Question

What is the value of ${\cos }^2{76}^0+{\cos }^2{16}^0-\cos {76}^0.\cos {16}^0$?

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

Answer 1

Answer: (D)

$\frac{3}{4}$

We have,

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

$={\cos }^2{76}^{\circ }+{\cos }^2{16}^{\circ }-\frac{1}{2}\times (2\cos {76}^{\circ }.\cos {16}^{\circ })$

$={\cos }^2{76}^{\circ }+1-{\sin }^2{16}^{\circ }-\frac{1}{2}\times (2\cos {76}^{\circ }.\cos {16}^{\circ })$

$={\cos }^2{76}^{\circ }-{\sin }^2{16}^{\circ }+1-\frac{1}{2}\times (2\cos {76}^{\circ }.\cos {16}^{\circ })$

We know that

${\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)$

And $2\cos A.\cos B=\cos (A+B)+\cos (A-B)$

Therefore,

$=\cos ({76}^{\circ }+{16}^{\circ })\cos ({76}^{\circ }-{16}^{\circ })+1-\frac{1}{2}[\cos ({76}^{\circ }+{16}^{\circ })+\cos ({76}^{\circ }-{16}^{\circ })]$

$=\cos \left({92}^{\circ }\right)\cos \left({60}^{\circ }\right)+1-\frac{1}{2}\times [\cos \left({92}^{\circ }\right)+\cos \left({60}^{\circ }\right)]$

$=\frac{1}{2}\times \cos \left({92}^{\circ }\right)+1-\frac{1}{2}\times \cos \left({92}^{\circ }\right)-\frac{1}{4}$

$=1-\frac{1}{4}$

$=\frac{3}{4}$

Hence, the required value is $\frac{3}{4}$.