Question

Evaluate:

`sin[cos^-1 (3/5)]`

Answer 1

Let x = `cos^-1 (3/5)`    .......(i)

∴ cos x = `3/5`

∴ sin x = `sqrt(1 - cos^2x)`

= `sqrt(1 - 9/25)`

= `sqrt(16/25)`

= `4/5`

⇒ x = `sin^-1 (4/5)`  .......(ii)

From (i) and (ii), we get

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

= `4/5`