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

Last updated date: 15th Jun 2024
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Hint: 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.

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 $\dfrac{2\pi }{3}$.