What is an algebraic expression for cot(arccos ((x+1)/2)?
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
To find the algebraic expression for cot(arccos ((x+1)/2)), we can start by breaking it down into steps.
Step 1: Let's find arccos ((x+1)/2).
To do that, we first need to understand that arccos (or inverse cosine) returns an angle whose cosine is equal to the given value. In this case, the given value is ((x+1)/2).
So, we have cos(arccos ((x+1)/2)) = ((x+1)/2).
Step 2: Now that we have cos(arccos ((x+1)/2)) = ((x+1)/2), we can find cot(arccos ((x+1)/2)).
The cotangent function is defined as the reciprocal of the tangent function, which is equal to the ratio of the adjacent side to the opposite side in a right triangle. So, cot(arccos ((x+1)/2)) will be equal to the adjacent side divided by the opposite side in the right triangle with the angle arccos ((x+1)/2).
However, finding the exact values of the adjacent and opposite sides is not straightforward because we have an algebraic expression as an argument.
Step 3: To simplify the expression further, we can use trigonometric identities.
By using the identity cos^2θ + sin^2θ = 1, we can represent sin^2θ in terms of cos^2θ.
Let's solve for sin^2θ:
sin^2θ = 1 - cos^2θ
Now, substitute θ with arccos ((x+1)/2):
sin^2(arccos ((x+1)/2)) = 1 - cos^2(arccos ((x+1)/2))
Since sin^2(arccos ((x+1)/2)) = sin(arccos ((x+1)/2))^2, we have:
(sin(arccos ((x+1)/2)))^2 = 1 - cos^2(arccos ((x+1)/2))
Since sin(arccos ((x+1)/2)) is positive in the first and second quadrants, we can take the positive square root:
sin(arccos ((x+1)/2)) = sqrt(1 - cos^2(arccos ((x+1)/2)))
sin(arccos ((x+1)/2)) = sqrt(1 - ((x+1)/2)^2)
Now we have the simplified expression for sin(arccos ((x+1)/2)).
Step 4: Finally, we can find the expression for cot(arccos ((x+1)/2)) using the formula cotθ = cosθ / sinθ:
cot(arccos ((x+1)/2)) = cos(arccos ((x+1)/2)) / sin(arccos ((x+1)/2))
cot(arccos ((x+1)/2)) = ((x+1)/2) / sqrt(1 - ((x+1)/2)^2)
Therefore, the algebraic expression for cot(arccos ((x+1)/2)) is ((x+1)/2) / sqrt(1 - ((x+1)/2)^2).