula. To find alpha, you multiply by two.

Example 5: Sum Formula for Cosine

cos 75 cos 15 + sin 75 sin 15

=cos (75-15) = cos 60 = 1/2

Example 6: Find the exact value of tan15 +tan30/1-tan15 (tan30)

= tan(15 + 30)

= tan(45)

=1

question time: can someone tell how to Rewriting a Sum or Difference as a Product??

Amy's Reflection #33

2010-04-10T01:32:00.002-05:00

okay here is a review of chapter 13. it's fairly easy...

13-1

1. sequence-list of numbers

2. two main types: 1). arithmetic-add or subtract 2).geometric-multiply

Formulas:

1. arithmetic-used to find a term: tn . t1 + (n-1)d

**n=term #, t1=first term, d=what you add, tn=term #

2. geometric: tn=t1 . r^(n-1)

**r=what you multiply by..

Examples:

1. Find the formula for the nth term of the arithmetic sequence: 3,5,7,...

tn = 3 + (n-1) (2)

tn = 3 +2n - 2

tn = 1 + 2

2. Find the formula for the nth term of the sequence: 3,4.5,6.75,..

**divide the 2nd term by the 1st term to find r

4.5/3 = 3/2 = r

tn = 3 . (3/2)^(n-1)

13-2

Formula for a sequence that involves the previous term: (an - 1)

Examples:

1. Find the recursive definition of: 81, 27, 9,3,...

an = an - 1/3

2. 1, 2, 6, 24, 120, 720, ....

n = 1: 1

n= 2: 2

n = 3: 6

an = n . an - 1

13 -3

Series-List of added or subtracted numbers

**Leave it as a list: do NOT add

Formulas:

1. Arithmetic: Sn = n(t1 + t2)/2

**Finds the sum of the first n terms

2. Geometric: Sn = t1 (1 -r^n)/1-r

Examples:

1. Find the sum of the first 25 terms of the series: 11 + 14 + 17 + 20 + ....

Sn = n (t1 + tn)/2

t25 = 11 + (24)(3)

Sn = 25 (11 + 83)/2

= 1175

2. Find the sum of the first 10 terms of the series: 2-6 + 18 - 54 +...

**This is a geometric sequence and that is because you have to add or subtract the same number for it to be an arithmetic sequence, got it??

r = -6/2 = -3

Sn = t1(1 - r^n)/1-r

= 2(1 -(-3)^10)/1 - (-3)

= 2(-59048)/2

= -29524

so i hope that helped to refreshen your minds..now for a question: can someone help with #7 on the study guide..i can't seem to remember how to do it :(

Alicia's 2nd Spring Break Blog

2010-04-06T15:04:00.002-05:00

Alrighty so here is my second reflection for the break.

These are the names and equations to remember for chapter 11:

*Limacon
r=a+b sin(theta) r=a+b cos(theta)

*Cardioid
r= a+ or -bsin (theta) r= a+ or - bcos (theta)

*Rose Curve
r=a sin(n theta) r=a cos (n theta)
n=how many petals there are

*Archimedes Spiral
r=a (theta)+b

*Logarithmic Spiral
r=ab^(theta)

*Common circle with center point at the pole
r=a sin (theta) r=a cos(theta)

When converting to recangular, use....

x=r cos (theta) or y=r sin (theta)

When converting to polar, use...

r= + or - √x² + y²

To take the tan inverse....
theta= tan^-1(y/x)

I could also use some help with recursive definitions!!

Alicia's 1st Spring Break Blog

2010-04-06T15:01:00.002-05:00

Okayy so I hope everyone had a great Easter!!! I am going to do both of my blogs now because I am leaving for Arizona Friday and won't be home until late Sunday night.....

Infinite Sequences and Series

*lim
n-infinity: if the degree of the top= the degree of the bottom, then the answer is the coefficients.

Example:

lim n^2+1/2n^2-3n = 1/2
n-infinity

*lim
n-infinity: if the degree of the top is > the degree of the bottom, then your answer is infinity

Example:

lim 7n^3/4n^2-5 = infinity
n-infinity

*lim
n-infinity: if the degree of the top is < the degree of the bottom, then your answer is 0

Example:

lim 5n^2/3n^3+7 = 0
n-infinity

***If no rules apply, then you have to use your calculator to find what the sequence is approaching.

13-5 Sums of Infinite Series

*they can only be found with a geometric serires where /r/<1

Formula: S= t1/1-r

Example: 9-6+4

r= -6/9= -2/3 geometric

/-2/3/<1

S= 9/(1-(-2/3))= 27/5

Example: Write .45 repeating as a fraction

45/100-1= 45/99= 5/11

I could use some help with remembering how to do sigma notation... Thanks!! :)

taylor easter reflection

2010-04-05T20:49:00.002-05:00

Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)

Area of Right Triangles
A=1/2bh

SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°

both law of sine and law of cosines are formulas used to solve for the components of a non right triangle

law of sine is a formula used to solve for the components of a triangle when there is a set or pair of information given meaning a length of a side on the opposite side of a given angle.

law of cosines is a formula used to solve for the components of a triangle when a set or pair of information is not given.

Area of Inscribed Shapes
A=nr²sinΘcosΘ

2010-04-04T23:26:00.002-05:00

This week we learned about Vectors:

-Find the components of the vector AB
to do this all you have to do is subtract the components of B from the components of A
-Find the magnitude
if you're dealing with two points, so to find the magnitude you're going to use the distance formula

-Find u + v
you have to add the components of u and v
-Find 2v + u
First you have to multiply the components of v by 2, then you add it to the components of u
then you add them together

so i think i got this, but something i need help with is everything from the past haha

Stephanie's Reflection

2010-04-04T21:05:00.003-05:00

Intersections of Lines
Solving a System’s Equations
A. Eliminate the variable
Solve for the variable
Plug back in
(#,#)

B. Solve for variable
Substitute
Plug back in

Point Slope Formula
y-y1=m(x-x1)
Slope Intercept Formula
y=mx+b
Standard Formula
ax+by=c
m=-a/b

1) 2x+5y=10
3x+4y=12
3(2x+5y=10)
2(3x+4y=12)
6x+15y=30
- 6x+8y=24
7y=6
y=6/7
2x+5(6/7)=10
2x+(34/7)=10
2x=5 5/7
x=20/7
(20/7,6/7)

2) y=3x+4
m=3
m,,=3

Breaks = Blogs? Noooo!

2010-04-04T20:13:00.046-05:00

We've only been off for a few days now, and I already feel like I forgot half of what we learned before break, so... I'm going to review some trig stuff.

SOHCAHTOA

Sin=opp/hyp
Cos=adj/hyp
Tan=opp/adj

Trig functions:

sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y

Law of sines:

(sinA)/a = (sinB)/b = (sinC)/c

Law of cosines:

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 -2(adjacent leg)(adjacent leg)cos(angle between)

taylor 28 march reflection

2010-03-30T19:23:00.002-05:00

so we have to start preparing for the trig exam

so away we go

**The Unit Circle

90 degrees, (0,1), pi/2

180 degrees, (-1,0), pi

270 degrees, (0,-1), 3pi/2

360 degrees, (1,0), 2pi

**6 Trig Functions

sin = y/r

cos = x/r

tan = y/x

csc = r/y

sec = r/x

cot = x/y

**Degrees & Radians

Degrees to radians= Degree * pi/180

Radians to degrees= Radians * 180/pi

**To solve coterminal angles, either add or subtract 360 to the angle

i need a review of the chapter with amplitude and the graphs and other information that goes along with a problem like that

Stephen's Reflection

2010-03-30T19:14:00.003-05:00

Ok so this week i think we are learning something new..idk but anyway i do remember trig functions really well because they are really east. there are six trig functions:

6 Trig Functions

sin = y/r

cos = x/r

tan = y/x

csc = r/y

sec = r/x

cot = x/y

theres a few things i do not understand which is law of sines, law of cosines, and sigma notations...

Lima Beans, Hearts, and Roses!

2010-03-28T21:12:00.019-05:00

Limacon - Looks like a lima bean!

r = a+b sin theta
r = a+b cos theta

Cardioid - Looks like a heart!

a-b sin theta
a-b cos theta

Rose - Looks like a ...rose.

r = a sin (number of petals) theta
r = a cos (number of petals) theta

Archimedes Spiral - The black and white spiral that hypnotizes people in the cartoons.

r = a theta +b

Logarithmic Spiral - Looks like a ...spiral.

r=a^theta b

Alicia's Reflection #32

2010-03-28T21:08:00.002-05:00

Alrighty so last week we finished taking our chapter tests. I think this week is ACT prep so hopefully we review ACT stuff in math. I am going to review some trig from Ch. 10.

Law of Sines:

sinA/a = sinB/b = sinC/c

Law of Cosines:
(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg)cos(angle between)

Example:
x= 6^2 + 5^2 -2(5)(6) cos 36
x=3.530

Here are some formulas:

Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2

tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β

sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α

I could use some help with sigma notation

Reflection

2010-03-28T21:00:00.002-05:00

So I'm thinking that since B-Rob is suppose to be coming back this week and originally was suppose to be coming back last week we will have A LOTTTT of work to do in order to catch up. I guess I could use it because I really do not remember too much...area of a non right triangle.

Area of a Non Right Triangle:

Formula:1/2(leg)(leg)sin(angle b/w)

So if you had 1/2(3)(6)sin(52) your answer would be: 7.100

Does anyone know what we're suppose to be doing when B-Rob comes back?

Amy's Reflection #32

2010-03-28T15:14:00.000-05:00

Stephanie's Reflection

2010-03-28T12:14:00.001-05:00

Limacon
r = a+b sin theta
r = a+b cos theta

Cardioid
a-b sin theta
a-b cos theta

Rose
r = a sin n theta
r = a cos n theta
n is how many petals

Archimedes Spiral
r = a theta +b

Logarithmic Spiral
r=a^theta b

Converting
polar to rectangular
x=r cos theta
y=r sin theta

rectangular to polar
r=+/- sqrt x^2 + y^2
theta is (x/y)

Trig Chart:

0°
sin0=0
cos0=1
tan0=undefined
sec0=1
cot0=0

30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3

45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1

60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2

90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

taylors reflection for 21 march

2010-03-23T08:10:00.002-05:00

there are two types of limit equations
the ones that use rules and the ones that use a calculator

the ones that use rules have simple hints to memorize for solving
the only ones that use rules are the polynomial equations problems

memorize this

((the rules))
t- top lead co
b- bottom lead co

t=b then coefficients
t>b then infinity
t
if you get a problem with a limit that is a polynomial equation
use the rules.
each and every time

the other type of problem is the one that calls for the use of a calculator
every single problem with limits that is not a polynommial equation calls for the use of a calculator

all you have to do is plug in for n three different times with
100
1000
10000
then plug into calculator
record what each outcome is and decipher what the numbers are headed toward which will then be your answer

what i need help on is memorizing the formulas for chapter eight
they are the only part of trig that i do not understand and since we are only going to have two review weeks before the trig exam i really could used some help...
i dont know if its chapter eight or chapter ten
but i need help on memorizing the formulas from the chapter in trig where you are substituting

Stephen's Reflection

2010-03-22T19:49:00.001-05:00

Ok so this week we dont have a teacher again so theres stuff i need help on. Anyway i really understand and remember trig stuff and trig functions. I understand the 6 trig functions and how to use them.

The trig functions are:

sin 0= y/r

cos 0= x/r

tan 0= y/x

csc 0= r/y

sec 0= r/x

cot 0= x/y

Waht i need help on is the formulas for hyperbolas and circles and stuff like that and i needa know what i need to find for each

alaina's blog, 21 march 2010

2010-03-21T21:30:00.000-05:00

law of sins :)

sinA/a=sinB/b=sinC/c

*only used when you have pairs, an angle and the side opposite of it.
*setting up a proportion.

Ex: a civil engineer wants to determine the distance from points A and B to an inaccessable point C. from direct measurement -- AB=25m,
first you would draw a diagram and lable EVERYTHING. Then, choose your pairs. Finally set up a proportion.

(sin50/25)=(sin20/B)
cross multiply--Bsin50=25sin20
divide by sin50
B=11.162m

you would follow the same process to find side "a".

I still don't understand integral coefficients if anyone wants to help.

Reflection

2010-03-21T20:23:00.000-05:00

SOHCAHTOA:
sin theta=opposite/hypotenuse
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent

SOHCAHTOA is used when either you have two sides of a right triangle and you need to find an angle or you have an angle and one side. Here's an example:

A right triangle has 3 angles: 90°, 30°, and 60°. The hypotenuse is x cm. The side opposite the 60° angle is 8 cm. What is the length of the hypotenuse?

You would use the sin formula and the equation would be sin(60)=8/x.
Then you would get .8660=8/x
You divide 8 by .8660 and get 9.2380
So the hypotenuse of the triangle would be 9.2380 cm.

So what exactly are we doing this week in this class?

Alicia's Reflection #31

2010-03-21T19:56:00.001-05:00

Alrighty so we took our exam on flatland which was easy for the most part. I am going to review some material on circles because I kind of forgot this chapter.

This is the standard form of a circle: (x-h)^2+(y-k)^2=r^2
The center of a circle is: (h,k)
The radiusis represented by: r

*Find the center and radius of the circle.

1.) (x-3)^2+(y+7)^2=19
center: (3,-7)
radius: squareroot of 19

*Find the intersection of the circle.

1.) x^2+4^2-25 and y=2x-2

a) y=2x-2
b) x^2=(2x-2)^2=25
c) x^2+4x^2-8x+4=25

5x^2-8x+4=25
5x^2-8x-21=0
5x^2-15x+7x-21=0
5x(x-3)+7(x-3)=0
(x-3)(5x+7)
x=3 x=-7/5
y=2(x)-2
2(3)-2=4
y=2(-7/5)-2=-24/5

(3,4) (-7/5,-24/5)

*Write in Standard form.

1.) Center: (4,3)
Radius: 2

(x-4)^2+(y-3)^2=4

Stephanie's Reflection

2010-03-21T19:48:00.000-05:00

Sine and Cosine Sum/Difference Formulas:
cos(alpha+/-beta)=cos alpha cos beta-/+sin alpha sin beta )
sin(alpha+/- beta)=sin alpha cos beta +/-cos alpha sin beta
sin x+sin y=2sin(x+y/2)cos(x-y/2)
sin x-sin y=2cos(x+y/2)sin(x-y/2)
cos x+cos y= 2cos(x+y/2)cos(x-y/2)
cos x-cos y=-2sin(x+y/2)sin(x-y/2)

Tangent Sum/Difference Formulas:
tan(alpha+beta)=tan alpha+tan beta/1-tan alpha tan beta
tan alpha-beta=tan alpha-tan beta/1+tan alpha tan beta

Double-Angle/Half-Angle Formulas:
sin 2α=2sinα cosα
cos 2α=cos2α-sin2α=1-2sin2α=2cos2α-1
tan 2α=2tanα/1-tan2α
sin(α/2)=+/-√(1-cosα/2) cos(α/2)= +/-√(1+cosα/2)
tan(α/2)= +/-√(1-cosα/1+cosα)=sinα/1+cosα= 1-cosα/sinα)

sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y

Trig Chart:
0°
sin0=0
cos0=1
tan0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

Reciprocal Relationships:
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ

Relationships with Negatives:
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
Tan -Θ= -tanΘ and cot -Θ= -cotΘ

Pythagorean Relationships:
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ

Cofunction Relationships:
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)

Amy's Reflection #31

2010-03-20T22:27:00.001-05:00

here's some stuff from chapter 11..

Imaginary Numbers are no longer "imaginary"

Rectangular form: a + bi

Polar form: z = r cos theta + r sin theta i (abbreviated z = r cis theta)

Examples:

1. Express 2 cis 50degrees in rectangular form

2 cos 50 + 2 sin 50 i

2. Express -1-2i in polar form

radius = +- sqrt of ((-1)^2 + (-2)^2)) = +- sqrt of (5)

theta = tan^-1(-2/-1)

theta = tan^-1(1)

*tangent is positive in the first and third quadrants, 63.435 and 243.435
*63 is positive for cosine so it goes with the positive sqrt of 5
*243 is negative for cosine so it goes with the negative sqrt of 5

z= sqrt of 5 cis 63.435

z= sqrt of 5 cos 63.435 + sqrt of 5 sin 63.435 i

z= negative sqrt of 5 cis 243.435

z= negative sqrt of 5 cos 243.435 + negative sqrt of 5 sin 243.435 i

De Moivre's Theorem: z^n = r^n cis(n)(theta)

Examples:

1. z=2cis20degrees Find z^2

z^2=2^2cis2(20degrees)

z^2=4cis40degrees

2. 4cis15degrees Find z^4

z^4=4^4cis4(15degrees)

z^4=256cis60degrees

Limacon
r=a+b sin(theta)
r=a+b cos(theta)

Cardioid
a-b sin(theta)
r=a-b cos(theta)

Rose
r=a sin(n theta)
r=a cos (n theta)

*n=how many petals

Archimedes Spiral
r=a theta+b

Logarithmic Spiral
r=a b^theta

Examples:
1. r=theta+2
2. r=2+3cos(theta)
3. r=5
4. r=3sin(4 theta)
5. r=1/2(3^theta)
6. r=2sin(theta)

1. archimedes spiral
2. limacon
3. circle with its center at the pole
4. rose with 4 petals
5. logarithmic spiral
6. circle that intersects with the pole

ok what i really dont understand is the first two sections..if someone could explain them to me that would be awesome..thanks..

Cemments

2010-03-18T10:23:00.002-05:00

Q. can someone help me with the trig chart?

A.Trig Chart:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

Q.I dont seem to remember the trig functions...can someone remind me what they are?

A.oh yeah sure man i got you...

The trig functions are:

sin 0= y/r

cos 0= x/r

tan 0= y/x

csc 0= r/y

sec 0= r/x

cot 0= x/y

taylors 17 March "comment" blog b/c there were no questions

2010-03-18T09:04:00.001-05:00

Im going to post some more review stuff since there were no comments

this is from around chapter nine

Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)

Area of Right Triangles
A=1/2bh

SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°

Area of Inscribed Shapes
A=nr²sinΘcosΘ

2010-03-17T21:07:00.001-05:00

Alrighty so we have our exam this week on Flatland. Basically just study the questions we did in class along with the vocabulary words. okay I am going to review some trig laws:

*Law of Sines: sin(opp. angle)/Leg =sin(Opp. angle)/Leg.

Example: triangle ABC where A=36 degrees a=3 and B=56 degrees. find b

sin36/3=sin56/x 3sin56/sin36= x

*Law of Cosines: (opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)

Example: for a triangle with C=36 degrees a=5 b=6

c^2=5^2+6^2-2(5)(6)cos36

c=Square root of(25+36-2(5)(6)cos36)
c= 3.53

*The area of non-right triangle:

1/2(leg)(leg)sin(angle between)

Example: Triangle ABC has sides a=5 b=3 and C=40 degrees

= (1/2)(5)(3)(sin(40))

Goodluck on the exam!!!! :)

taylor 15 march reflection

2010-03-16T08:08:00.002-05:00

Since we are focusing on review tests right now i figured id start reviewing on the more recent chapters because those will start becomming foggy to us as we review the early chapters.

Graph shapes and their formulas

Limacon
r=a+b sin(theta)
r=a+b cos(theta)

Cardioid
a-b sin(theta)
a-b cos(theta)

Rose
r=a sin(n theta)
r=a cos (n theta)
(n=how many petals {if n isodd[#=n] if n is even [#=2n]}

Archimedes Spiral
r=a theta+b

Logarithmic Spiral
r=a^theta b

CONVERTING

when going from polar to rectangular you plug into

X=rcos(theta)
Y=rsin(theta)

and work out until you get a x point and a y point

when going from rectangular to polar you plug into

r=+/- squareroot X^2 +Y^2
and
Theta= (Y/X)

once youve solved for both of these you"ll plug into (+r, theta) (-r, theta)

i need help with a review of the formulas from the triangle section if anyone can help please do!

Devin's Reflection

2010-03-15T19:58:00.001-05:00

Unit Circle:

90 degs. = (0,1) pi/2

180 degs. = (-1,0) pi2

70 degs. = (0,-1) 3pi/2

360 degs. = (1,0) 2pi

sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y

SOHCAHTOA:

S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)

C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)

T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)

*the hypotenuse is opposite the right angle.

*A= 1/2 bh*

*To find the area of a non right triangle use this formula:

*A= 1/2 (leg)(leg)SIN(angle b/w)

*When you have a non right triangle that has pairs, use the law of sines:

Sin A/a = Sin B/b= Sin C/c

*All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)

Devin's Reflection

2010-03-15T19:57:00.001-05:00

CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)

EG:
(x-4)^2+(y+2)^2=16
center:(4,-2) radius:4

x^2+y^2+12y+16x-5=0
x^2+16x+64+y^2+12y+36=5+64+36
(x+8)^2+(y+6)^2=105
center:(-8,-6) radius:square root of 105

ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph

EG:
x^2/4+y^2/1=1
1) (0,0)
2) x
3) +/-2 (2,0) (-2,0)
4) +/-1 (0,1) (0,-1)
5) 1=4-c^2 c=+/-square root of 3 (sr3,0) (-sr3,0)
6) 2 square root of 4 = 4
7) 2 square root of 1
8) graph

HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom

to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes

EG:
x/36-y/9=1
2) (0,0)
3) x
4) y
5) none
6) c^2=36+9 c^2=45 c=sr45 (sr45,0) (-sr45,0)
7) y=+/-square root of 5/square root of 6
8) +/-sr36 = +/-6 (6,0) (-6,0)
9) sketch

Devin's Reflection

2010-03-15T19:54:00.001-05:00

The standard form for a circle equation is (x-h)^2+(y-k)^@=r^2. The center is (h,k).

Examples: Find the center

a. (x-3)^2+(y+7)^2=19

center (3,-7)

b. x^2+y^2-6x+4y-12=0

x^2-6x+ (9)+y^2+4y+ (4)=12+9+4

(x-3)^2+(y+2)^2=25

center (3,-2)

The r stands for the radius of the circle. When the equation is not in standard equation you have to complete the square to put the equation in standard equation. You can determine the radius of a circle, by using the distance formula and if you are given the center and a point.

Example: Find radius

a.(x-3)^@+(y+7)^2=19

radius square root of (19)

To find the intersection of a line and a cirlce

1. solve the linear equation for y
2. substitute in circle equation
3. solve for x
4. plug x-value into get y-value

*If your x=value is imaginary, then there is no point of intersection.

With ellipses, the main thing to know is the steps. They are:

1. Find center
2. Find major axis - big denom.
3. Find vertex - square root of big denom.
4. Find other intercepts - square root of small denom
5. Find Focus
6. Find length of major axis
7. Find length of minor axis
8. Graph

Just like the ellipses, to sketch the hyperbolas you must follow the steps.
First find:

1. shape
2. center
3. major axis
4. minor axis
5. 0ther int.
6. vertex
7. focus
8. asymptotes
9. then sketch

To sketch a hyperbola,do the following

1. Draw a box using the vertex and square root of the other denom
2. Draw diagonals through box
3. Sketch a parabola on each vertex
4. Label focus and asymptotes

Devin's Reflection

2010-03-15T19:50:00.000-05:00

Parabolas:

how to find the axis of symmetry, vertex, focus, & directrx??

1.) to find the axis of symmetry: x = -b/2a

2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:

y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex

3.) to find the focus: 1/4p= the coefficient of x^2 and then add p

Note:

*If opens up, add to y value from vertex, if opens down, subtract

*If opens right, add to x value to vertex, if opens left, subtract)

4.) directrix: is p units behind the vertex

Note:

*If opens up, subtract; if opens down, add from y-value of vertex.
*If opens right, subtract x-value
*If opens left, add x-value

Example: x^2 + 1

~vertex:

x = -b/2a

x = 0/2(1) = 0

0^2 + 1 = 1

(0,1)

~Focus:

1/4p = 1

4p = 1

p = 1/4

(0, 1 + 1/4)

(0, 5/4)

~directrix:

y = 1 - 1/4

y = 3/4

Devin's Reflection

2010-03-15T19:48:00.001-05:00

TRIGONOMETRY

Angles

measured in degrees
to find minutes, multiply what is behind the decimal by 60
to find seconds, multiply what is behind the decimal by 0 and divide by 300 to get decimal
angles are measured in degrees and radian

radians = degrees times pi over 180

degrees = radians times 180 over pi

to find coterminal angles, add or subtract 360 degrees or 2 pi
must use degrees symbol if in degree or its wrong

if no degree symbol, its assumed that you are in radians

EG: 12.3 degrees... write as t

12 degrees .3 times 60

12 degrees 1.8 minutes

15.36 degrees

15 degrees .36 times 60

15 degrees 21 minutes .6 times 60

15 degrees 21 minutes 36 seconds

25 degrees 20 minutes 6 seconds

25 plus 20 divided by 60 plus 6 divided by 3600

25.335 degrees

15 degrees 26 minutes 15 seconds

15 plus 26 divided by 60 plus 15 divided by 3600

15.4375 degrees

36 degrees

36 divided by 180 pi

1/5 pi

pi/5

3 pi/4 times 180/pi

135 degrees

225 degrees

225 divided by 180 pi

5/4 pi

pi/9 times 80 divided by pi

20 degrees

2.4 times 180 divided by pi

137 degrees 30 minutes 35.535 seconds

245 degrees 15 minutes 300 seconds

2445 plus 15 divided by 60 plus 300 divided by 3600

245.3 degrees

Devin's Reflection

2010-03-15T19:47:00.000-05:00

1)s=rθ
2)k=1/2r^2θ (k=1/2rs)
r = radius, θ = angle, s = arc length, k = area of sector

EG: A sector of a circle has an arc length of 6cm and an area of 75cm^2. Find its radius and measure of its central angle.
s=6cm
k=75cm^2
r=?
θ=?
k=1/2rs
75=1/2r6
75=3r
r=25cm
s=r
6=25θ
θ=6/25

sinθ=y/r
cosθ=x/r
tanθ=y/x
cscθ=r/x
secθ=x/y
cotθ=x/y
r=√(x^2+y^2)

EG: sin180° = y/r
at 180°, y=0
0/1 = 0
thus sin180°=0

EG: cosπ/2 = x/r
π/2 is at 90°
at 90°, x=0
thus cosπ/2=0

Trig Chart:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)

1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°

EG: sin^-1(-√2/2) = 45°
45 + 180 = 225° = θ
I to IV
-45 + 360 = 315°
θ = 225°, 315°

Devin's Reflection

2010-03-15T19:44:00.001-05:00

The sin and csc both deal with the y factor and radius. The cos and sec both deal with the x factor and radius. The tan and cot both deal with the x factor along and the y factor. Radius is the square root of x(2) + y(2).

The trig functions are:

sin 0= y/r

cos 0= x/r

tan 0= y/x

csc 0= r/y

sec 0= r/x

cot 0= x/y

Unit Circle

2-(0,1)-90-pi/2 1-(1,0)-0,360-0,2pi

3-(-1,0)-180-pi 4-(0,-1)-270-3pi/2

The Trig Chart

0 sin 0=0 cos 0-1 tan 0=0

30 sin pi/6=1/2 cos pi/6= (3)/2 tan pi/6=(3)/3

45 sin pi/4=(2)/2 cos pi/4=(2)/2 tan pi/4=1

60 sin pi/3=(3)/2 cos pi/3= 1/2 tan pi/3=(3)

90 sin pi/2=1 cos pi/2=0 tan pi/2= underined

0 csc 0= undefined sec 0=1 cot 0= undefined

30 sin pi/6=2 sec pi/6=(2) cot pi/6=(3)

45 sin pi/4=(2) sec pi/4=(2) cot pi/4=1

60 sin pi/3=2(3)/3 sec pi/3=2 cot pi/3=(3)/3

90 sin pi/2=1 sec pi/2=undefined cot pi/2=0

Reference angles must be between 0 and 90, 0 and pi/2

1. Find which quadrant angle is in

2. Determine the sign in that quadrant (+ve or -ve)

3. Subtract 180 until the angle is between 0 and 90, 0 and pi/2

We also learned inverses.

To solve for an angle

1. simplify any expression

2. get the trig function by itself

3. take the trig inverse of both sides

4. use chart unit circle or calculator to find 1 angle

5. set up and find other quadrants with the same sign as the value.

Ex. sin -1((2)/2)=45 quadrants 1 and 2

To make from quadrant to quadrant:

1-3 make it -ve and add 360

1-2 make it -ve and add 180

1-4 add 360

2-4 add 180

Devin's Reflection

2010-03-15T19:43:00.001-05:00

Rational Root therom

Example: f(x)= 2x^3 + 3x^2 - 8 + 3

Step 1: find all possible roots..

p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2

*p is the leading constant term & q is the leading coefficient

possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2

Step 2: now you can plug all of the possible roots in your calculator to find the roots that work

the zero will be: 1, 1/2, -3

Step 3: use synthetic division to factor all of the roots that work

you should get: (x - 1) (2x^2 + 5x + 3)

Step 4: slove further

(this can be factored...)

= (x - 1) (2x^2 + 5x + 3)

= (x - 1) (2x - 1) (x + 3)

(set x = 0 )

x = 1, 1/2, -3