Prove $ \sin^4(x)$ identity using $z=cis(x)?$

Question

I cannot prove the $\sin^4(x)$ identity using $z= cis(x)$. I know that you have to use de Moivre's theorem and compare the real values of $z^4$ but I am stuck at this step: $$\sin^4(x)= -\cos^4(x) + 6\sin^2(x)\cos^2(x) + \cos(4x)$$

The identity is $\sin^4(x) = 1/8(\cos(4x) - 4\cos(2x) + 3)$

Answer 1

Note that

$$cos 4x+i\sin4x=(\cos x+i\sin x)^4$$

then use

$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$

which leads to

• $\sin {4x}= 4\cos^3 x\sin x-4\cos x\sin^3x$

• $\cos 4x= \cos^4 x-6\cos^2 x\sin^2 x+\sin^4 x$

from the last we can obtain $\sin^4x$ using that

• $\cos^2 x=1-\sin^2x$

• $\cos 2x=\cos^2x-\sin^2x=1-2\sin^2x$