Question

How do you calculate `cos (Arcsin -3/5)`?

Answer 1

`cos(sin^(-1)(-3/5))=4/5`

Explanation:

`cos(sin^(-1)(-3/5))=cos(-sin^(-1)(3/5))`
`=cos(sin^(-1)(3/5))`____(1)
Assume `sin^-1(3/5)=x`_____(2)

Substitute equation(2) in equation(1)
`cos(sin^(-1)(-3/5))=cosx`
`=sqrt(1-sin^2x)`
`=sqrt(1-(3/5)^2)` [since from equation 3]
`=4/5`

Which is solution.