Answer
$\dfrac{1}{2} [\sin{10x}+ \sin{6x}]$
Work Step by Step
$\cos{A} \sin{B} =\dfrac{1}{2}[\sin{(A+B)}- \sin{(A-B)}]$
$\cos{2x} \sin{8x} = \dfrac{1}{2} [\sin{10x} - \sin{(-6x)}]$
$\cos{2x} \sin{8x} = \dfrac{1}{2} [\sin{10x}+ \sin{6x}]$
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.