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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.

eyeballfrog
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Kevin
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3 Answers3

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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, $$ 1=\sin^2\theta+\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi\sin\theta$$

Julian Mejia
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  • thx a lot,but i wonder what leads u to the correct solution,feels like every step may lead to the wrong direction. – Kevin Apr 21 '19 at 04:28
  • My idea was start from $1=\sin^2\theta+\sin^2(\varphi+\psi)$. So, we have to prove that then $\sin^2(\varphi+\psi)-\sin^2\varphi+\sin^2\psi=2\sin\varphi\sin\psi\sin\theta$. and I just did it expanding the square. – Julian Mejia Apr 21 '19 at 04:40
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$$\begin{align*}& \sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi \\& = \sin^2 \theta + \sin^2 \phi + \sin^2 \left ( \frac {\pi} {2} - (\theta + \phi) \right ) + 2 \sin \theta \sin \phi \sin \left (\frac {\pi} {2} - (\theta+ \phi) \right ) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 (\theta + \phi) + 2 \sin \theta \sin \phi \cos (\theta + \phi) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi - {2 \cos \theta \cos \phi \sin \theta \sin \phi} + {2 \sin \theta \sin \phi \cos \theta \cos \phi} - 2 \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2 \theta \cos^2 \phi - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi + (1-\sin^2 \theta) \cdot (1 - \sin^2 \phi) - \sin^2 \theta \sin^2 \phi. \\ & = \sin^2 \theta + \sin^2 \phi +1 - \sin^2 \theta - \sin^2 \phi + \sin^2 \theta \sin^2 \phi - \sin^2 \theta \sin^2 \phi \\ & = 1. \end{align*}$$

Arnab Auddy
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little o
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$$F=\sin^2 \theta + \sin^2 \phi + \sin^2 \psi =1-(\cos^2 \theta -\sin^2 \phi) + \sin^2 \psi$$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$F=1-\cos(\theta+\phi)\cos(\theta-\phi)+\sin^2\psi$$

$$=1-\sin\psi[\cos(\theta-\phi)-\cos(\theta+\phi)]\text{ as } \theta + \phi + \psi = \dfrac\pi2,\cos(\theta+\phi)=\sin\psi$$

$$=1-\sin\psi(2\sin\theta\sin\phi)$$

lab bhattacharjee
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