[Math] Prove this identity: $\sin^4x = \frac{1}{8}(3 – 4\cos2x + \cos4x)$

algebra-precalculustrigonometry

The problem reads as follows.

Prove this identity: $$\sin^4x = \frac{1}{8}(3 – 4\cos2x + \cos4x)$$

I started with the right side and used double angles identities for $\cos2x$ and a sum and then then double angle identity for the $\cos4x…$ It all got messy and I hit a dead end. He doesn't give any hints and I'm pretty lost.

I'm thinking that I will eventually need a product-sum identity to get the $\dfrac{1}{8}….$ But I'm just confused how to get there…

Thanks in advance to anyone who can help!

Best Answer

Or better yet, use Euler's formula!

$$\sin^4 x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^4={1\over 16}\left(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix}\right)={1\over 8}\left(\frac{e^{4ix}+e^{-4ix}}{2}-4\frac{e^{2ix}+e^{-2ix}}{2}+3\right)={1\over 8}\left(\cos{4x}-4\cos{2x}+3\right)$$ as desired.

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