# Solid angle relation between sinθ dϕdθ and d(cos(θ))dϕ

I am a bit confused with regards to the concept of solid angle.

Why is the solid angle which is defined as $\sin \theta {\rm d}\phi\, {d\rm }\theta$ equal to $\sin\theta\,{\rm d}\theta {\rm d}\phi = {\rm d}\cos\theta{\rm d}\phi$

The identity

$$\sin\theta\,{\rm d}\theta {\rm d}\phi = {\rm d}\cos\theta{\rm d}{\phi}$$

comes from the fact that

$${\rm d}\cos\theta = \frac{{\rm d}\cos\theta}{{\rm d}\theta}{\rm d}\theta = -\sin \theta {\rm d}\theta$$

• but that doesn't account for the negative sign which is what bothered me about that ! – David Abraham Apr 27 '18 at 1:03
• @DavidAbraham Remember that the integral goes accompanied by the absolute value of the Jacobbian, so the minus sign disappears – caverac Apr 27 '18 at 1:14

use the following trick $$\frac {d(cos(θ))}{dt}=-\sin(\theta) \frac {d \theta}{dt}$$ multiply by $dt$ $${d(cos(θ))}=-\sin(\theta){d \theta}$$