# trig

posted by Sarah Kim

how can you confirm the identity

cos^4x = (1/8)(3+ 4cos2x+ cos4x)

and

sin4x = (4sinxcosx)(2cos^2x-1)

1. 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)]

2. 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:

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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