The value of sin^{2} theta + sin^{2}(theta + | vedclass

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

Question

(C) Given expression: $E = \sin^{2} 	heta + \sin^{2}(	heta + 60^{\circ}) + \sin^{2}(	heta - 60^{\circ})$Using the identity $\sin^{2} A = \frac{1 -

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) Given expression: $E = \sin^{2} 	heta + \sin^{2}(	heta + 60^{\circ}) + \sin^{2}(	heta - 60^{\circ})$Using the identity $\sin^{2} A = \frac{1 - \cos 2A}{2}$:$E = \frac{1 - \cos 2	heta}{2} + \frac{1 - \cos(2	heta + 120^{\circ})}{2} + \frac{1 - \cos(2	heta - 120^{\circ})}{2}$$E = \frac{1}{2} [3 - (\cos 2	heta + \cos(2	heta + 120^{\circ}) + \cos(2	heta - 120^{\circ}))]$Using $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$:$E = \frac{1}{2} [3 - (\cos 2	heta + 2 \cos 2	heta \cos 120^{\circ})]$Since $\cos 120^{\circ} = -\frac{1}{2}$:$E = \frac{1}{2} [3 - (\cos 2	heta + 2 \cos 2	heta (-1/2))]$$E = \frac{1}{2} [3 - (\cos 2	heta - \cos 2	heta)] = \frac{3}{2}$

The value of $\sin^{2} \theta + \sin^{2}(\theta + 60^{\circ}) + \sin^{2}(\theta - 60^{\circ}) =$

Similar Questions

If $\sin \alpha = -\frac{3}{5},$ where $\pi < \alpha < \frac{3\pi}{2},$ then $\cos \frac{\alpha}{2} = $

If $A, B, C, D$ are the angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D = $

If $\tan A = 2\tan B + \cot B,$ then $2\tan (A - B) = $

$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $

If $A, B, C$ are the angles of a triangle,then $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module