Using expansion of $\sin8\theta$ to find equation with given roots

Question

The question is to use the expansion of $\sin8\theta$ to find an equation with roots $+-\sin \frac{\pi}{8}$, $+-\sin \frac{3\pi}{8}$

The expansion of $\sin8\theta$ is:

$\cos\theta\sin\theta(-16\sin^6\theta+24\sin^4\theta-10\sin^2\theta+1)$

I can get to the correct equation using roots of polynomials:

$\alpha+\beta=\frac{-b}{a}$, $\alpha\beta=\frac{c}{a}$

$\sin^2 \frac{\pi}{8} +\sin^2 \frac{3\pi}{8}=\frac{-b}{a}$

$\sin^2 \frac{\pi}{8} \sin^2 \frac{3\pi}{8}=\frac{c}{a}$

$\therefore a=8, b=-8, c=1$

So the required equation is:

$8\sin^4\theta-8\sin^2\theta+1=0$

But I have not used the expansion of $\sin8\theta$. So how do I use the expansion to get the equation?

Answer 1

Using $\sin2A=2\sin A\cos A$ repeatedly

$$\sin8x=8\sin x\cos x\cos2x\cos4x$$

Now if $\sin8x=0,$

either $\sin x=0\implies x=m\pi$

or $\cos x=0\implies x=(2m+1)\dfrac\pi2\pmod{2\pi}$

or $\cos2x=0\implies x=(2m+1)\dfrac\pi4\pmod{2\pi}$

or $\cos4x=0\implies x=(2m+1)\dfrac\pi8,x\equiv\pm\dfrac\pi8,\pm\dfrac{3\pi}8\pmod{2\pi}$

Now $\cos4x=2\cos^22x-1=2(1-2\sin^2x)^2-1=1-8\sin^2x+8\sin^4x$

So, the equation whose roots are $\pm\sin\dfrac\pi8,\pm\sin\dfrac{3\pi}8$ will be $$1-8s^2+8s^4=0$$