# A chain rule for the angle

My question is fairly basic, namely can I do the substitution below? $$\frac{\mathrm{d}\theta(t)}{\mathrm{d}t} = -\frac{1}{\sin(\theta(t))}\frac{\mathrm{d}\cos(\theta(t))}{\mathrm{d}t}$$ If not, what corrections should I take into account? Since I am dealing with trigonometric functions there might be a sign-rule somwhere the I am missing.

Background information to the question: In reality I want to calculate the gradient of the angle $$\theta$$ between two vectors $$\vec{a}$$ and $$\vec{b}$$, with respect to the coordinates to one of the vectors. I use the chain rule because then I can use the relationship:$$\cos(\theta) = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|}$$ which allows me to easily calculate this derivative analytically. However, when I compare my result to a fully numerical calculation the results are not generally the same.

More details

In full I am looking for the gradient of a function $$f(d,r,\theta)$$, where the values of $$d$$, $$r$$ and $$\theta$$ are determined based on three vectors $$\vec{x}$$, $$\vec{y}$$ and $$\vec{z}$$. The values are given by: $$d=|\vec{x}-\vec{y}|$$, $$r = |\vec{z}-\vec{y}|$$ and $$\theta$$ by the angle between the vectors $$\vec{a} = \vec{x}-\vec{y}$$ and $$\vec{b}=\vec{z}-\vec{y}$$. The gradient I am looking at is with respect to the $$\vec{y}$$ vector, resulting in: $$\nabla_{\vec{y}}f = -\frac{\partial f}{\partial d}\frac{\vec{x}-\vec{y}}{|\vec{x}-\vec{y}|} -\frac{\partial f}{\partial r}\frac{\vec{z}-\vec{y}}{|\vec{z}-\vec{y}|} + \frac{\partial f}{\partial \theta}\nabla_{\vec{y}}\theta,$$ where $$\theta$$ is given by the formula above.

If I calculate the gradient using the direct derivative (original question) or fully numerical I seem to have different results.

• I think you might have to give more detail of what you are doing to get a useful answer. Oct 19, 2018 at 9:38
• @Mark Bennet, better?
– Nick
Oct 19, 2018 at 12:31

Yes, $$\frac{d cos(\theta(t))}{dt}= -sin(\theta)\frac{d\theta}{dt}$$ so, as long as $$sin(\theta(t))\ne 0$$, $$\frac{d\theta}{dt}= -\frac{1}{sin(\theta(t))}\frac{d cos(\theta(t))}{dt}$$
• Thanks, That was exactly what I was thinking as well. For the special cases of $\theta$ equal to 0 and $\pi$ i have exception rules, so that should not be an issue.