Question

How do I find the value of $\sin \left( \dfrac{4\pi }{3} \right)$?

Answer 1

Hint: In this question we have been given a trigonometric function of sine for which we have to find the value. Since there is no direct way to solve the given angle $\theta $ in expression, we will first write the angle in the form of $2$ simpler angles and then use the addition-subtraction of angles, expansion formula which is $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$ and then substitute the values of the angles and simplify to get the required solution.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow \sin \left( \dfrac{4\pi }{3} \right)$
Now since there is no direct way to get the solution, we will express it in the form of the addition of two angles. We can express $\dfrac{4\pi }{3}=\pi +\dfrac{\pi }{3}$ therefore, on substituting it in the expression, we get:
$\Rightarrow \sin \left( \pi +\dfrac{\pi }{3} \right)$
Now the above term is in the form of $\sin \left( a+b \right)$ therefore, we will expand the term using the formula $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$. On expanding, we get:
$\Rightarrow \sin \left( \pi \right)\cos \left( \dfrac{\pi }{3} \right)+\cos \left( \pi \right)\sin \left( \dfrac{\pi }{3} \right)$
Now we know that $\sin \left( \pi \right)=0$, $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$, $\cos \left( \pi \right)=-1$ and $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$.
On substituting the values in the expression, we get:
$\Rightarrow 0\times \dfrac{1}{2}+\left( -1 \right)\times \dfrac{\sqrt{3}}{2}$
On simplifying the values, we get:
$\Rightarrow 0-\dfrac{\sqrt{3}}{2}$
On simplifying further, we get:
$\Rightarrow -\dfrac{\sqrt{3}}{2}$, which is the required solution.

Note: It is to be remembered that the addition- subtraction formula for trigonometric functions is different for various trigonometric functions. The general rule for $\sin \left( \pi +\theta \right)$ should be remembered which is $\sin \left( \pi +\theta \right)=-\sin \theta $ therefore by using this formula, we can directly conclude that $\Rightarrow \sin \left( \pi +\dfrac{\pi }{3} \right)=-\sin \left( \dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$.
It is to be also remembered that the same formula applies for cosine function which means $\cos \left( \pi +\theta \right)=-\cos \theta $.