$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$ 
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.

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?
