trig
posted by Sarah Kim .
how can you confirm the identity
cos^4x = (1/8)(3+ 4cos2x+ cos4x)
and
sin4x = (4sinxcosx)(2cos^2x1)

trig 
Count Iblis
You use the identity:
exp(ix) = cos(x) + i sin(x)
You can then write:
cos(x) = [exp(ix)+ exp(ix)]/2
Take the fourth power:
cos^4(x) =
1/16 [exp(4 i x) + 4 exp(2 i x) + 6 +
4 exp(2 i x) + exp(4 i x)] =
1/8 {3 + [exp(4 i x) + exp(4 i x)]/2 + 4 [exp(2 i x) + exp(2 i x)]/2} =
1/8 [3 + cos(4 x) + 4 cos(2 x)] 
trig 
Count Iblis
sin(4x) can be expanded in the usual way by using the doubling formulas:
sin(4x) = 2 sin(2x) cos(2x) =
4 sin(x) cos(x)cos(2x)=
4 sin(x) cos(x) [2 cos^2(x)  1]
The general method also works, but it slightly more laborious in this simple case:
You start with
exp(ix) = cos(x) + i sin(x)
and take the fourth power of both sides and take the imaginary part:
Im[exp(4 i x)] = 4 cos^3(x)sin(x) 
4 cos(x) sin^3(x)
Im[exp(4 i x)] = sin(4 x), so we have:
sin(4 x) =
4 cos(x)sin(x)[cos^2(x) sin^2(x)]
Using that sin^2(x) = 1  cos^2(x) gives you the desired result.
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