Solve $\sin(5\theta)=1$, $0<\theta<2\pi$. Show that the roots of $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, $r=0,2,3,4$.

Question

This is a very interesting problem that I came across. I know it's got something to do with trigonometry identities, polynomials and complex numbers, but other than that, I'm not too sure how to approach this. Any guidance hints or help would be greatly appreciated. Thanks in advance. Here's the problem:

Solve the equation $\sin(5\theta)=1$ for $0<\theta<2\pi$ and hence show that the roots of the equation $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, where for $r=0,2,3,4$. Determine the exact value of $\sin\frac\pi{10}\sin\frac{3\pi}{10}$.

It might be useful to find $\sin(5\theta)$ in terms of $sin^n \theta$ and so on.

Answer 1

For $\sin 5\theta = 1$ and $\theta \in(0,2\pi)$, $\theta = \dfrac\pi{10}, \dfrac\pi2, \dfrac{9\pi}{10}, \dfrac{13\pi}{10}, \dfrac{17\pi}{10}.$

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To find $\sin 5x$ in terms of $\sin x$, consider $\cos 5x + i\sin 5x$:

$$\begin{align*} \cos 5x + i\sin 5x &= (\cos x + i\sin x)^5\\ i\sin 5x&= 5i\cos^4x\sin x + 10i^3 \cos^2x\sin^3 x + i^5\sin ^5x\\ \sin 5x &= 5\cos^4 x \sin x - 10\cos^2x\sin^3x + \sin^5x\\ &= 5(1-\sin^2x)^2\sin x - 10(1-\sin^2x)\sin^3x + \sin ^5 x\\ &=5(1-2\sin^2x+\sin^4x)\sin x- 10(1-\sin^2x)\sin^3x + \sin^5 x\\ &= 16\sin^5 x -20\sin^3x+5\sin x \end{align*}$$

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As the five $\theta$ values above satisfy $\sin 5\theta = 1$, they also satisfy $16\sin^5 \theta -20\sin^3\theta+5\sin \theta = 1$, and so the five $x = \sin \frac{(4r+1)\pi}{10}, r=0,1,2,3,4$ are roots of

$$\begin{align*} 16x^5-20x^3+5x &= 1\\ 16x^5-20x^3+5x -1 &= 0 \end{align*}$$

Taking the factor $(x-1)$ or $\left(x-\sin \frac{\pi}{2}\right)$ from the equation,

$$(x-1)(16x^4+16x^3-4x^2-4x+1) = 0$$

And so $x = \sin \frac{(4r+1)\pi}{10}, r=0,2,3,4$ satisfy

$$16x^4+16x^3-4x^2-4x+1 = 0$$

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$$\begin{align*} \sin\frac{9\pi}{10} &= \sin \frac\pi{10}\\ \sin\frac{13\pi}{10} &= -\sin \frac{3\pi}{10}\\ \sin\frac{17\pi}{10} &= -\sin \frac{3\pi}{10}\\ \end{align*}$$

By Vieta's formulae,

$$\begin{align*} \sin\frac{\pi}{10}\sin\frac{9\pi}{10}\sin\frac{13\pi}{10}\sin\frac{17\pi}{10} &= \frac1{16}\\ \sin\frac{\pi}{10}\sin\frac{\pi}{10}\left(-\sin \frac{3\pi}{10}\right)\left(-\sin \frac{3\pi}{10}\right) &= \frac1{16}\\ \sin \frac{\pi}{10}\sin \frac{3\pi}{10}&= \underline{\underline{\frac14}} \end{align*}$$

(the positive square root is taken)

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There is something left unproven: whether $\sin \frac{\pi}{10}$ and $-\sin \frac{3\pi}{10}$ are really both double roots. This may be confirmed by the $(4x^2+2x−1)^2$ factorisation by @dxiv above.

Answer 2

Think of the equation

$1=\sin 5\theta$

$=\sin3\theta cos2\theta+\cos3\theta sin2\theta$

$=(3\sin\theta-4\sin^3\theta)(1-2\sin^2\theta)+(4\cos^3\theta-3\cos\theta) 2\sin\theta \cos\theta$

$=(3\sin\theta-4\sin^3\theta)(1-2\sin^2\theta)+(4\cos^2\theta-3) 2\sin\theta (1-\sin^2\theta)$

$=(3\sin\theta-4\sin^3\theta)(1-2\sin^2\theta)+(1-4\sin^2\theta) 2\sin\theta (1-\sin^2\theta)$

Can you take it from here?

What equation do you get when you cancel out the one from both sides?

Answer 3

Use$$\sin5\theta=\sin{x}\cos4\theta+\cos{x}\sin4\theta=\sin{\theta}(2\cos^22\theta-1)+4\cos^2\theta\cos2x\sin{\theta}=$$ $$=\sin{\theta}(2\cos^22\theta-1+2(1+\cos2\theta)\cos2\theta))=\sin{\theta}(4\cos^22\theta+2\cos2\theta-1)$$ and $16x^4+16x^3-4x^2-4x+1=(4x^2+2x-1)^2$