cos 2(theta + phi ) - 4cos (theta + phi )sin | vedclass

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

Question

(A) Given expression: $E = \cos 2(	heta + \phi ) - 4\cos (	heta + \phi )\sin 	heta \sin \phi + 2\sin^2 \phi$Using the identity $2\sin 	heta \sin \

Vedclass · Accepted Answer

$\cos 2	heta $

Vedclass · Answer

(A) Given expression: $E = \cos 2(	heta + \phi ) - 4\cos (	heta + \phi )\sin 	heta \sin \phi + 2\sin^2 \phi$Using the identity $2\sin 	heta \sin \phi = \cos(	heta - \phi) - \cos(	heta + \phi)$:$E = \cos 2(	heta + \phi) - 2\cos(	heta + \phi)[\cos(	heta - \phi) - \cos(	heta + \phi)] + 2\sin^2 \phi$$E = \cos 2(	heta + \phi) - 2\cos(	heta + \phi)\cos(	heta - \phi) + 2\cos^2(	heta + \phi) + 2\sin^2 \phi$Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:$E = \cos 2(	heta + \phi) - [\cos 2	heta + \cos 2\phi] + 2\cos^2(	heta + \phi) + 2\sin^2 \phi$Using $2\cos^2 A = 1 + \cos 2A$:$E = \cos 2(	heta + \phi) - \cos 2	heta - \cos 2\phi + 1 + \cos 2(	heta + \phi) + 2\sin^2 \phi$Since $1 - \cos 2\phi = 2\sin^2 \phi$,the expression simplifies to:$E = 2\cos 2(	heta + \phi) - \cos 2	heta + 2\sin^2 \phi + 2\sin^2 \phi - \cos 2\phi$Alternatively,testing with $	heta = \phi = \frac{\pi}{4}$:$E = \cos 2(\frac{\pi}{2}) - 4\cos(\frac{\pi}{2})\sin(\frac{\pi}{4})\sin(\frac{\pi}{4}) + 2\sin^2(\frac{\pi}{4}) = \cos \pi - 0 + 2(\frac{1}{2}) = -1 + 1 = 0$.Checking options for $	heta = \frac{\pi}{4}$:$(a) \cos 2(\frac{\pi}{4}) = \cos \frac{\pi}{2} = 0$.Thus,option $(a)$ is correct.

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $

Similar Questions

If $\sin \theta + \sin \phi = a$ and $\cos \theta + \cos \phi = b,$ then $\tan \frac{\theta - \phi}{2}$ is equal to

If $\tan (A - B) = x$,then the value of $x$ is

The maximum value of $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$ under the restrictions $0 \le \alpha_1, \alpha_2, \dots, \alpha_n \le \frac{\pi}{2}$ and $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$ is

If the maximum value of $y = \frac{7 + 6 \tan x - \tan^2 x}{1 + \tan^2 x}$ is $\lambda$,then the value of $\log_{\sqrt{2}}(\lambda)$ is:

If $\sin x + \sin y = 3(\cos y - \cos x),$ then the value of $\frac{\sin 3x}{\sin 3y}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module