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

Find the value of sin⁡−1(sin⁡3π5){{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right) .

Explanation

Solution

For answering this question we will use the following two formulae as keysin⁡−1(sin⁡(π−θ))=sin⁡−1(sin⁡θ){{\sin }^{-1}}\left( \sin \left( \pi -\theta \right) \right)={{\sin }^{-1}}\left( \sin \theta \right)and sin⁡−1(sin⁡θ)=θ ∀θ∈[−π2,π2]{{\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(sin⁡3π5){{\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(sin⁡3π5){{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right) .
By observing it we can say that it is in the form of sin⁡−1(sin⁡θ){{\sin }^{-1}}\left( \sin \theta \right) whereθ>π2\theta >\dfrac{\pi }{2} .
The expression we have can be simply written as sin⁡−1(sin⁡(π−2π5)){{\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(sin⁡(π−θ)){{\sin }^{-1}}\left( \sin \left( \pi -\theta \right) \right) is given as sin⁡−1(sin⁡θ){{\sin }^{-1}}\left( \sin \theta \right) . We can use this here and write the given expression as sin⁡−1(sin⁡(π−2π5))=sin⁡−1(sin⁡2π5){{\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(sin⁡θ)=θ ∀θ∈[−π2,π2]{{\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(sin⁡2π5)=2π5 {{\sin }^{-1}}\left( \sin \dfrac{2\pi }{5} \right)=\dfrac{2\pi }{5}\text{ } .

Hence we can conclude that the value of sin⁡−1(sin⁡3π5){{\sin }^{-1}}\left( \sin \dfrac{3\pi }{5} \right) is 2π5\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(sin⁡(π−θ))=sin⁡−1(sin⁡θ){{\sin }^{-1}}\left( \sin \left( \pi -\theta \right) \right)={{\sin }^{-1}}\left( \sin \theta \right) and sin⁡−1(sin⁡θ)=θ ∀θ∈[−π2,π2]{{\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(cos⁡θ)=θ ∀θ∈[0,π]{{\cos }^{-1}}\left( \cos \theta \right)=\theta \text{ }\forall \theta \in \left[ 0,\pi \right] and for tangent function it is tan⁡−1(tan⁡θ)=θ ∀θ∈(−π2,π2){{\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(cot⁡θ)=θ ∀θ∈(0,π){{\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.