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Question: For the principal value, evaluate the following: \[{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \righ...

For the principal value, evaluate the following:
sin⁡−1(−32)+cos⁡−1(32){{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right).

Explanation

Solution

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 sin−1xsi{{n}^{-1}}x is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] and for cos⁡−1x{{\cos }^{-1}}x is [0,π]\left[ 0,\pi \right]. We can start solving this question by taking θ=sin⁡−1(−32)\theta ={{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right) and then find the principal value of θ\theta . Then we can proceed to cos⁡−1(32){{\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(−32)+cos⁡−1(32){{\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 sin−1xsi{{n}^{-1}}x the range is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
For cos⁡−1x{{\cos }^{-1}}x the range is [0,π]\left[ 0,\pi \right].
Let us suppose θ=sin⁡−1(−32)\theta ={{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right).
We know that sin⁡(−π3)=−32\sin \left( \dfrac{-\pi }{3} \right)=\dfrac{-\sqrt{3}}{2}.
So, by substituting the value of (−32)\left( \dfrac{-\sqrt{3}}{2} \right) in the above expression, we get as follows:
θ=sin−1(sin⁡−π3)\theta =si{{n}^{-1}}\left( \sin \dfrac{-\pi }{3} \right)
Since we know that sin−1sinx=xsi{{n}^{-1}}sinx=x, where x must lie between the interval [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
⇒θ=−π3\Rightarrow \theta =\dfrac{-\pi }{3}
Hence sin⁡−1(−32)=−π3{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)=\dfrac{-\pi }{3}.
Again, let us suppose θ=cos⁡−1(32)\theta ={{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right).
We know that cos⁡(π6)=32\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}.
So by substituting the value of (32)\left( \dfrac{\sqrt{3}}{2} \right) in the expression, we get as follows:
θ=cos⁡−1cos⁡π6\theta ={{\cos }^{-1}}\cos \dfrac{\pi }{6}
Since we know that cos⁡−1cosx⁡=x{{\cos }^{-1}}\operatorname{cosx}=x, where x must lie between the interval [0,π]\left[ 0,\pi \right].
⇒θ=π6\Rightarrow \theta =\dfrac{\pi }{6}
Hence cos⁡−1(32)=π6{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}.
Now substituting the values of sin−1(−32)=−π3si{{n}^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)=\dfrac{-\pi }{3} and cos⁡−1(32)=π6{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6} in the given expression we get as follows:
sin⁡−1(−32)+cos⁡−1(32)=−π3+π6=−2π+π6=−π6{{\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(−32)+cos⁡−1(32){{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) is equal to −π6\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(32)=π3{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3} which is wrong. So be careful while solving as cos⁡−1(32)=π6{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}.