Question

Evaluate `cos[pi/6 + cos^-1 (- sqrt(3)/2)]`

Answer 1

Let `cos^-1 ((-sqrt(3))/2)` = y

∴ cos y = `(-sqrt(3))/2`

= `- cos (pi/6)`

= `cos (pi - pi/6)`

= `cos  (5pi)/6`

The principal value branch of cos⁻¹ is [0, π] and `0 ≤ (5pi)/6 ≤ pi`.

∴ y = `(5pi)/6`

∴ `cos^-1 ((-sqrt(3))/2) = (5pi)/6`

∴ `pi/6 + cos^-1 ((-sqrt(3))/3)`

= `pi/6 + (5pi)/6`

= π

∴ `cos[pi/6 + cos^-1 (- sqrt(3)/2)]` = cos π = −1