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Question: How do you solve and find the value of \({{\sin }^{-1}}\left( \dfrac{1}{2} \right)\)?...

How do you solve and find the value of sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right)?

Explanation

Solution

To solve the given inverse trigonometric expression i.e. sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right) we are going to assume this expression as θ\theta and then take the sine on both the sides of this equation. After that, we will require this property that sin⁡(sin⁡−1)\sin \left( {{\sin }^{-1}} \right) is 1. And also, we need the information that sin⁡(π6)=12\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}.

Complete step by step answer:
The inverse trigonometric expression that we have to solve and find the value of:
sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right)
Now, let us assume that the above inverse trigonometric expression is equal to θ\theta so writing this statement in the mathematical form we get,
sin⁡−1(12)=θ{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\theta
Taking sine on both the sides we get,
sin⁡(sin⁡−1(12))=sin⁡θ\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\sin \theta
We know that there is an algebraic property that if a number or expression is multiplied by its inverse then we will get 1 so sin⁡(sin⁡−1)\sin \left( {{\sin }^{-1}} \right) is equal to 1 in the above expression.
12=sin⁡θ\dfrac{1}{2}=\sin \theta
We know that from the angles of sine that:
sin⁡π6=12\sin \dfrac{\pi }{6}=\dfrac{1}{2}
On comparing this trigonometric value with 12=sin⁡θ\dfrac{1}{2}=\sin \theta we get,
θ=π6\theta =\dfrac{\pi }{6}
And the general solution for this angle is equal to:
θ=2nπ±π6\theta =2n\pi \pm \dfrac{\pi }{6}
The value that “n” takes in the above equation is from 1, 2, 3……..

Note:
A concept that you can remember here is that the domain of sin⁡−1x{{\sin }^{-1}}x contains value from -1 to 1 so if you see any value of x which is outside this domain then the solution won’t exist for sin⁡−1x{{\sin }^{-1}}x. Sometimes, examiner gives the value of x as π2,π\dfrac{\pi }{2},\pi then as the values of these x values are 1.57 and 3.14 and these values are lying outside the domain of x which is from -1 to 1 so such values of sin⁡−1x{{\sin }^{-1}}x does not exist.
sin⁡−1(π2),sin⁡−1π{{\sin }^{-1}}\left( \dfrac{\pi }{2} \right),{{\sin }^{-1}}\pi
The above inverse trigonometric expressions won’t exist.