$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$ 
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.
 A: 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*}
A: Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2(A+60^\circ)+\cos^2A=1+\cos(A+60^\circ+A)\cos(A+60^\circ-A)=1+\dfrac{\cos(2A+60^\circ)}2$$
Using Werner's formula,
$$\cos(A+60^\circ)\cos A=\dfrac{\cos(A+60^\circ+A)+\cos(A+60^\circ-A)}2=?$$
Can you recognize $A$ here?
A: $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}=$$
$$=\frac{1}{2}\left(1+\cos152^{\circ}+1+\cos32^{\circ}-\cos60^{\circ}-\cos92^{\circ}\right)=$$
$$=\frac{3}{4}+\frac{1}{2}\left(\cos152^{\circ}+\cos32^{\circ}-\cos92^{\circ}\right)=$$
$$=\frac{3}{4}+\frac{1}{2}\left(2\cos92^{\circ}\cos60^{\circ}-\cos92^{\circ}\right)=\frac{3}{4}$$
