# Is arcsin sin rotated by 90°?

1. Jun 25, 2010

### zxh

Sry, noob, but i didn't find this anywhere.

2. Jun 25, 2010

### Martin Rattigan

3. Jun 25, 2010

### Staff: Mentor

The two functions are inverses of one another. In general, the graphs of y = f(x) and x = f-1(y) are identical, but the graphs of y = f(x) and y = f-1(x) are reflections of each other in the line y = x.

The situation is a little more complicated with y = sin(x) and y = arcsin(x) = sin-1(x) since the graph of the sine function isn't one-to-one (making the inverse not a function). The usual way around this is to restrict the domain of the sine function, defining y = Sin(x) = sin(x), with x restricted to the interval -pi/2 <= x <= pi/2.

4. Jun 25, 2010

### zxh

So the title would be true for -sin, viewed as a curve?

5. Jun 25, 2010

### Staff: Mentor

Are you asking whether y = arcsin(x) is the rotation by 90 deg of y = -sin(x)? If that's the question, then no.

If that isn't the question, then what are you asking?

6. Jun 25, 2010

### Martin Rattigan

No it's only a segment of the curve, but the graph of y=arcsin(x) would fit over y=-sin(x) if rotated 90° either way about the origin.

Last edited: Jun 25, 2010
7. Jun 25, 2010

### zxh

Thanks, that's what i was looking for. I'm not too concerned about range definitions.
I came to this looking for a trig definition of a (half) circle (not the pythagorean Sqrt(r-x^2)).
At first i was wondering why Cos(Sin(x)) (given that the 2 functions for a circle in a parametric plot are sinx and cosx) didn't work but it turns out it's
Cos(ArcSin(x)).

8. Jun 25, 2010

### disregardthat

You can show this algebraically. If you are familiar with complex numbers, this is easy.
1) multiply x+i (-sin x) with e^(i*pi/2) to rotate it by 90 degrees.
2) reflect x+i(-sin x) over the curve y=x to invert it.

9. Jun 25, 2010

### zxh

thanks, good one.