Trig question (1 Viewer)

[kaz1]

cos^2 theta = 2cos^2 (theta/2)

edit: Don't worry about it. I figured out how to do it.

 

[kaz1]

is that (2cos^2 (theta))/2 or 2cos^2 (theta/2)

2cos^2 (theta/2). Sorry about that.

 

[shaon0]

2cos^2 (theta/2). Sorry about that.

didn't we do this in class?
it was easy i figured it out second go..Question 11) was hard for a while but now it is easy.

 

[kaz1]

didn't we do this in class?

I forgot to copy it down.

 

[shaon0]

alrite i'll do it from memory...
Let A=theta/2
Therfore, cos^2(2A)=2cos^2(A)
cos^2(A)-sin^2(A)/2=cos^2(A)
(Allowable values are:0<=A<=180)
cos(A)-sin(A)/sqrt(2)=cos(A)
....forgot the rest
*working on Exercise 22c)*
128 degrees 168 degrees
or something like that i didn't copy it down either.

 

[namburger]

Formula:
(Cos x )^2 = (1 + cos 2x)/2

Therefore: 2cos^2(x/2) = 1 + cos x

1 + cos x = (Cos x )^2
(Cos x )^2 - cos x - 1 = 0
Solve quadratic