Question

${\cos }^2\left(76°\right)+{\cos }^2\left(16°\right)-\cos \left(76°\right)\cos \left(16°\right)=$

• A. $-\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $0$
• D. $\frac{3}{4}$

Answer 1

The correct option is D

$\frac{3}{4}$

Explanation for the correct option:

Simplify the series using Trigonometric identities

$$\begin{gathered} {\cos }^2\left(76°\right)+{\cos }^2\left(16°\right)-\cos \left(76°\right)\cos \left(16°\right) \\ =\frac{1}{2}[2{\cos }^2\left(76°\right)+2{\cos }^2\left(16°\right)-2\cos \left(76°\right)\cos \left(16°\right)] \\ =\frac{1}{2}[1+\cos \left(2\times 76°\right)+1+\cos \left(2\times 16°\right)-2\cos \left(76°\right)\cos \left(16°\right)][\because 2{\cos }^2\left(A\right)=1+\cos \left(2A\right)] \\ =\frac{1}{2}[2+\cos \left(152°\right)+\cos \left(32°\right)-\cos \left(76°+16°\right)-\cos \left(76°-16°\right)][\because 2\cos \left(A\right)\cos \left(B\right)=\cos \left(A+B\right)+\cos \left(A-B\right)] \\ =\frac{1}{2}[2+2\cos (\frac{152°+32°}{2}\left)\cos \right(\frac{152°-32°}{2})-\cos \left(92°\right)-\cos (60°\left)\right][\because \cos \left(C\right)+\cos \left(D\right)=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)] \\ =\frac{1}{2}[2+2\cos (92°\left)\cos \right(60°)-\cos \left(92°\right)-\cos (60°\left)\right] \\ =\frac{1}{2}[2+2\cos (92°)\times \frac{1}{2}-\cos \left(92°\right)-\frac{1}{2})] \\ =\frac{1}{2}[2+\cos (92°)-\cos \left(92°\right)-\frac{1}{2}\left)\right] \\ =\frac{1}{2}[2-\frac{1}{2}] \\ =\frac{1}{2}\times \frac{3}{2} \\ =\frac{3}{4} \end{gathered}$$

Therefore, Option(D) is correct.