Question

How do you evaluate sine, cosine, tangent of $\dfrac{-\pi }{6}$ without using a calculator?

Answer 1

We start solving the problem by making use of fact that $\pi$ radians = ${{180}^{\circ }}$ to convert $\dfrac{-\pi }{6}$ to degrees. We then find the value of $\sin \left( \dfrac{-\pi }{6} \right)$ by making use of facts that $\sin \left( -x \right)=-\sin x$ and $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$. We then find the value of $\cos \left( \dfrac{-\pi }{6} \right)$ by making use of facts that $\cos \left( -x \right)=\cos x$ and $\cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$. We then find the value of $\tan \left( \dfrac{-\pi }{6} \right)$ by making use of the facts that $\tan \left( -x \right)=-\tan x$ and $\tan \left( {{30}^{\circ }} \right)=\dfrac{1}{\sqrt{3}}$.

Complete step by step answer:
According to the problem, we are asked to find the value of sine, cosine, tangent of $\dfrac{-\pi }{6}$.
We know that $\pi$ radians = ${{180}^{\circ }}$. So, we get $\dfrac{-\pi }{6}=\dfrac{-{{180}^{\circ }}}{6}=-{{30}^{\circ }}$.
Now, let us find the value of $\sin \left( \dfrac{-\pi }{6} \right)$.
$\Rightarrow \sin \left( \dfrac{-\pi }{6} \right)=\sin \left( -{{30}^{\circ }} \right)$ ---(1).
We know that $\sin \left( -x \right)=-\sin x$. Let us use this result in equation (1).
$\Rightarrow \sin \left( \dfrac{-\pi }{6} \right)=-\sin \left( {{30}^{\circ }} \right)$ ---(2).
We know that $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$. Let us use this result in equation (2).
$\Rightarrow \sin \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{2}$.
Now, let us find the value of $\cos \left( \dfrac{-\pi }{6} \right)$.
$\Rightarrow \cos \left( \dfrac{-\pi }{6} \right)=\cos \left( -{{30}^{\circ }} \right)$ ---(3).
We know that $\cos \left( -x \right)=\cos x$. Let us use this result in equation (3).
$\Rightarrow \cos \left( \dfrac{-\pi }{6} \right)=\cos \left( {{30}^{\circ }} \right)$ ---(4).
We know that $\cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$. Let us use this result in equation (3).
$\Rightarrow \cos \left( \dfrac{-\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$.
Now, let us find the value of $\tan \left( \dfrac{-\pi }{6} \right)$.
$\Rightarrow \tan \left( \dfrac{-\pi }{6} \right)=\tan \left( -{{30}^{\circ }} \right)$ ---(5).
We know that $\tan \left( -x \right)=-\tan x$. Let us use this result in equation (5).
$\Rightarrow \tan \left( \dfrac{-\pi }{6} \right)=-\tan \left( {{30}^{\circ }} \right)$ ---(6).
We know that $\tan \left( {{30}^{\circ }} \right)=\dfrac{1}{\sqrt{3}}$. Let us use this result in equation (6).
$\Rightarrow \tan \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{\sqrt{3}}$.
$\therefore$ The values of sine, cosine, tangent of $\dfrac{-\pi }{6}$ are $\dfrac{-1}{2}$, $\dfrac{\sqrt{3}}{2}$, $\dfrac{1}{\sqrt{3}}$.

Note:
Whenever we get this type of problem, we first convert the radians to degrees to make the process of solving the problem easier. We can also solve this problem by making use of the facts that $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to find the values of $\cos \left( \dfrac{-\pi }{6} \right)$, $\tan \left( \dfrac{-\pi }{6} \right)$. We can also find the value of secant, cosecant and cotangent of $\dfrac{-\pi }{6}$ using the obtained angles. Similarly, we can expect problems to find the values of sine, cosine, tangent of $\dfrac{-\pi }{3}$.