The value of `sin^(-1)(cos(33pi)/5)` is `(3pi)/5` (b) `pi/(10)` (c) `pi/(10)` (d) `(7pi)/5`

Options

A. `(3pi)/(5)`
B. `(-7pi)/(5)`
C. `(pi)/(10)`
D. `(-pi)/(10)`

Answer

D

Explanation

We have,
`sin^(-1) (cos'(33pi)/(5)) = sin^(-1)[cos(6pi + (3pi)/(5))] = sin^(-1)[cos((3pi)/(5))] [:' cos(2npi + theta) = cos theta]`
` =sin^(-1)[cos(pi/2 + pi/10)]= sin^(-1)(-sin'(pi)/(10))`
`= -sin^(-1)(sin'(pi)/(10)) , [:' sin^(-1)(-x) = - sin^(-1)x]`
` = - (pi)/(10) [:' sin^(-1)(sinx) = x, x in ((-pi)/(2),(pi)/(2))]`