Use DeMoivre's Theorem to prove $ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$

Question

I need to prove the following equalities using DMT:

$ \cos 5x = 16 \cos^5x$$ - 20\cos^3x$$+5\cos x$

and

$ \sin 5x = 16 \sin^5x$$ - 20\sin^3x$$+5\sin x$

Can someone help me with this question?

(Attempt: $\cos5x+i\sin5x=(\cos x+i\sin x)^5$)

Answer 1

Use the short-hands $c= \cos x$, $s=\sin x$ and continue with $$\cos5x+i\sin5x=(c+is)^5 =c^5-10s^2c^3+5s^4c+i(s^5-10s^3c^2+5sc^4)$$ $$=c^5-10(1-c^2)c^3+5(1-c^2)^2c+i(s^5-10s^3(1-s^2)+5s(1-s^2)^2)$$ $$=16c^5-20c^3+5c+i(16s^5-20s^3+5s)$$

where $c^2+s^2=1$ is used. Thus,

$$ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$$

$$ \sin 5x = 16 \sin^5x - 20\sin^3x +5\sin x$$

Answer 2

By DeMoivre's Theorem:
$$\begin{equation}\begin{aligned} cos\,5x &= cos\,5x + i\,sin\,5x \\ &= (cos\,x + i\,sin\,x)^5 \\ &= cos^5\,x + 10\,cos^3\,sin^2\,x\,i + 5\,cos\,x\,sin^4\,x\,i^4\\ &= cos^5\,x - 10\,cos^3 x\,sin^2\,x + 5\,cos\,x\,sin^4\,x \\ &= cos^5\,x - 10\,cos^3\,x (1 - cos^2\,x) + 5\,cos\,x (1 - cos^2\,x)^2 \\ &= 16\,cos^5\,x - 20\,cos^3\,x + 5\,cos\,x \\ \end{aligned}\end{equation}$$