Question

How do I evaluate $\tan \left( \dfrac{\pi }{3} \right)$ without using a calculator?

Answer 1

To solve this equation, we need to know the relationship between the $\tan x,\sin x\And \cos x$, which states that, $\tan x=\dfrac{\sin x}{\cos x}$. We will use this relation to find the value of $\tan \left( \dfrac{\pi }{3} \right)$. Also, we should know the values of $\sin \left( \dfrac{\pi }{3} \right)\And \cos \left( \dfrac{\pi }{3} \right)$.

Complete step by step answer:
We know that the trigonometric ratios $\tan x,\sin x\And \cos x$ are related to each other in the following way, $\tan x=\dfrac{\sin x}{\cos x}$. As $\dfrac{\pi }{3}$ is a special angel, we know the values of $\sin \left( \dfrac{\pi }{3} \right)\And \cos \left( \dfrac{\pi }{3} \right)$. The value of $\sin \left( \dfrac{\pi }{3} \right)$ equals $\dfrac{\sqrt{3}}{2}$, and the value of $\cos \left( \dfrac{\pi }{3} \right)$ is $\dfrac{1}{2}$.
Using the relationship between the ratios, and these values, we can find the value of $\tan \left( \dfrac{\pi }{3} \right)$ as follows
$\tan x=\dfrac{\sin x}{\cos x}$
Substituting $x=\dfrac{\pi }{3}$, we get
$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\sin \left( \dfrac{\pi }{3} \right)}{\cos \left( \dfrac{\pi }{3} \right)}$
Substituting the values of $\sin \left( \dfrac{\pi }{3} \right)\And \cos \left( \dfrac{\pi }{3} \right)$ in the above equation, we get
$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}$
As the fraction in the numerator and the fraction in the denominator have the same denominator, we can cancel it out, by doing this we get
$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$
Thus, we get the value of $\tan \left( \dfrac{\pi }{3} \right)$.

Note:
We can also find the value of $\tan \left( \dfrac{\pi }{3} \right)$, if we know the value of the tangent of its half, that is the value of $\tan \left( \dfrac{\pi }{6} \right)$. The value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$.
We know the formula for $\tan (2x)$ is $\dfrac{2\tan x}{1-{{\tan }^{2}}x}$. As $\dfrac{\pi }{3}$ is twice of the $\dfrac{\pi }{6}$, we can use this formula to calculate the value of $\tan \left( \dfrac{\pi }{3} \right)$, using the value of $\tan \left( \dfrac{\pi }{6} \right)$, as follows
$\tan (2x)=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$
Substitute $x=\dfrac{\pi }{6}$ in the above formula, we get

& \Rightarrow \tan \left( 2\left( \dfrac{\pi }{6} \right) \right)=\dfrac{2\tan \left( \dfrac{\pi }{6} \right)}{1-{{\tan }^{2}}\left( \dfrac{\pi }{6} \right)} \\\ & \Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{2\tan \left( \dfrac{\pi }{6} \right)}{1-{{\tan }^{2}}\left( \dfrac{\pi }{6} \right)} \\\ \end{aligned}$$ Substituting the value of $$\tan \left( \dfrac{\pi }{6} \right)$$ as $$\dfrac{1}{\sqrt{3}}$$ in the above formula, we get $$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$$ $$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$$ To make the denominator of the fraction in numerator and denominator same, we multiply and divide the fraction in the numerator by $$\sqrt{3}$$, we get $$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\dfrac{2}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}{\dfrac{2}{3}}=\dfrac{\dfrac{2\sqrt{3}}{3}}{\dfrac{2}{3}}$$ Canceling the common factors from numerator and denominator, we get $$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$$