Question

Rewrite the following expression as an algebraic function of x

sin(arccos(x/2))

I know sine is y, which is opposite over hypotenuse. I also know that arccos is the inverse of cosine. I'm confused on what the question is asking and what to do with the x. Please help! Thanks

Answer 1

so arccos(?) is an angle, so let
arcos(x/2) = θ
that means cosθ = x/2
that is cosθ = adjacent / hypoenuse = x/2
so we are looking at a right-angles triangle with angle θ, adjacent = x, hypotenuse =2
Make a sketch,
then x^2 + y^2 = 4
y^2 = 4-x^2
y = √(4-x^2)
then
sin(arccos(x/2))
= sin θ
= opposite/hypotenuse

= √(4-x^2)/2