If sin(x+3alpha) + 3sin(x-alpha) = 0,then | vedclass

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

Question

(C) Given equation: $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$Expanding using $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$(\sin x \cos 3\alpha + \

Vedclass · Accepted Answer

$	an x = 	an^3 \alpha$

Vedclass · Answer

(C) Given equation: $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$Expanding using $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$(\sin x \cos 3\alpha + \cos x \sin 3\alpha) + 3(\sin x \cos \alpha - \cos x \sin \alpha) = 0$Using $\cos 3\alpha = 4\cos^3 \alpha - 3\cos \alpha$ and $\sin 3\alpha = 3\sin \alpha - 4\sin^3 \alpha$:$\sin x(4\cos^3 \alpha - 3\cos \alpha) + \cos x(3\sin \alpha - 4\sin^3 \alpha) + 3\sin x \cos \alpha - 3\cos x \sin \alpha = 0$$\sin x(4\cos^3 \alpha - 3\cos \alpha + 3\cos \alpha) + \cos x(3\sin \alpha - 4\sin^3 \alpha - 3\sin \alpha) = 0$$4\sin x \cos^3 \alpha - 4\cos x \sin^3 \alpha = 0$$4\sin x \cos^3 \alpha = 4\cos x \sin^3 \alpha$$\frac{\sin x}{\cos x} = \frac{\sin^3 \alpha}{\cos^3 \alpha}$$	an x = 	an^3 \alpha$

If $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$,then

Similar Questions

Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ given that $\cos x = -\frac{1}{3}$ and $x$ is in quadrant $III$.

If $\cos \theta = \frac{1}{2}\left( x + \frac{1}{x} \right)$,then $\frac{1}{2}\left( x^2 + \frac{1}{x^2} \right) = $

The value of $\cos ^4 x$ is

If $\sinh x = \frac{5}{12}$,then $\cosh \frac{x}{2} =$

Prove that: $\cos 4x = 1 - 8 \sin^2 x \cos^2 x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module