If $sin x + sin y = a$ and $cos x + cos y = b$ show that
$cos (x - y) = (a^{2} + b^{2} - 2) / 2$

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Solution

Squaring and adding the given relation
$({sin}^{2} x + {sin}^{2} y + 2 sin x sin y) + ({cos}^{2} x + {cos}^{2} y + 2 cos x cos y) = a^{2} + b^{2}$
or $2 (cos x cos y + sin x sin y) = a^{2} + b^{2} - 2$
or $cos (x - y)$ = $(a^{2} + b^{2} - 2) / 2$

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