a tricky trig questions from Step If $\theta + \phi + \psi = \pi/2$, show that $\sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi = 1$.
By taking $\theta = \phi =\pi/5$ in this equation, or otherwise, show that $\sin(\pi/10)$ satisfies the equation
$$8x^3 + 8x^2 − 1 = 0$$
I got stuck in the first part. I want to prove this by making connection
with $\sin(\theta + \phi + \psi)=1$,but I failed.
 A: If $\theta+\varphi+\psi=\pi/2$, then $$1=\sin^2\theta+\cos^2\theta=\sin^2\theta+\sin^2(\varphi+\psi)$$
But, since $\sin(\varphi+\psi)=\sin\varphi\cos\psi+\sin\psi\cos\varphi$, we have $$\sin^2(\varphi+\psi)=\sin^2\varphi\cos^2\psi+\sin^2\psi\cos^2\varphi+2\sin\varphi\cos\psi\sin\psi\cos\varphi.
$$
Since  $\sin^2\varphi\cos^2\psi=\sin^2\varphi(1-\sin^2\psi)$ and $\sin^2\psi\cos^2\varphi=\sin^2\psi(1-\sin^2\varphi)$. Replacing it on the previous equation, we get 
\begin{align}\sin^2(\varphi+\psi)&=\sin^2\varphi+\sin^2\psi-2\sin^2\psi\sin^2\varphi+2\sin\varphi\cos\psi\sin\psi\cos\varphi.\\
&=\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi(\cos\psi\cos\varphi-\sin\psi\sin\varphi)\\
&=\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi(\cos(\varphi+\psi))\\
&=\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi\sin\theta
\end{align}
Replacing this on the very first equation, we get what we wanted,
$$ 1=\sin^2\theta+\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi\sin\theta$$
A: $$\begin{align*}&      \sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi 
\\& = \sin^2 \theta + \sin^2 \phi + \sin^2 \left ( \frac {\pi} {2} - (\theta + \phi) \right ) + 2 \sin \theta \sin \phi \sin \left (\frac {\pi} {2} - (\theta+ \phi) \right ) \\  & = \sin^2 \theta + \sin^2 \phi + \cos^2 (\theta + \phi) + 2 \sin \theta \sin \phi \cos (\theta + \phi) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi - {2 \cos \theta \cos \phi \sin \theta \sin \phi} + {2 \sin \theta \sin \phi \cos \theta \cos \phi}  - 2 \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + (1-\sin^2 \theta) \cdot (1 - \sin^2 \phi) - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi +1 - \sin^2 \theta - \sin^2 \phi + \sin^2 \theta \sin^2 \phi  - \sin^2 \theta \sin^2 \phi \\ & = 1. \end{align*}$$ 
A: $$F=\sin^2 \theta + \sin^2 \phi + \sin^2 \psi =1-(\cos^2 \theta -\sin^2 \phi) + \sin^2 \psi$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=1-\cos(\theta+\phi)\cos(\theta-\phi)+\sin^2\psi$$ 
$$=1-\sin\psi[\cos(\theta-\phi)-\cos(\theta+\phi)]\text{ as } \theta + \phi + \psi = \dfrac\pi2,\cos(\theta+\phi)=\sin\psi$$
$$=1-\sin\psi(2\sin\theta\sin\phi)$$
