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

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

To rewrite the given expression as an algebraic function of x, we will make use of the trigonometric identity:

sin(arccosθ) = √(1 - θ^2)

In our case, θ = x/2.

Substitute this value into the identity:

sin(arccos(x/2)) = √(1 - (x/2)^2)

Now, let's simplify further:

sin(arccos(x/2)) = √(1 - x^2/4)

Since x is a variable, we can rewrite the expression as an algebraic function:

f(x) = √(1 - x^2/4)

Therefore, the algebraic function of x is f(x) = √(1 - x^2/4).