Prove the equality : $\sqrt{33 − 16\sqrt{3}\sin80^\circ}= 1+8\sin10^\circ$

Prove the equality :
$\sqrt{33 − 16\sqrt{3}\sin80^\circ}= 1+8\sin10^\circ$

First, note that

$\begin{align*}\cos^2 20^\circ&=\cos20^\circ(2\cdot\dfrac{1}{2}\cos20^\circ)\\&=\cos20^\circ(2\cos60^\circ\cos20^\circ)\\&=\cos20^\circ((\cos(60^\circ+20^\circ)+\cos(60^\circ-20^\circ))\\&=\cos20^\circ(\cos80^\circ+\cos40^\circ)\\&=\cos20^\circ\cos80^\circ+\cos20^\circ\cos40^\circ\\&=\dfrac{1}{2}(\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))+\dfrac{1}{2}(\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ))\\&=\dfrac{1}{2}(\cos100^\circ+\cos60^\circ)+\dfrac{1}{2}(\cos60^\circ+\cos20^\circ)\\&=\dfrac{1}{2}\left(\cos(90^\circ+10^\circ)+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\cos 20^\circ\right)\\&=\dfrac{1}{2}(-\sin 10^\circ)+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{2}\cos 20^\circ\\&=\dfrac{1}{2}(\cos20^\circ-\sin10^\circ+1)\end{align*}$

$\therefore 2\cos^2 20^\circ=\cos20^\circ-\sin10^\circ+1$

Next, we work from the LHS of the given equality and see that

$\begin{align*}\sqrt{33 − 16\sqrt{3}\sin80^\circ}&=\sqrt{33-32\left(\dfrac{\sqrt{3}}{2}\right)\sin 80^\circ}\\&=\sqrt{33-32\sin 60^\circ\sin 80^\circ}\\&=\sqrt{33-16(2\sin 60^\circ\sin 80^\circ)}\\&=\sqrt{33-16\left(-\cos( 60^\circ+80^\circ)+\cos( 80^\circ-60^\circ)\right)}\\&=\sqrt{33-16\left(-\cos(140^\circ)+\cos(20^\circ)\right)}\\&=\sqrt{33-16\left(-\cos(180^\circ-40^\circ)+\cos(20^\circ)\right)}\\&=\sqrt{33-16\left(\cos(40^\circ)+\cos(20^\circ)\right)}\\&=\sqrt{33-16\cos(40^\circ)-16\cos(20^\circ)}\\&=\sqrt{33-16(2\cos^2(20^\circ)-1)-16\cos(20^\circ)}\\&=\sqrt{33-16(\cos20^\circ-\sin10^\circ+1-1)-16\cos(20^\circ)}\\&=\sqrt{33-32\cos20^\circ+16\sin10^\circ}\\&=\sqrt{33-32(1-2\sin^2 10^\circ)+16\sin10^\circ}\\&=\sqrt{1+16\sin10^\circ+64\sin^2 10^\circ)}\\&=\sqrt{(1+8\sin10^\circ)^2}\\&=1+8\sin10^\circ\end{align*}$

and we're hence done.