The exact value of cos^2 73^circ + cos^2 47^c | vedclass

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(C) Let the expression be $E = \cos^2 73^\circ + \cos^2 47^\circ + \cos 73^\circ \cos 47^\circ$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(C) Let the expression be $E = \cos^2 73^\circ + \cos^2 47^\circ + \cos 73^\circ \cos 47^\circ$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we have:$E = \frac{1 + \cos 146^\circ}{2} + \frac{1 + \cos 94^\circ}{2} + \cos 73^\circ \cos 47^\circ$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $\cos 73^\circ \cos 47^\circ = \frac{\cos 120^\circ + \cos 26^\circ}{2} = \frac{-1/2 + \cos 26^\circ}{2} = -1/4 + \frac{\cos 26^\circ}{2}$.Substituting this back:$E = 1 + \frac{\cos 146^\circ + \cos 94^\circ}{2} - 1/4 + \frac{\cos 26^\circ}{2}$.Using $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:$\cos 146^\circ + \cos 94^\circ = 2 \cos(120^\circ) \cos(26^\circ) = 2(-1/2) \cos 26^\circ = -\cos 26^\circ$.Thus,$E = 1 - 1/4 - \frac{\cos 26^\circ}{2} + \frac{\cos 26^\circ}{2} = 3/4$.

The exact value of $\cos^2 73^\circ + \cos^2 47^\circ + (\cos 73^\circ \cdot \cos 47^\circ)$ is

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