Problem source

If $\theta+\phi+\psi=\tfrac{1}{2}\pi,$ show that 
\[
\sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi=1.
\]
By taking $\theta=\phi=\tfrac{1}{5}\pi$ in this equation, or otherwise, show that $\sin\tfrac{1}{10}\pi$ satisfies the equation 
\[
8x^{3}+8x^{2}-1=0.
\]

Solution source

\begin{align*}
S &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi \\
&= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}(\tfrac\pi2-\theta-\phi)+2\sin\theta\sin\phi\sin(\tfrac\pi2-\theta-\phi) \\
&= \sin^{2}\theta+\sin^{2}\phi+\cos^{2}(\theta+\phi)+2\sin\theta\sin\phi\cos(\theta+\phi) \\
&= \sin^{2}\theta+\sin^{2}\phi+\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right)^2+2\sin\theta\sin\phi\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right) \\
&= \sin^{2}\theta+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi-\sin^2 \theta \sin^2 \phi  \\
&= \sin^{2}\theta(1-\sin^2 \phi)+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\
&= \sin^{2}\theta\cos^2 \phi+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\
&= \sin^{2}\phi+\cos^2 \phi \\
&= 1
\end{align*}

Suppose $\theta = \phi = \tfrac15 \pi, \psi = \tfrac1{10}\pi$. Also let $s = \sin \tfrac1{10}$

\begin{align*}
1 &= 2\sin^2 \tfrac15 \pi + \sin^2 \tfrac1{10} \pi + 2 \sin^2\tfrac15 \pi \sin \tfrac1{10} \pi \\
&= 8\sin^2  \tfrac1{10} \pi \cos^2  \tfrac1{10} \pi + \sin^2  \tfrac1{10} \pi + 8 \sin^2  \tfrac1{10} \pi \cos^2  \tfrac1{10} \pi \sin  \tfrac1{10} \pi \\
&= 8\sin^2  \tfrac1{10} \pi(1- \sin^2  \tfrac1{10} \pi) + \sin^2  \tfrac1{10} \pi + 8 \sin^2  \tfrac1{10} \pi (1-\sin^2  \tfrac1{10} \pi) \sin  \tfrac1{10} \pi \\
&= 8s^2(1-s^2)+s^2 + 8s^2(1-s^2)s \\
&= -8 s^5 - 8 s^4 + 8 s^3 + 9 s^2 
\end{align*}

Therefore $s$ is a root of $8s^5+8s^4-8s^3-9s^2+1 = 0$, but notice that

\begin{align*}
8s^5+8s^4-8s^3-9s^2+1 &= (s-1)(8 s^4 + 16 s^3 + 8 s^2 - s - 1 ) \\
&= (s-1)(s+1)(8s^3+8s^2-1)
\end{align*}

Therefore since $\sin \tfrac{1}{10} \pi \neq \pm 1$ it must be a root of $8x^3+8x^2-1=0$