Question

Find the principal value of $${\sin }^{-1}(-{\displaystyle \frac{\sqrt{3}}{2}})$$

Answer 1

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

Complete step-by-step answer:
In this question, we have to find the principal value of $${\sin }^{-1}(-{\displaystyle \frac{\sqrt{3}}{2}})$$.
First of all, let us consider the expression given in the question,
$$E={\sin }^{-1}(-{\displaystyle \frac{\sqrt{3}}{2}})$$
We know that, $${\sin }^{-1}(-x)=-{\sin }^{-1}\left(x\right)$$. By using this in the above expression, we get,
$$E=-{\sin }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)....\left(i\right)$$
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 $$[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$$.
From the table of general trigonometric ratios, we get,
$$\sin \left({\displaystyle \frac{\pi }{3}}\right)={\displaystyle \frac{\sqrt{3}}{2}}$$
By taking $${\sin }^{-1}$$ on both the sides, we get,
$${\sin }^{-1}\sin \left({\displaystyle \frac{\pi }{3}}\right)={\sin }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)$$
We know that for $${\displaystyle \frac{-\pi }{2}}\le x\le {\displaystyle \frac{\pi }{2}},{\sin }^{-1}\sin \left(x\right)=x$$. So, we get,
$${\displaystyle \frac{\pi }{3}}={\sin }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)....\left(ii\right)$$
Now by substituting the value of $${\sin }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)$$ in equation (i), we get,
$$E={\displaystyle \frac{-\pi }{3}}$$
Hence, we get the principal value of $${\sin }^{-1}(-{\displaystyle \frac{\sqrt{3}}{2}})$$ as $${\displaystyle \frac{-\pi }{3}}$$.

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 $$[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$$. For example, we know that $$\sin \left({\displaystyle \frac{2\pi }{3}}\right)$$ is also equal to $${\displaystyle \frac{\sqrt{3}}{2}}$$ but we never take $${\sin }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)$$ as $${\displaystyle \frac{2\pi }{3}}$$ because $${\displaystyle \frac{2\pi }{3}}$$ 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.