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Question: How do you find \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right)\]?...

How do you find sin⁡−1(sin⁡(5π6)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right)?

Explanation

Solution

Here, we will first evaluate the sine function. We will write the given angle as a difference of two angles and then apply the trigonometric function to simplify the expression. We will then substitute the value of the obtained angle to find the value of the sine function. We will then substitute this value in the given expression and simplify it using the range of sine function to find the required value.

Formula Used:
We will use the following formulas:
Trigonometric Identity: sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta
Trigonometric Ratio: sin⁡π6=12\sin \dfrac{\pi }{6} = \dfrac{1}{2}
Trigonometric Identity: sin⁡−1(12)=π6+2nπ{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} + 2n\pi

Complete Step by Step Solution:
We are given a trigonometric function sin⁡−1(sin⁡(5π6)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right).
First, we will find the value of sin⁡(5π6)\sin \left( {\dfrac{{5\pi }}{6}} \right).
sin⁡(5π6)=sin⁡(π−π6)\sin \left( {\dfrac{{5\pi }}{6}} \right) = \sin \left( {\pi - \dfrac{\pi }{6}} \right)
We know that Trigonometric Identity sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta since it lies in the second quadrant. So, we get
⇒sin⁡(5π6)=sin⁡(π6)\Rightarrow \sin \left( {\dfrac{{5\pi }}{6}} \right) = \sin \left( {\dfrac{\pi }{6}} \right)
Now substituting the value sin⁡π6=12\sin \dfrac{\pi }{6} = \dfrac{1}{2} in the above equation, we get
⇒sin⁡(5π6)=12\Rightarrow \sin \left( {\dfrac{{5\pi }}{6}} \right) = \dfrac{1}{2} ……………………………..(1)\left( 1 \right)
Now, we will find the value of sin⁡−1(sin⁡(5π6)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right).
By substituting equation (1)\left( 1 \right) in the expression sin⁡−1(sin⁡(5π6)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right), we get
sin⁡−1(sin⁡(5π6))=sin⁡−1(12){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
Now we know that sin⁡−1(12)=π6+2nπ,n∈Z{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} + 2n\pi ,n \in {\bf{Z}}. So we can write above equation as
⇒sin⁡−1(sin⁡(5π6))=π6+2nπ\Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right) = \dfrac{\pi }{6} + 2n\pi
We know that sin⁡−1θ{\sin ^{ - 1}}\theta always lies in [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right].
⇒sin⁡−1(sin⁡(5π6))=π6∈[−π2,π2]\Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right) = \dfrac{\pi }{6} \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]

Therefore, the solution for sin⁡−1(sin⁡(5π6)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right) is π6\dfrac{\pi }{6} which lies in the domain [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right].

Note:
We know that Trigonometric Equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation that is always true for all the variables. We should know that we have many trigonometric identities that are related to all the other trigonometric equations. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right-angled triangle with respect to any of its acute angle. They are used to find the relationships between the sides of a right-angle triangle. Also, the value of the inverse trigonometric ratio should always lie in the domain of the trigonometric ratio.