Question

For the principal values, evaluate the following:

sin^-1[cos{cosec^-1(-2)}]

Answer 1

First of all we need to find the principal value for cosec^–1(–2)

Let,

As we know cosec(–θ) = –cosecθ

The range of principal value of cosec^–1 is 

\([\frac{-\pi}2,\frac{\pi}2] \) - {0} and cosec\((\frac{-\pi}6)=-2\)

Therefore, the principal value of cosec^–1(–2) is \(\frac{-\pi}6.\)

\(\therefore\) Now, the question changes to

sin^-1[cos\(\frac{-\pi}6\)]

Cos(–θ) = cos(θ)

\(\therefore\) we can write the above expression as

sin^-1[cos\(\frac{-\pi}6\)]

Let,

The range of principal value of sin^–1 is \([\frac{-\pi}2,\frac{\pi}2] \)

and sin\((\frac{\pi}3)=\frac{\sqrt3}2\)

Therefore, the principal value of \(sin^{-1}(\frac{\sqrt3}2)\) is \(\frac{\pi}3.\) 

Hence, the principal value of the given equation is

\(\frac{\pi}3.\)