Question

Find the value of ${{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right)$ .

Answer 1

Hint: For answering this question we will use the following two formulae as key${{\sin }^{-1}}\left( \sin \left( \pi -\theta \right) \right)={{\sin }^{-1}}\left( \sin \theta \right)$and ${{\sin }^{-1}}\left( \sin \theta \right)=\theta \text{ }\forall \theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ . Simplify the expression we have from the question that is ${{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right)$ and derive its value.

Complete step by step answer:
Now considering from the question we have the expression ${{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right)$ .
By observing it we can say that it is in the form of ${{\sin }^{-1}}\left( \sin \theta \right)$ where$\theta >\dfrac{\pi }{2}$ .
The expression we have can be simply written as ${{\sin }^{-1}}\left( \sin \left( \pi -\dfrac{2\pi }{5} \right) \right)$ .
From the basic concept of inverse trigonometric functions we know that the value of ${{\sin }^{-1}}\left( \sin \left( \pi -\theta \right) \right)$ is given as ${{\sin }^{-1}}\left( \sin \theta \right)$ . We can use this here and write the given expression as ${{\sin }^{-1}}\left( \sin \left( \pi -\dfrac{2\pi }{5} \right) \right)={{\sin }^{-1}}\left( \sin \dfrac{2\pi }{5} \right)$ which is simplified before in this form for our convenience.
Now from the basic concept we know that the formulae relating to the concept of inverse trigonometric functions ${{\sin }^{-1}}\left( \sin \theta \right)=\theta \text{ }\forall \theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ .
Using this formula discussed above we can simply write the expression we have as ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{5} \right)=\dfrac{2\pi }{5}\text{ }$ .

Hence we can conclude that the value of ${{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right)$ is $\dfrac{2\pi }{5}$.

Note: While answering this type of question we should have a sure shot about the formulae. In this case those are ${{\sin }^{-1}}\left( \sin \left( \pi -\theta \right) \right)={{\sin }^{-1}}\left( \sin \theta \right)$ and ${{\sin }^{-1}}\left( \sin \theta \right)=\theta \text{ }\forall \theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ . There are many other similar formulae like for cosine function it is ${{\cos }^{-1}}\left( \cos \theta \right)=\theta \text{ }\forall \theta \in \left[ 0,\pi \right]$ and for tangent function it is ${{\tan }^{-1}}\left( \tan \theta \right)=\theta \text{ }\forall \theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ and for cotangent function it is ${{\cot }^{-1}}\left( \cot \theta \right)=\theta \text{ }\forall \theta \in \left( 0,\pi \right)$ and for secant function it is ${{\sec }^{-1}}\left( \sec \theta \right)=\theta \text{ }\forall \theta \in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$ and for cosecant function it is ${{\csc }^{-1}}\left( \csc \theta \right)=\theta \text{ }\forall \theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}$ . And similarly there are many other formulae which you can find in many books you can refer to.