<s>[INST] Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By considering the equation $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$
I have completed part (a). I am stuck on part (b) however.
My Workings:
By Factor formulae, $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{1}{2} (\cos(\frac{2\pi}{5}) + \cos(\frac{\pi}{5}))$
Considering roots of $z^5 = 1$. The sum of the roots equals 0.
So $\cos(\frac{2\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{6\pi}{5}) + \cos(\frac{8\pi}{5}) + \cos(\frac{10\pi}{5})=0$
$\cos(\frac{2\pi}{5}) = \cos(\frac{8\pi}{5})$, $\cos(\frac{4\pi}{5}) = \cos(\frac{6\pi}{5}) = -\cos(\frac{\pi}{5})$
So $2\cos(\frac{2\pi}{5})-2\cos(\frac{\pi}{5}) = -1$
$\cos(\frac{2\pi}{5})-\cos(\frac{\pi}{5}) = -\frac{1}{2}$
Been on stuck on this problem for an hour. Please help me.
 [/INST] If $\cos(5\theta) = 0$, then $z = e^{i\theta}$, must satisfy $z^5 = \pm i$. Note that the two possibilities correspond to conjugates of each other, which won't change the real part, so let's just assume $z^5 = i$. The roots of $z^5 = i$ are:
$$e^{i\pi/10}, e^{5i\pi/10} \color{red}{( = i)}, e^{9\pi i/10}, e^{13\pi i/10} \color{red}{( = e^{-7\pi i/10})}, e^{17\pi i/10} \color{red}{( = e^{-3\pi i/10})}.$$
So, the roots of the polynomial $\cos(5\theta) = 0$ in terms of $\cos(\theta)$ will be the real parts of the above, specifically:
$$\cos(\pi/10), 0, \cos(9\pi/10),\cos(7\pi/10), \cos(3\pi10),$$
remembering that $\cos$ is even. It's also worth noting that, since $\cos(\pi - x) = -\cos(x)$, we can further simplify to:
$$0, \pm \cos(\pi/10), \pm \cos(3\pi/10).$$
Using Vieta on $\cos(5\theta)/\cos(\theta)$ (to remove the zero root), we can therefore see that the product of the non-zero roots is $\frac{5}{16}$. That is,
$$\frac{5}{16} = (-\cos^2(\pi/10))(-\cos^2(3\pi/10)) = (\cos(\pi/10)\cos(3\pi/10))^2.$$
Obviously $\cos(\pi/10)$ and $\cos(3\pi/10)$ are positive, so
$$\cos(\pi/10)\cos(3\pi/10) = \frac{\sqrt{5}}{4}.$$
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