3left[ sin^4left( frac{3pi}{2} - alpha right) | vedclass

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

Question

(B) Given expression: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \al

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given expression: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right]$Using trigonometric reduction formulas:$\sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha$,$\sin(3\pi + \alpha) = -\sin \alpha$,$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$,$\sin(5\pi - \alpha) = \sin \alpha$Substituting these values:$= 3\left[ (-\cos \alpha)^4 + (-\sin \alpha)^4 \right] - 2\left[ (\cos \alpha)^6 + (\sin \alpha)^6 \right]$$= 3(\cos^4 \alpha + \sin^4 \alpha) - 2(\cos^6 \alpha + \sin^6 \alpha)$Using identities $\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$ and $\sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha$:$= 3(1 - 2\sin^2 \alpha \cos^2 \alpha) - 2(1 - 3\sin^2 \alpha \cos^2 \alpha)$$= 3 - 6\sin^2 \alpha \cos^2 \alpha - 2 + 6\sin^2 \alpha \cos^2 \alpha$$= 3 - 2 = 1$

$3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right] = $

Similar Questions

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

If $3 \cos \theta + 4 \sin \theta = 5$,then the value of $3 \sin \theta - 4 \cos \theta$ is:

If $\cos \theta, \sin \theta$ and $\cot \theta$ are in geometric progression,then $\sin ^9 \theta+\sin ^6 \theta+3 \sin ^5 \theta+\sin ^3 \theta+\sin ^2 \theta=$

If $P = \sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} + \sin \frac{8 \pi}{7}$ and $Q = \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}$,then the point $(P, Q)$ lies on the circle of radius

If $\sin x + \cos x = a$,where $a \in [-\sqrt{2}, \sqrt{2}] - \{-1, 1\}$,then $\sum_{n=1}^{\infty} (\sin^n x + \cos^n x)$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module