cos^2 76^{circ} + cos^2 16^{circ} - cos 76^{c | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(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}$$E = 1 + \frac{1}{2} (\cos 152^{\circ} + \cos 32^{\circ}) - \cos 76^{\circ} \cos 16^{\circ}$Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:$\cos 152^{\circ} + \cos 32^{\circ} = 2 \cos 92^{\circ} \cos 60^{\circ} = 2 \cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:$\cos 76^{\circ} \cos 16^{\circ} = \frac{1}{2} (\cos 92^{\circ} + \cos 60^{\circ}) = \frac{1}{2} \cos 92^{\circ} + \frac{1}{4}$Substituting these back:$E = 1 + \frac{1}{2} \cos 92^{\circ} - (\frac{1}{2} \cos 92^{\circ} + \frac{1}{4})$$E = 1 - \frac{1}{4} = \frac{3}{4}$

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

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