Amy's Blog Topic Response

**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

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

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

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

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

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

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)