cos ^3left(frac{pi}{8}right) cos left(frac{3 | vedclass

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

Question

(A) Let the given expression be $E = \cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}}$

Vedclass · Answer

(A) Let the given expression be $E = \cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right)$.Using the identity $\cos \left(\frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{8}\right)$ and $\sin \left(\frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)$,we substitute these into the expression:$E = \cos ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) + \sin ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$Factor out $\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$:$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \left[ \cos ^2\left(\frac{\pi}{8}\right) + \sin ^2\left(\frac{\pi}{8}\right) \right]$Using the Pythagorean identity $\sin ^2 A + \cos ^2 A = 1$:$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \cdot (1) = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$Multiply and divide by $2$ to use the double angle formula $\sin (2A) = 2 \sin A \cos A$:$E = \frac{1}{2} \left[ 2 \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \right] = \frac{1}{2} \sin \left( 2 \cdot \frac{\pi}{8} \right) = \frac{1}{2} \sin \left(\frac{\pi}{4}\right)$Since $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:$E = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2 \sqrt{2}}$.

$\cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right) = $

Similar Questions

If $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then the value of $\cos ^2 48^{\circ} - \sin ^2 12^{\circ}$ is:

$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8} = ?$

The number of real solutions of the equation $(\sin x - x)(\cos x - x^2) = 0$ is

If $\sin x + \sin y = p$ and $\cos x + \cos y = q$,then $\sec(x + y) = $

The smallest integer $n$ such that $\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 47^{\circ} \sin 48^{\circ}}+\ldots+\frac{1}{\sin 133^{\circ} \sin 134^{\circ}}=\frac{1}{\sin \left(n^{\circ}\right)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module