The value of cos^2 frac{pi}{12} + cos^2 frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given expression: $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$We know that $\cos^2 	heta = \frac{1 + \cos(2	heta)}{2}

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) Given expression: $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$We know that $\cos^2 	heta = \frac{1 + \cos(2	heta)}{2}$.Using this identity:$= \frac{1 + \cos(\frac{\pi}{6})}{2} + \cos^2(\frac{\pi}{4}) + \frac{1 + \cos(\frac{5\pi}{6})}{2}$$= \frac{1}{2} + \frac{1}{2}\cos(\frac{\pi}{6}) + (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} + \frac{1}{2}\cos(\frac{5\pi}{6})$$= 1 + \frac{1}{2} + \frac{1}{2}(\cos \frac{\pi}{6} + \cos \frac{5\pi}{6})$Since $\cos(\frac{5\pi}{6}) = -\cos(\frac{\pi}{6})$:$= \frac{3}{2} + \frac{1}{2}(\cos \frac{\pi}{6} - \cos \frac{\pi}{6})$$= \frac{3}{2} + 0 = \frac{3}{2}$.

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

Similar Questions

If $A, B, C, D$ are the angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D = $

If $x = \sec \theta + \tan \theta ,$ then $x + \frac{1}{x} = $

If $\sin \theta + \cos \theta = m$ and $\sec \theta + \csc \theta = n$,then $n(m + 1)(m - 1) = $

$A$ telegraph post is bent at a point above the ground due to a storm. Its top just touches the ground at a distance of $10 \sqrt{3} \text{ m}$ from its foot and makes an angle of $30^{\circ}$ with the horizontal. Then,the height (in meters) of the telegraph post is:

If the sides of a right-angled triangle are $\{\cos2\alpha + \cos2\beta + 2\cos(\alpha + \beta )\}$ and $\{\sin2\alpha + \sin2\beta + 2\sin(\alpha + \beta )\}$,then the length of the hypotenuse is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module