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Question: Find the principal value of \[{{\sin }^{-1}}\left( \cos \left( \dfrac{2\pi }{3} \right) \right)\]....

Find the principal value of sin⁡−1(cos⁡(2π3)){{\sin }^{-1}}\left( \cos \left( \dfrac{2\pi }{3} \right) \right).

Explanation

Solution

Hint: In this question we are given to find the principal value of inverse trigonometric function. To solve this question first we need to know that the range of sin⁡−1{{\sin }^{-1}} is between (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right). Then equate the expression to’θ\theta ’ then apply the basic trigonometric identity of cos⁡θ=sin⁡(π2−θ)\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)and Simplify the expression.

Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. The principal value of sin⁡−1x{{\sin }^{-1}}xfor x>0x>0 , is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x.
The principal value of sin⁡−1{{\sin }^{-1}} lies between the range of (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right).
Hence, the principal value of the given function will be between the range(−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right).
Now, we have been given the function,sin⁡−1(cos⁡(2π3)){{\sin }^{-1}}\left( \cos \left( \dfrac{2\pi }{3} \right) \right).
Let we will take the principal value of sin⁡−1(cos⁡(2π3)){{\sin }^{-1}}\left( \cos \left( \dfrac{2\pi }{3} \right) \right) as θ\theta . Thus, we get,
θ=sin⁡−1(cos⁡2π3)\theta ={{\sin }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)
We know the basic representation of cosine function as,
cos⁡θ=sin⁡(π2−θ),\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right), where θ∈(−∞,∞)\theta \in (-\infty ,\infty )
Now we will put θ=2π3\theta =\dfrac{2\pi }{3}in the above expression. Hence we get,

& \cos \dfrac{2\pi }{3}=\sin \left( \dfrac{\pi }{2}-\dfrac{2\pi }{3} \right)=\sin \left( \dfrac{3\pi -4\pi }{6} \right)=\sin \left( -\dfrac{\pi }{6} \right) \\\ & \cos \dfrac{2\pi }{3}=\sin \left( -\dfrac{\pi }{6} \right). \\\ \end{aligned}$$ Now we will replace the value of $$\cos \dfrac{2\pi }{3}$$ in the equation represented by $$\theta $$. $$\begin{aligned} & \theta ={{\sin }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)={{\sin }^{-1}}\sin \left( -\dfrac{\pi }{6} \right) \\\ & \theta =-\dfrac{\pi }{6} \\\ \end{aligned}$$ Thus, we got the principal value of the expression as $$\left( -\dfrac{\pi }{6} \right)$$. Hence, the principal value of $${{\sin }^{-1}}\left( \cos \left( \dfrac{2\pi }{3} \right) \right)$$ is $$\left( -\dfrac{\pi }{6} \right)$$ . Note: To solve this question students should be familiar with the domain and range of the sine functions as well as the domain and range of the inverse sine functions. They should also know the basic trigonometric identities to solve the problem easily. We have to understand by looking at the question that we have to convert the cosine term to sine term first. So this way, for every question, students must first decide the operation to be done and then look for suitable formulas, identities instead of trying to apply all the formulas and complicating it.