I need to show the steps to prove this identity:
$$\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$$
I know that $\cos{2x}=\cos^2{x}-\sin^2{x}$. From there I do not know what to do.
The solution should look like:
$$\sin^4{x}=sin^4{x}$$
I need to prove the right side equals the left side.
Best Answer
If you know $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix}+e^{-ix}}{2}$, then just expand using the binomial theorem: $$\sin^4 x = \frac{e^{4ix} - 4e^{3ix}e^{-ix}+6e^{2ix}e^{-2ix}-4e^{ix}e^{-3ix}+e^{-4ix}}{(2i)^4}$$ $$=\frac{e^{4ix}+e^{-4ix} -4e^{2ix}-4e^{-2ix} + 6}{16}$$ $$=\frac{\frac{e^{4ix}+e^{-4ix}}{2} -4\frac{e^{2ix}+e^{-2ix}}{2} + 3}{8}$$ $$=\boxed{\frac{\cos 4x -4\cos 2x + 3}{8}}$$ as desired.