<s>[INST] $\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$ 
Find the value of $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$$

I did it like this
$$\cos^2 76^{\circ}+\cos^2 16^{\circ} = \cos(76^{\circ}+16^{\circ}) \, \cos(76^{\circ}-16^{\circ}).$$
So the expression is $$\cos 92^{\circ} \cos 60^{\circ}-\cos 76^{\circ} \, \cos16^{\circ}.$$
I couldn't simplify after that.
 [/INST] Let $c= \cos 16^\circ$ and $s=\sin 16^\circ$. Cosine addition gives ($76^\circ=60^\circ+16^\circ$)
\begin{eqnarray*}
\cos(76^\circ)=\frac{c-\sqrt{3}s}{2}
\end{eqnarray*}
So the expression is 
\begin{eqnarray*}
\cos^2(76^\circ)+\cos^2(16^\circ)-\cos(76^\circ)\cos(16^\circ) =\left(\frac{c-\sqrt{3}s}{2}\right)^2+c^2-c\left(\frac{c-\sqrt{3}s}{2}\right) \\=\frac{c^2-2\sqrt{3}cs+3s^2+4c^2-2c^2+2\sqrt{3}cs}{4}=\color{red}{\frac{3}{4}}.
\end{eqnarray*}
</s>