Rewrite cos(arcsin(v)) as an algebraic expression in v.

Question

i know that the answer is squroot of 1-v^2

but i don't know the process of how to come to this answer. Thanks for the help!

Answer 1

let v = sinθ

draw a right triangle, and at one angle θ, label the opposite side = v, and the hypotenuse = 1. so, sinθ = v.

Now, the other leg, adjacent to θ, is √(1-v^2).

cosθ = adjacent/hypotenuse = √(1-v^2)