Using $\sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta$ to show that $w=\sin^2(\pi/5)$ satisfies $16w^2-20w+5=0$

Question

Using $\sin5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$, how do I deduce that the equation $16w^2 - 20w + 5 = 0$ is satisfied by $w = \sin^2(\pi/5)$?

Answer 1

Set $s=\sin\frac \pi5$, so that $w=s^2$. As $\sin\bigl(5\cdot\frac \pi 2\bigr)=\sin\pi=0$, $s$ is a root of the polynomial $$16s^5-20s^3+5s=s(16s^4-20s^2+5).$$ Furthermore, $s\ne 0$, so $\;16s^4-20s^2+5=0$, i.e. $\;16w^2-20w+5=0$.

Answer 2

Rewrite the given equation

$$\sin5θ = 16\sin^5θ - 20\sin^3θ + 5\sinθ$$

as

$$ 16\sin^4θ - 20\sin^2θ + 5=\frac {\sin 5\theta}{\sin\theta} $$

Plug in $\theta = \frac {\pi}{5}$ and the right-hand-side becomes zero,

$$ 16\sin^4 \frac {\pi}{5} - 20\sin^2\frac {\pi}{5} + 5 = \frac{\sin{\pi}}{\sin (\pi/5)}=0$$

Hence, $w = \sin^2(\pi/5)$ satisfies $16w^2 - 20w + 5 = 0$