Answer
$5[\sin{8x} + \sin{2x}]$
Work Step by Step
$\sin{A} \cos{B} =\dfrac{1}{2}[\sin{(A+B)} + \sin{(A-B)}]$
$10 \sin{5x} \cos{3x} = 5[\sin{8x} + \sin{2x}]$
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