Question

The value of ${\cos }^2{76}^{\circ }+{\cos }^2{16}^{\circ }-\cos {76}^{\circ }.\cos {16}^{\circ }$ is

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

Answer 1

The correct option is D $\frac{3}{4}$

Let ${\cos }^2{76}^{\circ }+{\cos }^2{16}^{\circ }-\cos {76}^{\circ }\cos {16}^{\circ }=I$

$I={\cos }^2({60}^{\circ }+{16}^{\circ })+{\cos }^2{16}^{\circ }-\cos ({60}^{\circ }+{16}^{\circ })\cos {16}^{\circ }$

Using $\cos (A+B)=\cos A\cos B-\sin A\sin B$

$I={(\cos {60}^{\circ }\cos {16}^{\circ }-\sin {60}^{\circ }\sin {16}^{\circ })}^2+co{s}^2{16}^{\circ }-(\cos {60}^{\circ }\cos {16}^{\circ }-\sin {60}^{\circ }\sin {16}^{\circ })\cos {16}^{\circ }$

$=\frac{{\cos }^2{16}^{\circ }}{4}+\frac{3{\sin }^2{16}^{\circ }}{4}-\frac{\sqrt{3}\cos {16}^{\circ }\sin {16}^{\circ }}{2}+{\cos }^2{16}^{\circ }-\frac{co{s}^2{16}^{\circ }}{2}+\frac{\sqrt{3}\cos {16}^{\circ }\sin {16}^{\circ }}{2}$

$=\frac{3{\cos }^2{16}^{\circ }}{4}+\frac{3{\sin }^2{16}^{\circ }}{4}$

$=\frac{3}{4}$