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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.

Michael Rozenberg
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user466377
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3 Answers3

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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*}

Donald Splutterwit
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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?

lab bhattacharjee
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$$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}=$$ $$=\frac{1}{2}\left(1+\cos152^{\circ}+1+\cos32^{\circ}-\cos60^{\circ}-\cos92^{\circ}\right)=$$ $$=\frac{3}{4}+\frac{1}{2}\left(\cos152^{\circ}+\cos32^{\circ}-\cos92^{\circ}\right)=$$ $$=\frac{3}{4}+\frac{1}{2}\left(2\cos92^{\circ}\cos60^{\circ}-\cos92^{\circ}\right)=\frac{3}{4}$$

Michael Rozenberg
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