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Question: Find the principal value of \[{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)\]...

Find the principal value of sin⁡−1(−32){{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)

Explanation

Solution

First of all, use sin⁡−1(−x)=−sin⁡−1x{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x. Now consider the range of the principal value of sin⁡−1x{{\sin }^{-1}}x. Now find the angle θ\theta in this range at which sin⁡θ=32\sin \theta =\dfrac{\sqrt{3}}{2} or the value of sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) to get the desired result.

Complete step-by-step answer:
In this question, we have to find the principal value of sin⁡−1(−32){{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right).
First of all, let us consider the expression given in the question,
E=sin⁡−1(−32)E={{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)
We know that, sin⁡−1(−x)=−sin⁡−1(x){{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}\left( x \right). By using this in the above expression, we get,
E=−sin⁡−1(32)....(i)E=-{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)....\left( i \right)
Now, let us draw the table for trigonometric ratios of general angles.

Now we know that the range of principal value of sin⁡−1(x){{\sin }^{-1}}\left( x \right) lies between [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
From the table of general trigonometric ratios, we get,
sin⁡(π3)=32\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}
By taking sin⁡−1{{\sin }^{-1}} on both the sides, we get,
sin⁡−1sin⁡(π3)=sin⁡−1(32){{\sin }^{-1}}\sin \left( \dfrac{\pi }{3} \right)={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)
We know that for −π2≤x≤π2,sin⁡−1sin⁡(x)=x\dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2},{{\sin }^{-1}}\sin \left( x \right)=x. So, we get,
π3=sin⁡−1(32)....(ii)\dfrac{\pi }{3}={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)....\left( ii \right)
Now by substituting the value of sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) in equation (i), we get,
E=−π3E=\dfrac{-\pi }{3}
Hence, we get the principal value of sin⁡−1(−32){{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) as −π3\dfrac{-\pi }{3}.

Note: In this question, first of all, students must take care that the value of the angle must lie in the range of sin⁡−1x{{\sin }^{-1}}x which is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]. For example, we know that sin⁡(2π3)\sin \left( \dfrac{2\pi }{3} \right) is also equal to 32\dfrac{\sqrt{3}}{2} but we never take sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) as 2π3\dfrac{2\pi }{3} because 2π3\dfrac{2\pi }{3} does not lie in the range of sin⁡−1x{{\sin }^{-1}}x. In the case of inverse trigonometric functions, students should remember the range and domain of various functions as they are very useful while solving the questions.