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

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(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 -

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$\frac{3}{2}$

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

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