If 180^{circ}

Question

(A) Given expression: $E = \sqrt{4 \sin^{4} 	heta + \sin^{2} 2	heta} + 4 \cos^{2} \left(\frac{\pi}{4} - \frac{	heta}{2}ight)$.Using $\sin 2	heta

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given expression: $E = \sqrt{4 \sin^{4} 	heta + \sin^{2} 2	heta} + 4 \cos^{2} \left(\frac{\pi}{4} - \frac{	heta}{2}ight)$.Using $\sin 2	heta = 2 \sin 	heta \cos 	heta,$ we have $\sin^{2} 2	heta = 4 \sin^{2} 	heta \cos^{2} 	heta$.So,$\sqrt{4 \sin^{4} 	heta + 4 \sin^{2} 	heta \cos^{2} 	heta} = \sqrt{4 \sin^{2} 	heta (\sin^{2} 	heta + \cos^{2} 	heta)} = \sqrt{4 \sin^{2} 	heta} = 2 |\sin 	heta|$.Using $2 \cos^{2} A = 1 + \cos 2A,$ the second term is $2 [1 + \cos(\frac{\pi}{2} - 	heta)] = 2(1 + \sin 	heta)$.Thus,$E = 2 |\sin 	heta| + 2 + 2 \sin 	heta$.Since $180^{\circ} Therefore,$E = 2(-\sin 	heta) + 2 + 2 \sin 	heta = 2$.

If $180^{\circ} < \theta < 270^{\circ},$ then the value of $\sqrt{4 \sin^{4} \theta + \sin^{2} 2\theta} + 4 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$ is

Similar Questions

If $\tan \theta = \frac{a}{b},$ then $\frac{{\sin \theta }}{{{{\cos }^8}\theta }} + \frac{{\cos \theta }}{{{{\sin }^8}\theta }} = $

If $1+\cos^{2} \theta = 3 \sin \theta \cos \theta$,then the integral value of $\cot \theta$ is $(0 < \theta < \frac{\pi}{2})$.

Let $A, B$ and $C$ be the angles of a plane triangle and $\tan \frac{A}{2} = \frac{1}{3}, \tan \frac{B}{2} = \frac{2}{3}$. Then $\tan \frac{C}{2}$ is equal to

$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $

If angle $\theta$ is divided into two parts such that the tangent of one part is $k$ times the tangent of the other and $\phi$ is their difference,then $\sin \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module