# Time derivative of rotating vector

Wardub
Trying to teach myself physics and I've run into a problem I don't quite understand.

"The magnitude of dA/dt can be found by the following geometrical argument. The change of A in the time interval t to Δt is"

ΔA = A(t + Δt) - A(t)

And then somehow it gets to
A| = 2Asin(Δ$\theta$/2)

I just can't see how those are equal. Especially where theta/2 comes in.
This comes out of Kleppner pages 25 and 26, if that helps. Supposedly there's a sketch, but I can't see one.

$$|\Delta A| =A \sqrt{(1-\cos(\theta)^2 + \sin(\theta)^2} =A \sqrt{1-2\cos(\theta) + \cos(\theta)^2 + \sin(\theta)^2} =A \sqrt{2-2\cos(\theta)} = 2A \sin(\frac{\theta}{2})$$
$$\sin(\frac{\theta}{2}) = \sqrt{\frac{1-\cos(\theta)}{2}}$$