Dustin's Final Reflection

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

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

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

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:

1. check identities
2. algebra
3. 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

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

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

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

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

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

**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

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

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.

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

Taylor final reflection

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

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!

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

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

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

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

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

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 :)

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

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

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

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

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

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

Taylor reflection 2 may 2010

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

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

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

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

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

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

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!!!