# Prove this trigonometry equation: sin &#x2061;<!-- ⁡ --> 40 &#x2218;<!-- ∘ -->

Prove this trigonometry equation: $\mathrm{sin}{40}^{\circ }\cdot \mathrm{sin}{50}^{\circ }$ is equal to $\frac{1}{2}\mathrm{cos}{10}^{\circ }$
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Ellie Meyers
Taking L.H.S.
$sin40°sin50°$ or $sin50°sin40°$
We know the identity, $sinAsinB=\frac{1}{2}\left(cos\left(A-B\right)-cos\left(A+B\right)\right)$
Here we consider A=50° and B=40°
Then, $sin50°sin40°=\frac{1}{2}\left(cos\left(50°-40°\right)-cos\left(50°+40°\right)\right)$
$=\frac{1}{2}\left(cos\phantom{\rule{thinmathspace}{0ex}}10°-cos\phantom{\rule{thinmathspace}{0ex}}90°\right)$
$=\frac{1}{2}\left(cos\phantom{\rule{thinmathspace}{0ex}}10°-0\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\because cos\phantom{\rule{thinmathspace}{0ex}}90°=0\right)$
$=\frac{1}{2}cos\phantom{\rule{thinmathspace}{0ex}}10°$
L.H.S.=R.H.S.
###### Not exactly what you’re looking for?
measgachyx5q9
Here,
$L.H.S=sin40°.sin50°$
$=\frac{2}{2}.sin40°.sin50°$
$=\frac{1}{2}\left(cos10°-cos90°\right)$
$=\frac{1}{2}cos10°=R.H.S$
Proved.