Math 2nd question
posted by Alex .
Express as a single sine or cosine function (note: this is using double angle formulas)
g) 8sin^2x4
I just don't get this one. I know it's got something to do with the 12sin^2x double angle formula. It's the opposite though? :S
h) 12sin^2 (π/4x/2)
= 1sin^2(π/4x/2)sin^2(π/4x/2)
= cos^2(π/4x/2)sin^2(π/4x/2)
I got all the way up to cos (π/4  x/2 + π/4  x/2)
The answer is supposed to be sin x. I have no clue how they got that.

g) 8sin^2x4
= 4(2sin^2 x  1)
= 4(cos 2x)
= 4cos 2x
h) 12sin^2 (π/4x/2)
= cos 2(π/4x/2)
= cos (π/2x)
= cos(π/2)cosx + sin(π/2)sinx
= 0(sinx) + 1(sinx)
= sinx 
Thanks a lot!
But can you please explain how you got 4(2sin^2 x  1) for g and cos 2(π/4x/2) for h?
I think I might understand h because 1sin^2x = cosx but wouldn't it just be cos(π/4x/2) cos(π/4x/2)? 
"But can you please explain how you got 4(2sin^2 x  1)"
I took out a common factor of 4
"..and cos 2(π/4x/2) for h"
Ok, let's work it in reverse.
You know that cos 2A = cos^2 A  sin^2 A, or cos 2A = 1  2sin^2 A ,right?
so I simply let A = (π/4x/2)
then 2A = 2(π/4x/2)
= π/2  x