Question

Find principal value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)+{{\cos }^{-1}}\left( 0 \right)$.

Answer 1

We will first find the principal value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$. Then we will find the principal value of ${{\cos }^{-1}}\left( 0 \right)$. We will add these two values to get the required answer. The principal value of an inverse trigonometric function at a point $x$ is the value of the inverse function at the point $x$, which lies in the range of the principal branch.

Complete step by step answer:
We know that the principle interval of ${{\sin }^{-1}}\theta$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Let ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=x$.
Therefore, we now know that $\sin x=\dfrac{1}{2}$. This implies that $x=\dfrac{\pi }{6}$. Since $\dfrac{\pi }{6}\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, we can say that $x=\dfrac{\pi }{6}$ is the principle value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$.
Now, we will find the principal value of ${{\cos }^{-1}}\left( 0 \right)$. We know that the principal interval of ${{\cos }^{-1}}\theta$ is $\left[ 0,\pi \right]$. Let ${{\cos }^{-1}}\left( 0 \right)=y$. Therefore, we have $\cos y=0$. So, we get $y=\dfrac{\pi }{2}$. Since $\dfrac{\pi }{2}\in \left[ 0,\pi \right]$, we conclude that $y=\dfrac{\pi }{2}$ is the principle value of ${{\cos }^{-1}}\left( 0 \right)$.
Now, we can substitute the values of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\cos }^{-1}}\left( 0 \right)$ to find the required answer in the following manner,
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)+{{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{6}+\dfrac{\pi }{2}=\dfrac{\pi +3\pi }{6}=\dfrac{4\pi }{6}=\dfrac{2\pi }{3}$.

Therefore, the answer is $\dfrac{2\pi }{3}$.

Note: There are multiple angles which have the same values for the trigonometric functions. The concept of principle value is essential so that values can be standardized for solving questions. Apart from trigonometry, this concept also arises in certain functions that involve complex numbers. Similar to trigonometric functions, there are identities and relations involving inverse trigonometric functions. We should be familiar with these identities as it will be useful for simplifying equations involving inverse trigonometric functions.