Question

Evaluate: `cos(2cos^-1x + sin^-1x)` at `x = 1/5`.

Answer 1

cos (2 cos^–1x + sin^–1x) = cos [cos^–1x + cos^–1x + sin^–1x]

= cos [cos^–1x + (cos^–1x + sin^–1x)]

= `cos [cos^-1x + π/2]`

= `cos (π/2 + cos^-1x)`

= – sin (cos^–1x)   ...`[∵ cos(π/2 + θ) = -sin θ]`

= `-sin(sin^-1 sqrt(1 - x^2))`   ...`[∵ cos^-1x = sin^-1 sqrt(1 - x^2)]`

= `-sqrt(1 - x^2)`

= `-sqrt(1 - (1/5)^2`   ...`[∵ x = 1/5 ("Given")]`

= `-sqrt(1 - 1/25)`

= `-sqrt((25 - 1)/25)`

= `-sqrt(24/25)`

= `-(2sqrt(6))/5`