Polar Coordinates: Arc length of two overlapping curves

[Badgerspin]

This question may be something of a dumb one. I feel I should know this, but well, I don't.

I'm being asked to find the perimeter inside of the curve r=15sin(theta) and outside of r = 1

Setting up the equation I can do. If it were just an indefinite integral, this would be cake. My challenge right now is finding the angle in which to compute the problem. From where to where? Let x = theta

15sin(x) = 1
Sin(x) = 1/15

For what value of theta would I get 1/15? I can get the numeric value by taking the arcsin, but I need to be able to show it in the format (pi/#, or perhaps ((#pi)/#).

While I'm on that note, for future reference, is there any easy way to compute something like this where I'm being asked oddball angles?

 

[fzero]

For what value of theta would I get 1/15? I can get the numeric value by taking the arcsin, but I need to be able to show it in the format (pi/#, or perhaps ((#pi)/#).

While I'm on that note, for future reference, is there any easy way to compute something like this where I'm being asked oddball angles?

In general you cannot express \sin^{-1}x in terms of rational multiples of \pi. You can't express \sin \phi in elementary terms either for general \phi either. That why these functions are referred to by the term transcendental.

In this particular problem, it looks like the appearances of \sin^{-1}(1/15) cancel out in the final result, so you never have to worry about it.