# How to solve 2cos 150°-3sin 90° +tan 210°

How to solve $2\mathrm{cos}150°-3\mathrm{sin}90°+\mathrm{tan}210°$
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unessodopunsep
To find the related angles of given ones, and their (+) or (-) signs, use the unit circle.
$\mathrm{cos}150°=\mathrm{cos}\left(180°-30°\right)=-\mathrm{cos}30°=-\frac{\sqrt{3}}{2}\left(30°-60°-90°,\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}1-2-\sqrt{3}\mathrm{△}\right),\mathrm{sin}90°=1$, and
$\mathrm{tan}210°=\mathrm{tan}\left(180°+30°\right)=\mathrm{tan}30°=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$ (30°-60°-90° triangle)
The given equation is restated as follows:
$2\mathrm{cos}150°-3\mathrm{sin}90°+\mathrm{tan}210°=-2\mathrm{cos}30°-3\mathrm{sin}90°+\mathrm{tan}30°$
$=-2\frac{\sqrt{3}}{2}-3\left(1\right)+\frac{\sqrt{3}}{3}=-\frac{9+2\sqrt{3}}{3}$ (= approx. -4.1547)