# Math 2nd question

Express as a single sine or cosine function (note: this is using double angle formulas)
g) 8sin^2x-4
I just don't get this one. I know it's got something to do with the 1-2sin^2x double angle formula. It's the opposite though? :S

h) 1-2sin^2 (π/4-x/2)
= 1-sin^2(π/4-x/2)-sin^2(π/4-x/2)
= cos^2(π/4-x/2)-sin^2(π/4-x/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.

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1. g) 8sin^2x-4
= 4(2sin^2 x - 1)
= 4(-cos 2x)
= -4cos 2x

h) 1-2sin^2 (π/4-x/2)
= cos 2(π/4-x/2)
= cos (π/2-x)
= cos(π/2)cosx + sin(π/2)sinx
= 0(sinx) + 1(sinx)
= sinx

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2. Thanks a lot!
But can you please explain how you got 4(2sin^2 x - 1) for g and cos 2(π/4-x/2) for h?

I think I might understand h because 1-sin^2x = cosx but wouldn't it just be cos(π/4-x/2) cos(π/4-x/2)?

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3. "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/4-x/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/4-x/2)
then 2A = 2(π/4-x/2)
= π/2 - x

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