Question

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

Answer 1

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

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

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

Note: In this question, first of all, students must take care that the value of the angle must lie in the range of \[{{\sin }^{-1}}x\] which is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]. For example, we know that \[\sin \left( \dfrac{2\pi }{3} \right)\] is also equal to \[\dfrac{\sqrt{3}}{2}\] but we never take \[{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] as \[\dfrac{2\pi }{3}\] because \[\dfrac{2\pi }{3}\] does not lie in the range of \[{{\sin }^{-1}}x\]. 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.