The value of cos ^2 76^{circ}+cos ^2 16^{circ | vedclass

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

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$\frac{3}{4}$

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(C) Let $E = \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$.Using the identity $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we have:$E = \frac{1+\cos 152^{\circ}}{2} + \frac{1+\cos 32^{\circ}}{2} - \cos 76^{\circ} \cos 16^{\circ}$$E = 1 + \frac{1}{2}(\cos 152^{\circ} + \cos 32^{\circ}) - \cos 76^{\circ} \cos 16^{\circ}$Using $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$:$\cos 152^{\circ} + \cos 32^{\circ} = 2\cos 92^{\circ}\cos 60^{\circ} = 2\cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$Also,$\cos 92^{\circ} = \cos(180^{\circ} - 88^{\circ}) = -\cos 88^{\circ} = -\sin 2^{\circ}$ (not helpful here).Alternatively,use $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:$E = \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\frac{1}{2}(\cos 92^{\circ} + \cos 60^{\circ})$$E = \frac{1+\cos 152^{\circ}}{2} + \frac{1+\cos 32^{\circ}}{2} - \frac{1}{2}\cos 92^{\circ} - \frac{1}{4}$$E = 1 + \frac{1}{2}(\cos 152^{\circ} + \cos 32^{\circ} - \cos 92^{\circ}) - \frac{1}{4}$Since $\cos 152^{\circ} + \cos 32^{\circ} = \cos 92^{\circ}$,the expression simplifies to:$E = 1 + \frac{1}{2}(\cos 92^{\circ} - \cos 92^{\circ}) - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4}$.

The value of $\cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$ is equal to

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