1 problem found
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: \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 \(8x^3+8x^2-1=0\)