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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