What is an algebraic expression for cot(arccos ((x+1)/2)?

Answer 1

Let's visualize what we have here:
recall that cos Ø = adjacent/hypotenuse

and arccos((x+1)/2) mean we have an angle Ø so that
cos Ø = (x+1)/2

so construct a right-angled triangle with base of x+1 and hypotenuse 2 and base angle of Ø
Let the opposite side be y
y^2 + (x+1)^2 = 2^2
y^2 = 4 - x^2 - 2x - 1
y = √(3 - x^2 - 2x)

then cotØ = (x+1)/√(3 - x^2 - 2x)

now let's consider the restriction on x
remember that in cosØ = (x+1)/2
-1 ≤ (x+1)/2 ≤ 1
-2 ≤ x+1 ≤ 2
-3 ≤ x ≤ 1 , but when x = 1, we would be dividing by zero in cotØ = (x1)/√(3 - x^2 - 2x) , so
-3 ≤ x < 1

cot(arccos ((x+1)/2) = (x+1)/√(3 - x^2 - 2x) , -3 ≤ x < 1