cos^-1(cos15pi/6)

Not much calculus here, but here goes:

by definition
cos(cos^-1(x)) = x
cos^-1(cos(x)) = x

for x in suitable ranges. Now, for cos^-1(x) the function takes on principal values between 0 and π.

So, cos^-1(cos(15π/6)) = cos^-1(cos 5π/2) = cos^-1(0) = π/2.

To find the value of cos^-1(cos(15π/6)), we need to understand the inverse cosine function and apply the concept.

The inverse cosine function, denoted as cos^⁻¹(x) or arccos(x), returns the angle whose cosine is equal to x. In other words, if cos(θ) = x, then cos^⁻¹(x) = θ.

In this case, we need to find the angle θ whose cosine is equal to the value cos(15π/6).

To simplify the angle, we need to convert it from radians to degrees. Since π radians is equal to 180 degrees, we can rewrite 15π/6 as (15/6) * π = 2.5π.

So, we are looking for the angle θ whose cosine is equal to cos(2.5π).

To find this angle, follow these steps:
1. Determine the reference angle: Since cosine is positive in the first and fourth quadrants, we need to find the positive angle that has the same cosine value. So, the reference angle is 2.5π - 2π = 0.5π.

2. Determine the quadrants: The angle 2.5π lies in the second and third quadrants because π/2 < 2.5π < 3π/2.

3. Evaluate the angle:
- In the second quadrant, the cosine value is negative. Hence, cos^⁻¹(cos(2.5π)) = -0.5π.
- In the third quadrant, the cosine value is also negative. Hence, cos^⁻¹(cos(2.5π)) = -0.5π.

So, cos^⁻¹(cos(15π/6)) can be written as either -0.5π or -180 degrees.