# Solving polynomial equation using known trigonometric identity

I recently did an exam paper in which the following question was asked:

Prove $$\sin(5\theta)=16\sin^{5}(\theta)-20\sin^{3}(\theta)+5\sin(\theta)$$ Hence find all the solutions to: $$16x^{5}-20x^{3}+5x-1=0$$

The first part of the question is simple application of de-Moivre's theorem:

$$\sin(5\theta)=\Im(e^{5i\theta})=\Im\left(\left(e^{i\theta}\right)^{5}\right)$$

And using the binomial formula to expand $(\cos(\theta)+i\sin(\theta))^{5}$ gives the identity fairly trivially. However, for the second part of the formula, using the substitution $x=\sin(\theta)$ gives $\sin(5\theta)=1 \implies \theta=\frac{\pi}{10}$, thus: $x=\frac{1}{4}(-1+\sqrt{5})$, which is one of the answers (I checked using Mathematica), however, I'm not sure how to get the rest of the answers?

• If $\sin 5\theta=1,$ aren't there multiple values of $\theta$ that provide a solution, perhaps $\theta =\frac{\pi}{10}+k\frac{2\pi}{5}$? – abiessu Jun 7 '14 at 12:43

We have

$$\sin(5\theta)=1\iff 5\theta\equiv \frac\pi2\mod 2\pi\iff\theta=\frac\pi{10}+\frac{2k\pi}5,\quad k\in\Bbb Z$$ and denote for $k=0,\ldots,4,\; \theta_k=\frac\pi{10}+\frac{2k\pi}5$ and if we have $k\neq p\in\{0,\ldots,4\}$ then $\sin(\theta_k)\ne\sin(\theta_p)$ so $\sin(\theta_k)$ for $k=0,\ldots,4$ are $5$ distinct roots of the given equation. By the fundamental theorem of algebra we know that they are the all roots of the equation.

• Well done! ${}{}{}{}$ – Namaste Jun 8 '14 at 12:01

$$\sin5x=1=\sin\frac\pi2\implies5x=2n\pi+\frac\pi2=\frac\pi2(4n+1)$$ where $n$ is any integer

$$\implies x=\frac{\pi(4n+1)}{10}$$ where $0\le n\le4$

For $n=1,x=\frac\pi2\implies \sin x=1$

So, the rest $4$ roots will be from $$\frac{16x^5-20x^3+5x-1}{x-1}=0\iff 16x^4+16x^3-4x^2-4x+1=0$$

Now, $\displaystyle 16x^4+16x^3-4x^2-4x+1=(4x^2)^2+2\cdot4x^2(2x)+(2x)^2+2\cdot4x^2(-1)+2\cdot2x(-1)+(-1)^2=(4x^2+2x-1)^2$ which is evident from

$$\frac\pi{10}(\text{ with } n=0) =\pi-\frac{9\pi}{10}(\text{ with } n=2)\implies\sin\frac{9\pi}{10}=\sin\frac{\pi}{10}$$

Same for $n=3,n=4$

So, $\displaystyle\sin\frac{9\pi}{10}=\sin\frac{\pi}{10}, \sin\frac{13\pi}{10}=\sin\frac{17\pi}{10}$ are the roots of $4x^2+2x-1=0$

Observe that $\displaystyle\sin\frac{\pi}{10}>0, \sin\frac{13\pi}{10}=\sin\left(\pi+\frac{3\pi}{10}\right)=-\sin\frac{3\pi}{10}<0$