Question

Find the principal value of each of the following:
${{\cos }^{-1}}\left( \sin \dfrac{4\pi }{3} \right)$.

Answer 1

Hint:Substitute the value of $\sin \left( \dfrac{4\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$. Find the value of angle for which its cosine is $-\dfrac{\sqrt{3}}{2}$ in the range of angle $\left[ 0,\pi \right]$. Assume this angle as $\theta$ and write the above expression as ${{\cos }^{-1}}\left( \cos \theta \right)$. Now, simply remove the function ${{\cos }^{-1}}$ and cosine 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.
We know that, $\sin \left( \dfrac{4\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$, as the angle is in third quadrant therefore, its sine is negative. So, substituting this value in the expression: ${{\cos }^{-1}}\left( \sin \dfrac{4\pi }{3} \right)$, we get, ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$.
Now let us come to the question. We have to find the principal value of ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$.
We know that the range of ${{\cos }^{-1}}x$ is between 0 and $\pi$ including these two values. So, we have to select such a value of the angle that must lie in the range from 0 to $\pi$ and its cosine is $\left( -\dfrac{\sqrt{3}}{2} \right)$.
We know that, the value of cosine is $\left( -\dfrac{\sqrt{3}}{2} \right)$ when the angle Is $\dfrac{2\pi }{3}$, which lies in the range of 0 and $\pi$. Therefore, the expression ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$ can be written as:
${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)={{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$
We know that,
${{\cos }^{-1}}\left( \cos x \right)=x$, when ‘x’ lies in the range of 0 and $\pi$.
Since, $\dfrac{2\pi }{3}$ lies in the range $\left[ 0,\pi \right]$. Therefore,
${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)=\dfrac{2\pi }{3}$
Hence, the principal value of ${{\cos }^{-1}}\left( \sin \dfrac{4\pi }{3} \right)$ is $\dfrac{2\pi }{3}$.

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 cosine is $\left( -\dfrac{\sqrt{3}}{2} \right)$ but we have to remember the range in which cosine 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.