Answer
$\sin 160^{\mathrm{o}}-\sin 44^{\mathrm{o}}$
Work Step by Step
Product-to-Sum:
$\sin A \displaystyle \cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$
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$2\sin 58^{\mathrm{o}}\cos 102^{\mathrm{o}}=$
$=2(\displaystyle \frac{1}{2}[\sin(58^{\mathrm{o}}+102^{\mathrm{o}})+\sin(58^{\mathrm{o}}-102^{\mathrm{o}})])$
$=\sin 160^{\mathrm{o}}+\sin(-44^{\mathrm{o}})$
$=\sin 160^{\mathrm{o}}-\sin 44^{\mathrm{o}}$
