# Finding the center by area of a range of a sine wave

1. Jun 29, 2011

### Puggley

hey.

I used to be quite talented at math, but I've let my talent deteriorate in the ~10 years since my last calculus class.

I have a problem which I'm sure I would have devoured easily back in my school days, but am having trouble with now.

I am considering two points on a normal (0 to 2pi, amplitude of 1) sine wave, p1 and p2. What I want to do is select p3 in between them so that the area of the region between p1 and p3 is equal to the area of the region between p3 and p2.

THANX
Puggley

2. Jun 29, 2011

### micromass

Hi Puggley!

You'll need to find the t that solves

$$\int_{p_1}^t{\sin(x)dx}=\int_t^{p_2}{\sin(x)dx}$$

(Note: I'm calculating the orientated area here, that is: point below the x-axis count as negative area. If you don't want this, you'll need to take the absolute value).

Calculating the integrals gives us

$$[-\cos(x)]_{p_1}^t=[-\cos(x)]_t^{p_2}$$

And thus, you need to find x such that

$$\cos(p_1)-\cos(t)=\cos(t)-\cos(p_2)$$

And thus

$$\cos(t)=\frac{\cos(p_1)+\cos(p_2)}{2}$$

Hence, we take

$$t=arccos\left(\frac{\cos(p_1)+\cos(p_2)}{2}\right)$$

3. Jun 29, 2011

### I like Serena

I'd like to add the second possible solution $2\pi - t$ and choose the one that is between p1 and p2.