How did they get from 2nd last to last step?
$$-\Bigg(\pi - \tan^{-1}\bigg({\frac{1}{\sqrt{3}}}\bigg)\Bigg)=-\frac{5 \pi}{6}$$
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It's $\arctan\frac1{\sqrt 3}$, not $\arctan\frac13$. – J. M. ain't a mathematician Nov 21 '11 at 10:50
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There is a mistake: $\frac{\sqrt{5}}{\sqrt{15}} = \frac{1}{\sqrt 3}$. After you fix it, you see that $\tan^{-1}\frac{1}{\sqrt 3} = \frac{\pi}{6}$.
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...and if you don't know why $\tan^{-1}\frac{1}{\sqrt 3} = \frac{\pi}{6}$, you'll want to remember the 30-60-90 triangle... – J. M. ain't a mathematician Nov 21 '11 at 10:54
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