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There was question I came upon, and I was stumped. The question was: evaluate the sine ,cosine, and tangent of the angle without using a calculator.

I was given $-\pi/6$. I know that sine is $-1/2$ and cosine is $\sqrt{3}/2$.

Normally I know that $\tan = \sin/\cos$. And doing so gives $-1/\sqrt{3}$. The problem was that the actual answer is $-\sqrt{3}/3$, and I can't seem to figure out how that can be possible

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  • $\begingroup$ $-\frac 1{\sqrt 3} = -\frac {\sqrt 3}{3}$. Those are both the same answer. $\endgroup$ – fleablood Nov 29 '17 at 7:09
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Those are both the same number and the same answer.

$-\frac 1{\sqrt{3}} = -\frac 1{\sqrt{3}} * \frac {\sqrt{3}}{\sqrt{3}} = - \frac {\sqrt 3}{3}$.

It's a standard practice (no-one really knows why or has a universal explaination why) for many texts and classes to consider radical signs in the denominator of a fraction to be taboo and require to be "deradicalized".

I wouldn't worry. You did get the right answer.

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$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt 3}$. To see this, rationalise the denominator of the latter by multiplying top and bottom by $\sqrt 3$.

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