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Question: Compute the value of the following expression: \[{{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin...

Compute the value of the following expression:
cos⁡−1(12)−2sin⁡−1(−12){{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( \dfrac{-1}{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 cos⁡−1x{{\cos }^{-1}}x is [0,π]\left[ 0,\pi \right] and for sin−1xsi{{n}^{-1}}x is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]. We can start solving this question by taking θ=cos⁡−1(12)\theta ={{\cos }^{-1}}\left( \dfrac{1}{2} \right) and then find the principal value of θ\theta . Then we can proceed to sin⁡−1(−12){{\sin }^{-1}}\left( \dfrac{-1}{2} \right) in a similar way.

Complete step-by-step answer:
We have been given to evaluate the trigonometric expression cos⁡−1(12)−2sin⁡−1(−12){{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( \dfrac{-1}{2} \right).
Now we know that the principal value means the value which lies between the defined range of inverse trigonometric functions.
For cos⁡−1x{{\cos }^{-1}}x the range is [0,π]\left[ 0,\pi \right].
For sin−1xsi{{n}^{-1}}x the range is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
Let us suppose θ=cos⁡−1(12)\theta ={{\cos }^{-1}}\left( \dfrac{1}{2} \right)
We know that cos⁡(π3)=12\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}.
So, by substituting the value of 12\dfrac{1}{2} in the above expression, we get as follows:
θ=cos⁡−1(cos⁡π3)\theta ={{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right)
Since we know that cos⁡−1cos⁡x=x{{\cos }^{-1}}\cos x=x, where x must lie between the interval [0,π]\left[ 0,\pi \right].
⇒θ=π3\Rightarrow \theta =\dfrac{\pi }{3}
Hence cos⁡−1(12)=π3{{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}.
Again, let us suppose θ=sin−1(−12)\theta =si{{n}^{-1}}\left( \dfrac{-1}{2} \right).
We know that sin(−π6)=−12sin\left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{2}.
So by substituting the value of −12\dfrac{-1}{2} in the expression, we get as follows:
θ=sin⁡−1sin⁡−π6\theta ={{\sin }^{-1}}\sin \dfrac{-\pi }{6}
Since we know that sin⁡−1sinx⁡=x{{\sin }^{-1}}\operatorname{sinx}=x, where x must lie between the interval [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
⇒θ=−π6\Rightarrow \theta =\dfrac{-\pi }{6}
Hence sin−1(−12)=−π6si{{n}^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{-\pi }{6}
Now substituting the values of cos⁡−1(12)=π3{{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3} and sin−1(−12)=−π6si{{n}^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{-\pi }{6} in the given expression we get as follows:
cos⁡−1(12)−2sin−1(−12)=π3−2(−π6)=π3+2π6=π3+π3=2π3{{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2si{{n}^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{\pi }{3}-2\left( \dfrac{-\pi }{6} \right)=\dfrac{\pi }{3}+\dfrac{2\pi }{6}=\dfrac{\pi }{3}+\dfrac{\pi }{3}=\dfrac{2\pi }{3}
Therefore, the value of the given expression, cos⁡−1(12)−2sin−1(−12){{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2si{{n}^{-1}}\left( \dfrac{-1}{2} \right) is equal to 2π3\dfrac{2\pi }{3}.

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 we forget the ‘2’ multiplied by sin⁡−1(−12){{\sin }^{-1}}\left( \dfrac{-1}{2} \right) in the given expression and we just substitute the values of sin⁡−1(−12){{\sin }^{-1}}\left( \dfrac{-1}{2} \right) and we get the incorrect answer.Students should remember the properties of inverse trigonometric functions and trigonometric standard angles to solve these types of questions.