# Find the value of \tan(\frac{\pi}{3})

Find the value of $$\displaystyle{\tan{{\left({\frac{{\pi}}{{{3}}}}\right)}}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Neil Dismukes
Well, $$\displaystyle{\tan{{\left({\frac{{\pi}}{{{3}}}}\right)}}}={\frac{{{\sin{{\left({\frac{{\pi}}{{{3}}}}\right)}}}}}{{{\cos{{\left({\frac{{\pi}}{{{3}}}}\right)}}}}}}={\frac{{{\frac{{\sqrt{{{3}}}}}{{{2}}}}}}{{{\frac{{{1}}}{{{2}}}}}}}={\frac{{\sqrt{{{3}}}}}{{{2}}}}{\left({\frac{{{2}}}{{{1}}}}\right)}=\sqrt{{{3}}}$$
However, you can find it in another way:
Just think of it as a $$\displaystyle{\tan{{60}}}°$$ and then draw a $$\displaystyle{30}°-{60}°-{90}°$$ tringle
And $$\displaystyle{\tan{{60}}}°$$ will be equal to $$\displaystyle{\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}$$ in reference to $$\displaystyle{60}°$$ angle. Thus, $$\displaystyle{o}{p}{p}{o}{s}{i}{t}{e}=\sqrt{{{3}}}$$ and $$\displaystyle{a}{d}{j}{a}{c}{e}{n}{t}={1}$$. Hence,
$$\displaystyle{\tan{{60}}}°={\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}={\frac{{\sqrt{{{3}}}}}{{{1}}}}=\sqrt{{{3}}}$$
###### Not exactly what you’re looking for?
twineg4
Use the Unit Circle and examine it at $$\displaystyle{\frac{{\pi}}{{{3}}}}$$
And determine the tangent, if we know the point $$\displaystyle{\left({\left\lbrace{\frac{{\sqrt{{{3}}}}}{{{2}}}};{\left\lbrace{\frac{{{1}}}{{{2}}}}\right)}\right.}\right.}$$, thinking of it as a slope of the line in the unit circle.
$$\displaystyle{\tan{{\frac{{\pi}}{{{3}}}}}}={\frac{{{\frac{{\sqrt{{{3}}}}}{{{2}}}}-{0}}}{{{\frac{{{1}}}{{{2}}}}-{0}}}}=\sqrt{{{3}}}$$
nick1337

$$\tan(\frac{\pi}{3})=\sqrt{3}$$