Question

Write the value of ${\cos }^{2}76°+{\cos }^{2}16°-\cos 76°\cos 16°$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\cos }^{2}76°+{\cos }^{2}16°-\cos 76°\cos 16° \\ =\frac{1}{2}\left[1+\cos 2\left(76\right)°+1+\cos 2\left(16\right)°-\cos \left(76+16\right)°-\cos \left(76-16\right)°\right] \\ \left[\because 2{\cos }^2\mathrm{\theta }=1+\cos 2\mathrm{\theta } \\ \mathrm{and}2\cos A\cos B=\cos \left(A\mathit{+}B\right)+\cos \left(A\mathit{-}B\right)\right] \\ =\frac{1}{2}\left[2+\cos 152°+\cos 32°-\cos 92°-\frac{1}{2}\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos \left(180-152°\right)+\cos 32°-\cos 92°\right] \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+2\sin \frac{92°+32°}{2}\sin \frac{92°-32°}{2}\right] \\ \left[\cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+2\sin \frac{124°}{2}\sin \frac{60°}{2}\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+2\sin 62°\sin 30°\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+2\sin 62°\times \frac{1}{2}\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+\sin 62°\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+\sin \left(90-28\right)°\right] \\ =\frac{1}{2}\left[\frac{3}{2}-\cos 28°+\cos 28°\right] \\ =\frac{3}{4} \end{gathered}$$