Evaluate: cos^2 76^circ + cos^2 16^circ - cos | vedclass

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(D) We use the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.Let $E = \cos^2 76^\circ + \cos^2 16

Vedclass · Accepted Answer

$3/4$

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(D) We use the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.Let $E = \cos^2 76^\circ + \cos^2 16^\circ - \cos 76^\circ \cos 16^\circ$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$:$E = \frac{1 + \cos 152^\circ}{2} + \frac{1 + \cos 32^\circ}{2} - \cos 76^\circ \cos 16^\circ$$E = 1 + \frac{1}{2}(\cos 152^\circ + \cos 32^\circ) - \cos 76^\circ \cos 16^\circ$Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:$\cos 152^\circ + \cos 32^\circ = 2 \cos 92^\circ \cos 60^\circ = 2 \cos 92^\circ (\frac{1}{2}) = \cos 92^\circ$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:$\cos 76^\circ \cos 16^\circ = \frac{1}{2}(\cos 92^\circ + \cos 60^\circ) = \frac{1}{2}(\cos 92^\circ + \frac{1}{2}) = \frac{1}{2} \cos 92^\circ + \frac{1}{4}$.Substituting these back:$E = 1 + \frac{1}{2}(\cos 92^\circ) - (\frac{1}{2} \cos 92^\circ + \frac{1}{4})$$E = 1 - \frac{1}{4} = \frac{3}{4}$.

Evaluate: $\cos^2 76^\circ + \cos^2 16^\circ - \cos 76^\circ \cos 16^\circ$

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