Question

For the principal value, evaluate the following:

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

Answer 1

`sin^-1[cos{2\text(cosec)^-1(-2)}]=sin^-1[cos{2\text(cosec)^-1\text(cosec- )pi/6}]`

`=sin^-1[cos{-pi/3}]`

`=sin^-1[cos(pi/3)]`

`=sin^-1(1/2)`

`=sin^-1(sin π/6)`

`=pi/6`