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

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Question

(C) Let $E = \sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ}$.Using the identity $2\sin^2 	heta = 1 - \cos 2	heta$,we have:$E

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) Let $E = \sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ}$.Using the identity $2\sin^2 	heta = 1 - \cos 2	heta$,we have:$E = \frac{1}{2} [ (1 - \cos 152^{\circ}) + (1 - \cos 32^{\circ}) ] - \sin 76^{\circ} \sin 16^{\circ}$$E = 1 - \frac{1}{2} [ \cos 152^{\circ} + \cos 32^{\circ} ] - \sin 76^{\circ} \sin 16^{\circ}$Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:$\cos 152^{\circ} + \cos 32^{\circ} = 2 \cos 92^{\circ} \cos 60^{\circ} = 2 \cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$Also,using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$\sin 76^{\circ} \sin 16^{\circ} = \frac{1}{2} [ \cos 60^{\circ} - \cos 92^{\circ} ] = \frac{1}{2} [ \frac{1}{2} - \cos 92^{\circ} ] = \frac{1}{4} - \frac{1}{2} \cos 92^{\circ}$Substituting these back into the expression for $E$:$E = 1 - \frac{1}{2} [ \cos 92^{\circ} ] - [ \frac{1}{4} - \frac{1}{2} \cos 92^{\circ} ]$$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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