Question

Find the value of the following:

`tan^-1 [2 cos (2  sin^-1  1/2)]`

Answer 1

Let `sin^-1  1/2 = x`.

Then, `sin x = 1/2 = sin (pi/6)`.

∴ `sin^-1  1/2 = pi/6`

∴ `tan^-1 [2  cos (2 sin^-1  1/2)]`

= `tan^-1 [2  cos (2 xx pi/6)]`

= `tan^-1 [2  cos  pi/3]`

= `tan^-1 [2 xx 1/2]`

= tan⁻¹ 1

= `pi/4`