Question

The value of ${\cos }^2\left(\frac{\pi }{12}\right)+{\cos }^2\left(\frac{\pi }{4}\right)+{\cos }^2\left(\frac{5\pi }{12}\right)$ is

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3+\sqrt{3}}{2}$
• D. $\frac{2}{3+\sqrt{3}}$

Answer 1

The correct option is A

$\frac{3}{2}$

Explanation for correct option

Given trigonometric function is ${\cos }^2\left(\frac{\pi }{12}\right)+{\cos }^2\left(\frac{\pi }{4}\right)+{\cos }^2\left(\frac{5\pi }{12}\right)$

${\cos }^2\left(\frac{\pi }{12}\right)={\left(\cos \left(\frac{\pi }{2}-\frac{5\pi }{12}\right)\right)}^2={\left(\sin \left(\frac{5\pi }{12}\right)\right)}^2$

Therefore,

$$\begin{gathered} {\cos }^2\left(\frac{\pi }{12}\right)+{\cos }^2\left(\frac{\pi }{4}\right)+{\cos }^2\left(\frac{5\pi }{12}\right) \\ ={\sin }^2\left(\frac{5\pi }{12}\right)+{\cos }^2\left(\frac{5\pi }{12}\right)+{\cos }^2\left(\frac{\pi }{4}\right) \\ =1+{\left(\frac{1}{\sqrt{2}}\right)}^2\left[\because {\sin }^2\left(A\right)+{\cos }^2\left(A\right)=1\&\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right] \\ =1+\frac{1}{2} \\ =\frac{3}{2} \end{gathered}$$

Hence, the correct option is $\mathrm{Option}\left(\mathrm{A}\right)$