Question

Find the value of the expression \[\sin \left( -{{330}^{\circ }} \right)\]

Answer 1

Hint:Use the trigonometric identity $\sin \left( -\theta \right)=-\sin \theta $ to get the given relation in simplified form. Convert the angle ${{330}^{\circ }}$ by ${{180}^{\circ }}$ $\left( {{330}^{\circ }}={{180}^{\circ }}\times 2-{{30}^{\circ }} \right)$ . Now use the identity $\sin \left( 2\pi -\theta \right)=-\sin \theta $ to get the value of the given expression in the problem.

Complete step-by-step answer:
Let us suppose the value of the given expression in the problem is ‘A’. So, we can write an equation as
$A=\sin \left( -{{330}^{\circ }} \right)...........\left( i \right)$
As, we know the trigonometric identity
$\sin \left( -x \right)=-\sin x............\left( ii \right)$
So, we can use the result of equation (ii) with the equation (i), so, we can re-write the equation (i) as
$\begin{align}
  & A=\sin \left( -{{330}^{\circ }} \right)=-\sin {{330}^{\circ }} \\
 & A=-\sin {{330}^{\circ }}................\left( iii \right) \\
\end{align}$
Now, as the angle involved with the equation (iii) is greater than ${{90}^{\circ }}$ and we do not have known values of trigonometric functions for angles greater than ${{90}^{\circ }}$ . So, we can express ${{330}^{\circ }}$ as
$\begin{align}
  & {{330}^{\circ }}={{360}^{\circ }}-{{30}^{\circ }} \\
 & \Rightarrow {{330}^{\circ }}={{180}^{\circ }}\times 2-{{30}^{\circ }}.............\left( iv \right) \\
\end{align}$
We can write ${{180}^{\circ }}$ and ${{30}^{\circ }}$ in radian form using the relation
${{180}^{\circ }}=\pi \text{ }radian...............\left( v \right)$
So, we can re-write the expression (iv) by replacing ${{180}^{\circ }}$ by $\pi $ radian and ${{30}^{\circ }}$ by $\dfrac{\pi }{6}$ .
So, we get
${{330}^{\circ }}=2\pi -\dfrac{\pi }{6}............\left( vi \right)$
Now, we can get equation (iii) using the above equation as
$A=-\sin \left( 2\pi -\dfrac{\pi }{6} \right)..............\left( vii \right)$
As we know the trigonometric identity of $\sin \left( 2\pi -\theta \right)$ can be given as:
$\sin \left( 2\pi -\theta \right)=-\sin \theta ...........\left( viii \right)$
So, we can get value of A from equation (vii) with the help of equation (viii) by putting $\theta =\dfrac{\pi }{6}$
So, we get
$\begin{align}
  & A=-\left( -\sin \dfrac{\pi }{6} \right) \\
 & A=\sin \dfrac{\pi }{6} \\
\end{align}$
We know the value of $\sin \dfrac{\pi }{6}$ is given as $\dfrac{1}{2}$ .So, we get value of A as
$A=\dfrac{1}{2}$
Hence, we get the value of the given expression in the problem i.e.
$\sin \left( -{{330}^{\circ }} \right)$ is $\dfrac{1}{2}$ .

Note: One can apply the trigonometric identity with the given expression directly without changing the angles in radian form. It is done in the solution because we are more familiar with the trigonometric relation given in solution in radian form only. So, don’t confuse with that part of the solution, you can evaluate the given expression by using degree form of angle as well.
One may confuse the identity $\sin \left( -x \right)=-\sin x$ by the other relations for other trigonometric functions. For future reference, other trigonometric relations are given as –
\[\begin{align}
  & \cos \left( -x \right)=\cos x,\tan \left( -x \right)=-\tan x \\
 & \csc \left( -x \right)=-\csc x,\sec \left( -x \right)=\sec x \\
 & \cot \left( -x \right)=-\cot x \\
\end{align}\].
Here we can observe that $\cos x$ and $\sec x$ both are even functions and other trigonometric functions are odd functions.