<s>[INST] Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
 [/INST] Generalization :
$$I=\cos^2A+\cos^2B-\cos A\cos B$$
$$=\cos^2A-\sin^2B+1-\frac{\cos(A-B)+\cos(A-B)}2$$
$$=\cos(A-B)\cos(A+B)+1-\frac{\cos(A-B)+\cos(A+B)}2$$
$$I=\cos(A-B)\left(\cos(A+B)-\frac12\right)+1-\frac{\cos(A+B)}2$$
If $\displaystyle\cos(A+B)=\frac12=\cos60^\circ\iff A+B=2n\pi\pm60^\circ$ where $n$ is any integer
$I$ becomes $\displaystyle1-\frac{\dfrac12}2=\frac34$
$$\text{Again, }I=\cos(A+B)\left(\cos(A-B)-\frac12\right)+1-\frac{\cos(A-B)}2$$
What if $\displaystyle\cos(A-B)=\frac12 ?$
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