Problem source

By applying de Moivre's theorem to $\cos5\theta+\mathrm{i}\sin5\theta,$
expanding the result using the binomial theorem, and then equating
imaginary parts, show that 
\[
\sin5\theta=\sin\theta\left(16\cos^{4}\theta-12\cos^{2}\theta+1\right).
\]
Use this identity to evaluate $\cos^{2}\frac{1}{5}\pi$, and deduce that $\cos\frac{1}{5}\pi=\frac{1}{4}(1+\sqrt{5}).$

Solution source

\begin{align*}
&& (\cos \theta + i \sin \theta)^n &= \cos n \theta + i \sin n \theta \\
n = 5: && \cos 5 \theta + i \sin 5 \theta &= (\cos \theta + i \sin \theta)^5 \\
\textrm{Im}: && \sin 5 \theta &= \binom{5}{1}\cos^4 \theta \sin \theta + \binom{5}{3} \cos^2 \theta (- \sin^3 \theta) + \binom{5}{5} \sin^5 \theta \\
&&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta \sin^2 \theta+\sin^4 \theta) \\
&&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta (1-\cos^2 \theta)+(1-\cos^2 \theta)^2) \\
&&&= \sin \theta((5+10+1)\cos^4 \theta +(-10-2)\cos^2 \theta + 1) \\
&&&= \sin \theta(16\cos^4 \theta -12\cos^2 \theta + 1) \\
\end{align*}

Suppose $\theta= \frac{\pi}{5}$, then $\sin 5 \theta = 0, \sin \theta \neq 0$, therefore if $c = \cos \theta$ we must have

\begin{align*}
&& 0 &= 16c^4-12c^2+1 \\
\Rightarrow && c^2 &= \frac{3 \pm \sqrt{5}}{8} \\
&&&= \frac{6\pm 2\sqrt{5}}{16} \\
&&&= \frac{(1 \pm \sqrt{5})^2}{16} \\
\Rightarrow && c &= \pm \frac{1 \pm \sqrt{5}}{4}
\end{align*}

Since $c > 0$ we either have $\cos \frac15 \pi = \frac{1+\sqrt{5}}4$ or  $\cos \frac15 \pi = \frac{\sqrt{5}-1}4$, however $\sqrt{5}-1 < 1.5$ and so $\frac{\sqrt{5}-1}{4} < \frac12 = \cos \frac13 \pi$ we must have $\cos \frac15 \pi = \frac{1+\sqrt{5}}4$