Simplify: arcsin (sin x), pi/2 <= x <=pi?

Cool question...it really ensures that you UNDERSTAND the arcsine function!

For the graph of y = sin x, you can put in ANY value of x you want. So the domian is (-inf, inf). You can only get OUT [-1, 1], so that's your range. That's because sine ALWAYS returns a value between -1 and 1, inclusive...you can't go beyond the UNIT of the unit circle. The amplitude of the graph is 1. There's lots of ways to say it, but they all boil down to this: 1.01 = sin x has NO solution in the real number system!

For the inverse graph x = sin y, you're NOT a function anymore! It's a graph of a vertical squiggle. So you have to cut it off. The domain is still [-1, 1], because |x| can't be greater than 1 for x = sin y to have a solution. However, we cut the range: the range is now [-pi/2, pi/2] This is the NEW graph of the FUNCTION x = sin y with those restrictions. We define arcsin x = y to be this graph as well.

So, from pi/2 to pi, your function (sin x) goes DOWN from +1 to 0. It hits ALL The same values from 0 to pi/2, but in the reverse order. In other words, sin x = sin (pi - x) You can see this with drawing the angles x and pi-x on the unit circle: they go UP the same amount when they hit the unit circle.

So if you took the arcsine of this value, you'd get back the FIRST quadrant angle x, not the second quadrant angle pi-x. Thus, your answre is pi-x.