The Mauve Theorem

Relevant wiki: De Moivre's Theorem

$\begin{aligned} \cos x & = \frac {\sqrt{16-\pi}}4 \\ & = \sqrt{1-\left(\frac \pi 4 \right)^2} & \small \color{#3D99F6}{\text{Since }\cos x = \sqrt{1-\sin^2 x}} \\ \implies \sin x & = \frac \pi 4 = a \end{aligned}$

Therefore, we have

$\begin{aligned} (\cos 8 + i \sin 8)^a & = e^{8ia} & \small \color{#3D99F6}{\text{By De Moivre's theorem}} \\ & = e^{8i \cdot \frac \pi 4} \\ & = e^{2\pi i} \\ & = \cos (2\pi) + i\sin (2\pi) \\ & = \boxed{1} \end{aligned}$ .

$\cos x = \large\frac{\sqrt{16 - \pi^2}}{4}$ $\therefore \cos^2 x = \Big(\frac{16 - \pi^2}{16}\Big)$ $$ So $\sin^2 x = \frac{16 - (16 - \pi^2)}{16}$ $$ $\sin^2 x = \frac{\pi^2}{16}$ $$ $\therefore \sin x = \frac{\pi}{4}$ $$ So $a = \frac{\pi}{4}$ $$ Hence, $\Big(\cos 8 + i\sin 8 \Big)^a = \Big(\cos 8 + i\sin 8 \Big)^\frac{\pi}{4}$ $$ $= \cos \frac{8\pi}{4} + i\sin{\frac{8\pi}{4}}$ by De Moivre's Theorem. $$ We can clearly see that this is $\cos 2\pi + i\sin 2\pi = 1 + 0. = \boxed{1.}$