interdicoxd

2021-12-31

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

Neil Dismukes

Beginner2022-01-01Added 37 answers

Well, $\mathrm{tan}\left(\frac{\pi}{3}\right)=\frac{\mathrm{sin}\left(\frac{\pi}{3}\right)}{\mathrm{cos}\left(\frac{\pi}{3}\right)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\left(\frac{2}{1}\right)=\sqrt{3}$

However, you can find it in another way:

Just think of it as a$\mathrm{tan}60\xb0$ and then draw a $30\xb0-60\xb0-90\xb0$ tringle

And$\mathrm{tan}60\xb0$ will be equal to $\frac{opposite}{adjacent}$ in reference to $60\xb0$ angle. Thus, $opposite=\sqrt{3}$ and $adjacent=1$ . Hence,

$\mathrm{tan}60\xb0=\frac{opposite}{adjacent}=\frac{\sqrt{3}}{1}=\sqrt{3}$

However, you can find it in another way:

Just think of it as a

And

twineg4

Beginner2022-01-02Added 33 answers

Use the Unit Circle and examine it at $\frac{\pi}{3}$

And determine the tangent, if we know the point$(\{\frac{\sqrt{3}}{2};\left\{\frac{1}{2}\right)$ , thinking of it as a slope of the line in the unit circle.

$\mathrm{tan}\frac{\pi}{3}=\frac{\frac{\sqrt{3}}{2}-0}{\frac{1}{2}-0}=\sqrt{3}$

And determine the tangent, if we know the point

nick1337

Expert2022-01-08Added 777 answers

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