Question

Find the value of the following:

`cos^(-1) (cos  (13pi)/6)`

Answer 1

We know that cos^–1 (cos x) = x if `x ∈ [0, pi]`, which is the principal value branch of cos^–1x.

Here, `(13pi)/6 !in [0,pi]`.

Now, `cos^(-1) (cos  (13pi)/6)` can be written as:

`cos^(-1) (cos  (13pi)/6) `

= `cos^(-1) [cos(2pi + pi/6)]`

= `cos^(-1) [cos(pi/6)]`, where `pi/6 ∈ [0, pi]`

∴ `cos^(-1) (cos  (13pi)/6) `

= `cos^(-1)[cos (pi/6)] `

= `pi/6`