Question: Use an expression for $\frac{\sin(5\theta)}{\sin(\theta)}$ , ($\theta \neq k \pi)$ , k an integer to find the roots of the equation $x^4-3x^2+1=0$ in trigonometric form?
What I have done
By using demovires theorem and expanding
$$ cis(5\theta) = (\cos(\theta) + i\sin(\theta))^5$$
$$ \cos(5\theta) + i \sin(5\theta) = \cos^5(\theta) - 10\cos^3(\theta)\sin^2(\theta) + 5\cos(\theta)\sin^4(\theta) +i(5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta)$$
Considering only $Im(z) = \sin(5\theta)$
$$ \sin(5\theta) = 5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta) $$
$$ \therefore \frac{\sin(5\theta)}{\sin(\theta)} = \frac{5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta)}{\sin(\theta)}$$
$$ \frac{\sin(5\theta)}{\sin(\theta)} = 5\cos^4(\theta) -10\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta) $$
How should I proceed , I'm stuck trying to incorporate what I got into the equation..
HINT:
Using Prosthaphaeresis Formula,
$$\sin5x-\sin x=2\sin2x\cos3x=4\sin x\cos x\cos3x$$
If $\sin x\ne0,$ $$\dfrac{\sin5x}{\sin x}-1=4\cos x\cos3x=4\cos x(4\cos^3x-3\cos x)=(4\cos^2x)^2-3(4\cos^2x)$$
OR replace $\sin^2x$ with $1-\cos^2x$ in your $$ 5\cos^4x-10\cos^2x\sin^2x + \sin^4x$$
Now if $\sin5x=0,5x=n\pi$ where $n$ is any integer
$x=\dfrac{n\pi}5$ where $n\equiv0,\pm1,\pm2\pmod5$
So, the roots of $\dfrac{\sin5x}{\sin x}=0\implies x=\dfrac{n\pi}5$ where $n\equiv\pm1,\pm2\pmod5$
But $$\dfrac{\sin5x}{\sin x}=(4\cos^2x)^2-3(4\cos^2x)+1$$