Question

Find the value of ${\cos ^2}\dfrac{\pi }{{12}} + {\cos ^2}\dfrac{\pi }{4} + {\cos ^2}\dfrac{{5\pi }}{{12}}$.
A. $\dfrac{3}{2}$
B. $\dfrac{2}{3}$
C. $\dfrac{{3 + \sqrt 3 }}{2}$
D. $\dfrac{2}{{3 + {\kern 1pt} \sqrt 3 }}$

Answer 1

Hint: In order to solve the question, first convert ${\cos ^2}\theta $ to ${\sin ^2}\theta $. Then use the identities to further simplify the equation. Finally apply the trigonometric values to find the answer.

Formula Used:
${\cos ^2}\left( {\dfrac{\pi }{2} - \theta } \right) = {\sin ^2}\theta $
${\cos ^2}\theta + {\sin ^2}\theta = 1$
$\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$

Complete step by step solution:
Given that ${\cos ^2}\dfrac{\pi }{{12}} + {\cos ^2}\dfrac{\pi }{4} + {\cos ^2}\dfrac{{5\pi }}{{12}}$.
Here it is asked to find the value of the equation.
That is
${\cos ^2}\dfrac{\pi }{{12}} + {\cos ^2}\dfrac{\pi }{4} + {\cos ^2}\dfrac{{5\pi }}{{12}}$
$ = {\cos ^2}\dfrac{\pi }{{12}} + {\cos ^2}\dfrac{\pi }{4} + {\cos ^2}\left( {\dfrac{\pi }{2} - \dfrac{\pi }{{12}}} \right)$
$ = {\cos ^2}\dfrac{\pi }{{12}} + {\cos ^2}\dfrac{\pi }{4} + {\sin ^2}\dfrac{\pi }{{12}}$. . . . . . (since ${\cos ^2}\left( {\dfrac{\pi }{2} - \theta } \right) = {\sin ^2}\theta $)
$ = \left( {{{\cos }^2}\dfrac{\pi }{{12}} + {{\sin }^2}\dfrac{\pi }{{12}}} \right) + {\cos ^2}\dfrac{\pi }{4}$
$ = 1 + {\cos ^2}\dfrac{\pi }{4}$. . . . . . (since ${\cos ^2}\theta + {\sin ^2}\theta = 1$)
$ = 1 + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2}$. . . . . . (since $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$)
$ = 1 + \dfrac{1}{2}$
$ = \dfrac{3}{2}$

Option ‘A’ is correct

Additional information
The trigonometric functions are also known as the functions of angle of a triangle. In this type of question we are converting the angles from cosine to sine with the help of trigonometric identity to obtain a simplified form of the given equation.

Note: To solve this type of questions one must have the knowledge of trigonometric identities. Here we are using the identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$. Also one needs to have the knowledge of the signs of the trigonometric functions in four quadrants.