<s>[INST] 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.
 [/INST] $$\begin{align*}&      \sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi 
\\& = \sin^2 \theta + \sin^2 \phi + \sin^2 \left ( \frac {\pi} {2} - (\theta + \phi) \right ) + 2 \sin \theta \sin \phi \sin \left (\frac {\pi} {2} - (\theta+ \phi) \right ) \\  & = \sin^2 \theta + \sin^2 \phi + \cos^2 (\theta + \phi) + 2 \sin \theta \sin \phi \cos (\theta + \phi) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi - {2 \cos \theta \cos \phi \sin \theta \sin \phi} + {2 \sin \theta \sin \phi \cos \theta \cos \phi}  - 2 \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + (1-\sin^2 \theta) \cdot (1 - \sin^2 \phi) - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi +1 - \sin^2 \theta - \sin^2 \phi + \sin^2 \theta \sin^2 \phi  - \sin^2 \theta \sin^2 \phi \\ & = 1. \end{align*}$$ 
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