cos 47 cos13 - sin 47 sin13

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Solution

$\cos 47 ° . \cos 13 ° - \sin 47 ° . \sin 13 ° = \cos (47 ° + 13 °) [as, \cos (A + B) = \cos A . \cos B - \sin A . \sin B] = \cos 60 ° = \frac{1}{2}$

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