sin ^2 76^{circ}+sin ^2 16^{circ}-sin 76^{cir | vedclass

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(C) ધારો કે $E = \sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ}$.નિત્યસમ $2\sin^2 	heta = 1 - \cos 2	heta$ નો ઉપયોગ કરતા:$E =

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$\frac{3}{4}$

Vedclass · Answer

(C) ધારો કે $E = \sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ}$.નિત્યસમ $2\sin^2 	heta = 1 - \cos 2	heta$ નો ઉપયોગ કરતા:$E = \frac{1}{2} [ (1 - \cos 152^{\circ}) + (1 - \cos 32^{\circ}) ] - \sin 76^{\circ} \sin 16^{\circ}$$E = 1 - \frac{1}{2} [ \cos 152^{\circ} + \cos 32^{\circ} ] - \sin 76^{\circ} \sin 16^{\circ}$$\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$ નો ઉપયોગ કરતા:$\cos 152^{\circ} + \cos 32^{\circ} = 2 \cos 92^{\circ} \cos 60^{\circ} = 2 \cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$વળી,$2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ નો ઉપયોગ કરતા:$\sin 76^{\circ} \sin 16^{\circ} = \frac{1}{2} [ \cos 60^{\circ} - \cos 92^{\circ} ] = \frac{1}{2} [ \frac{1}{2} - \cos 92^{\circ} ] = \frac{1}{4} - \frac{1}{2} \cos 92^{\circ}$આ કિંમતો $E$ માં મૂકતા:$E = 1 - \frac{1}{2} [ \cos 92^{\circ} ] - [ \frac{1}{4} - \frac{1}{2} \cos 92^{\circ} ]$$E = 1 - \frac{1}{2} \cos 92^{\circ} - \frac{1}{4} + \frac{1}{2} \cos 92^{\circ} = 1 - \frac{1}{4} = \frac{3}{4}$

$\sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ} = $

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