If $sin x + sin y = 3 (cos y - cos x)$ , prove that $sin 3 x + sin 3 y = 0$

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Solution

$3 cos x + sin x = 3 cos y - sin y . . . (1)$
Put $3 = r cos α, 1 = r sin α,$
$∴ r = \sqrt{10}, tan α = \frac{1}{3}$
$∴ r cos (x - α) = y cos (y - α)$
$∴ (x - α) = \pm (y - α)$
$∴ x = - y$
or $x = y + 2 α$
Clearly $x = - y$ satisfies (1)
$sin 3 x = sin (- 3 y) = - sin 3 y$ or $sin 3 x + sin 3 y = 0$

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