Question

Find the principal value of ${{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)$.

Answer 1

Hint: Find the value of angle for which its sine is $\left( -\dfrac{1}{\sqrt{2}} \right)$ in the range of angle $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. Assume this angle as $\theta $ and write the above expression as ${{\sin }^{-1}}\left( \sin \theta \right)$. Now, simply remove the function ${{\sin }^{-1}}$ and sin and write the value of $\theta $ as the principal value.

Complete step-by-step answer:
Since, none of the six trigonometric functions are one-to-one, they are restricted in order to have inverse functions. Therefore, the ranges of the inverse functions are proper subsets of the domains of the original functions.
Now let us come to the question. We have to find the principal value of ${{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)$.
We know that, the range of ${{\sin }^{-1}}x$ is between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$ including these two values. So, we have to select such value of the angle that must lies between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$ and its sine is \[\dfrac{-1}{\sqrt{2}}\].
We know that, the value of sine is \[\dfrac{-1}{\sqrt{2}}\] when the angle is $-\dfrac{\pi }{4}$, which lies between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. Clearly we can see that this angle lies in the 4th quadrant and therefore its sine is negative. Therefore, the expression ${{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)$ can be written as:
${{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{2}} \right)={{\sin }^{-1}}\left( \sin \dfrac{-\pi }{4} \right)$
We know that,
${{\sin }^{-1}}\left( \sin x \right)=x$, when ‘x’ lies between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$.
Since, $-\dfrac{\pi }{4}$ lies between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. Therefore,
${{\sin }^{-1}}\left( \sin \dfrac{-\pi }{4} \right)=\dfrac{-\pi }{4}$
Hence, the principal value of ${{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)$ is $\dfrac{-\pi }{4}$.

Note: One may note that there is only one principal value of an inverse trigonometric function. We know that at many angles the value of sin is $\left( -\dfrac{1}{\sqrt{2}} \right)$ but we have to remember the range in which sin inverse function is defined. We have to choose such an angle which lies in the range and satisfies the function. So, there can be only one answer.