Solve: $sin 3 x cos 4 x$ .

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Solution

$sin 3 x cos 4 x$
$= \frac{1}{2} (2 sin 3 x cos 4 x)$
$= \frac{1}{2} (sin 7 x + sin (- x)) [2 sin A cos B = sin (A + B) + sin (A - B)]$
$= \frac{1}{2} (sin 7 x - sin x)$
$∴ sin 3 x cos 4 x = \frac{1}{2} (sin 7 x - sin x)$ .

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