Question

Find the value of $ \sin \left( -\dfrac{11\pi }{3} \right) $ .

Answer 1

Hint: We have to find the sine of a negative angle. We will use the identity $ \sin \left( -x \right)=-\sin x $ . Then we will rewrite the angle as a difference of two values. After that, we will use the formula for the difference of angles. This formula is given as $ \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B $ . The values of the trigonometric functions for the standard angles will be used to obtain the required answer.

Complete step by step answer:
We know that $ \sin \left( -x \right)=-\sin x $ . Therefore, we can write the trigonometric function with negative angle in the following manner,
 $ \sin \left( -\dfrac{11\pi }{3} \right)=-\sin \left( \dfrac{11\pi }{3} \right) $
Now, we can write $ 11\pi =12\pi -1\pi $ . Substituting this in the above expression, we get
 $ \begin{align}
  & \sin \left( -\dfrac{11\pi }{3} \right)=-\sin \left( \dfrac{12\pi -\pi }{3} \right) \\
 & \Rightarrow \sin \left( -\dfrac{11\pi }{3} \right)=-\sin \left( \dfrac{12\pi }{3}-\dfrac{\pi }{3} \right) \\
 & \therefore \sin \left( -\dfrac{11\pi }{3} \right)=-\sin \left( 4\pi -\dfrac{\pi }{3} \right)....(i) \\
\end{align} $
We know that if we have a difference of angles, then the trigonometric function can be written using the following formula,
 $ \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B $
Substituting $ A=4\pi $ and $ B=\dfrac{\pi }{3} $ in the above formula, we get the following
 $ \sin \left( 4\pi -\dfrac{\pi }{3} \right)=\sin 4\pi \cos \dfrac{\pi }{3}-\cos 4\pi \sin \dfrac{\pi }{3} $
Let us look at the sine and cosine values for angles $ 4\pi $ and $ \dfrac{\pi }{3} $ . We know that $ \sin n\pi =0 $ for any integer $ n $ . We also know that $ \cos n\pi =1 $ when $ n $ is an even integer and $ \cos n\pi =-1 $ when $ n $ is an odd integer. So, $ \sin 4\pi =0 $ and $ \cos 4\pi =1 $ . We know that $ \sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2} $ and $ \cos \dfrac{\pi }{3}=\dfrac{1}{2} $ . Substituting these values in the above equation, we get
 $ \begin{align}
  & \sin \left( 4\pi -\dfrac{\pi }{3} \right)=0\times \dfrac{1}{2}-1\times \dfrac{\sqrt{3}}{2} \\
 & \therefore \sin \left( 4\pi -\dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2} \\
\end{align} $
Substituting this value in equation $ (i) $ , we get
 $ \begin{align}
  & \sin \left( -\dfrac{11\pi }{3} \right)=-\left( -\dfrac{\sqrt{3}}{2} \right) \\
 & \therefore \sin \left( -\dfrac{11\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\
\end{align} $

Note:
 We should be familiar with the values of trigonometric functions of standard angles for such type of questions. Instead of using the formula for $ \sin \left( A-B \right) $ , we can solve this question by using another method. We can determine the quadrant in which the angle $ \left( 4\pi -\dfrac{\pi }{3} \right) $ lies. We can see that it lies in the fourth quadrant. In the fourth quadrant, only the cosine values are positive. Therefore, we get that $ \sin \left( 4\pi -\dfrac{\pi }{3} \right)=-\sin \dfrac{\pi }{3}=-\dfrac{\sqrt{3}}{2} $ , which is the same value we found using the formula for $ \sin \left( A-B \right) $ .