Sin(2sec^-1 u/9)

Answer 1

I'll use arcsec(x) as sec^-1(x) and arccos(x) = cos^-1(x) to avoid confusion.

Using the double angle identity: sin2x = 2sin(x)cos(x) we can convert this to:

2sin(arcsec(u/9))cos(arcsec(u/9)) but arcsec(u/9) = arccos(9/u) substituting, we get:

2sin(arccos(9/u))cos(arccos(9/u))

Let θ = arccos(9/u) then it becomes 2sinθcosθ and the resulting triangle would look like this:

Using the Pythagorean Theorem we solve for the other side to get:

Then sinθ = √(u² - 81)/u and cosθ = 9/u so plugging that in to the 2sinθcosθ it becomes:

2√(u² - 81)/u * 9/u or 18√(u² - 81)/u² or: