#### Answer

$\sin 225^{o}+\sin 55^{o}$

#### Work Step by Step

Product-to-Sum:
$\displaystyle \cos A\sin B=\frac{1}{2}[\sin(A+B)-\sin(A-B)]$
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$2\cos 85^{o}\sin 140^{o}=$
$=2(\displaystyle \frac{1}{2}[\sin(85^{o}+140^{o})-\sin(85^{o}-140^{o})])$
$=\sin 225^{o}-\sin(-55^{o})$
$=\sin 225^{o}+\sin 55^{o}$