Question

For the principal value, evaluate the following:
\[{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\].

Answer 1

Hint:For the above question we will have to know about the principal value of an inverse trigonometric function is a value that belongs to the principal branch of range of function. We know that the principal branch of range for \[si{{n}^{-1}}x\] is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] and for \[{{\cos }^{-1}}x\] is \[\left[ 0,\pi \right]\]. We can start solving this question by taking \[\theta ={{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)\] and then find the principal value of \[\theta \]. Then we can proceed to \[{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] in a similar way.

Complete step-by-step answer:
We have been given to evaluate the trigonometric expression \[{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\].
Now we know that the principal value means the value which lies between the defined range of inverse trigonometric functions.
For \[si{{n}^{-1}}x\] the range is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].
For \[{{\cos }^{-1}}x\] the range is \[\left[ 0,\pi \right]\].
Let us suppose \[\theta ={{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)\].
We know that \[\sin \left( \dfrac{-\pi }{3} \right)=\dfrac{-\sqrt{3}}{2}\].
So, by substituting the value of \[\left( \dfrac{-\sqrt{3}}{2} \right)\] in the above expression, we get as follows:
\[\theta =si{{n}^{-1}}\left( \sin \dfrac{-\pi }{3} \right)\]
Since we know that \[si{{n}^{-1}}sinx=x\], where x must lie between the interval \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].
\[\Rightarrow \theta =\dfrac{-\pi }{3}\]
Hence \[{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)=\dfrac{-\pi }{3}\].
Again, let us suppose \[\theta ={{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\].
We know that \[\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\].
So by substituting the value of \[\left( \dfrac{\sqrt{3}}{2} \right)\] in the expression, we get as follows:
\[\theta ={{\cos }^{-1}}\cos \dfrac{\pi }{6}\]
Since we know that \[{{\cos }^{-1}}\operatorname{cosx}=x\], where x must lie between the interval \[\left[ 0,\pi \right]\].
\[\Rightarrow \theta =\dfrac{\pi }{6}\]
Hence \[{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}\].
Now substituting the values of \[si{{n}^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)=\dfrac{-\pi }{3}\] and \[{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}\] in the given expression we get as follows:
\[{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{-\pi }{3}+\dfrac{\pi }{6}=\dfrac{-2\pi +\pi }{6}=\dfrac{-\pi }{6}\]
Therefore, the value of the given expression \[{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] is equal to \[\dfrac{-\pi }{6}\].

Note: Be careful while finding the principal value of the inverse trigonometric function and do check it once that the value must lie between the principal branch of range of the function. Sometimes by mistake we take the value of \[{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3}\] which is wrong. So be careful while solving as \[{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}\].