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

$\cos 1^\circ + \cos 2^\circ + \cos 3^\circ + \dots + \cos 180^\circ = $

The value of $\cos y \cos \left( \frac{\pi}{2} - x \right) - \cos \left( \frac{\pi}{2} - y \right) \cos x + \sin y \cos \left( \frac{\pi}{2} - x \right) + \cos x \sin \left( \frac{\pi}{2} - y \right)$ is zero,if

For a positive integer $n$,let ${f_n}(\theta ) = \left( {\tan \frac{\theta }{2}} \right)\,(1 + \sec \theta )\,(1 + \sec 2\theta )\,(1 + \sec 4\theta ) \dots (1 + \sec {2^n}\theta ).$ Then

If $A, B, C$ are acute positive angles such that $A + B + C = \pi$ and $\cot A \cot B \cot C = K$,then

If $\tan 80^{\circ} = a$ and $\tan 47^{\circ} = b$,then $\tan 37^{\circ}$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module