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

interdicoxd 2021-12-31 Answered
Find the value of \(\displaystyle{\tan{{\left({\frac{{\pi}}{{{3}}}}\right)}}}\)

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Expert Answer

Neil Dismukes
Answered 2022-01-01 Author has 3622 answers
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,
\(\displaystyle{\tan{{60}}}°={\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}={\frac{{\sqrt{{{3}}}}}{{{1}}}}=\sqrt{{{3}}}\)
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twineg4
Answered 2022-01-02 Author has 3655 answers
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.
\(\displaystyle{\tan{{\frac{{\pi}}{{{3}}}}}}={\frac{{{\frac{{\sqrt{{{3}}}}}{{{2}}}}-{0}}}{{{\frac{{{1}}}{{{2}}}}-{0}}}}=\sqrt{{{3}}}\)
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nick1337
Answered 2022-01-08 Author has 9672 answers

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

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