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 the product-to-sum formula $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.Given exp

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$3/4$

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(D) We use the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and the product-to-sum formula $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.Given expression: $E = \cos^2 76^\circ + \cos^2 16^\circ - \cos 76^\circ \cos 16^\circ$Multiply by $2/2$ to simplify:$E = \frac{1}{2} [2\cos^2 76^\circ + 2\cos^2 16^\circ - 2\cos 76^\circ \cos 16^\circ]$Using $2\cos^2 	heta = 1 + \cos 2	heta$:$E = \frac{1}{2} [(1 + \cos 152^\circ) + (1 + \cos 32^\circ) - (\cos(76^\circ + 16^\circ) + \cos(76^\circ - 16^\circ))]$$E = \frac{1}{2} [2 + \cos 152^\circ + \cos 32^\circ - \cos 92^\circ - \cos 60^\circ]$Since $\cos 152^\circ = \cos(180^\circ - 28^\circ) = -\cos 28^\circ$ and $\cos 60^\circ = 1/2$:$E = \frac{1}{2} [2 - \cos 28^\circ + \cos 32^\circ - \cos 92^\circ - 1/2]$Alternatively,using $\cos 152^\circ + \cos 32^\circ = 2\cos(\frac{152+32}{2})\cos(\frac{152-32}{2}) = 2\cos 92^\circ \cos 60^\circ = 2\cos 92^\circ (1/2) = \cos 92^\circ$.Substituting this back:$E = \frac{1}{2} [2 + \cos 92^\circ - \cos 92^\circ - 1/2]$$E = \frac{1}{2} [3/2] = 3/4$.

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

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