How to interpret coefficients of angular terms in a regression model?

Question

Given the model $$Y = a + \beta_1 x_1 + \beta_2 x_2 + \beta_3 \cos(\angle) + \beta_4 \sin(\angle) + \text{ errors}$$ resulting from a negative binomial regression, how can I explain the $\beta_3 \cos((\angle) + \beta_4 \sin(\angle)$ terms given that they are the two predictors emerging from the angle of inclination $\angle$ (a variable measured in the raw dataset)?

Answer 1

The question is a good one: when including an angular variable $w$ among the explanatory variables in any regression model by incorporating its sine and cosine, how does one interpret the two estimated coefficients?

Let's write this mathematically just to make things perfectly clear. Suppose you model a response $Y$ in the form

$$Y = X\beta + \alpha_1 \cos(w) + \alpha_2 \sin(w) + \text{ random error}$$

and (somehow) you estimate the parameters as $b, a_1, a_2$ to obtain the fit

$$\hat Y(X,w) = Xb + a_1 \cos(w) + a_2\sin(w) + \text{ residuals}.$$

What do the $a_i$ mean when you're interested in the relationship between $w$ and $Y$?

Typically, we express such relationships differentially by stating what happens to $\hat Y$ when (hypothetically) the value of $w$ is changed a tiny bit, say to $w+\delta$ for a small angle $\delta.$ We can work this out trigonometrically. Note that all the change occurs in just the two angular terms of the model ($Xb$ is not involved), so let's focus on how the angular terms differ.

When angles are expressed in radians we may feel free to treat $\sin(\delta)$ as approximately $\delta$ and $\cos(\delta)$ as approximately equal to $1,$ because the relative error made in either instance is proportional to $\delta^2,$ which is truly tiny. Trigonometry tells us that

$$\cos(w+\delta) - \cos(w) = \cos(w)\cos(\delta) - \sin(w)\sin(\delta) - \cos(w) \approx - \sin(w)\delta$$

and

$$\sin(w+\delta) - \sin(w)=\sin(w)\cos(\delta) + \cos(w)\sin(\delta) - \sin(w) \approx \cos(w)\delta.$$

Consequently the difference in the fit induced by changing the angle $w$ by a small amount of $\delta$ radians is $$\hat Y(X,w+\delta) - \hat Y(X,w) \approx (-a_1\sin(w) + a_2\cos(w))\,\delta.$$

To interpret this further, it's handy to re-express the $a_i$ in polar coordinates where $a_1 = r\cos(\phi)$ and $a_2=r\sin(\phi)$ for some non-negative number $r$ and angle $\phi.$ This can always be done by taking $r = \sqrt{a_1^2+a_2^2}$ and then computing $\phi$ as the arctangent of the vector $(a_1, a_2)$ or, equivalently, as the argument of the complex number $a_1 + a_2\mathbf{i}.$ The change in response can now be expressed as

$$\hat Y(X,w+\delta) - \hat Y(X,w) \approx (-a_1\sin(w) + a_2\cos(w))\ \delta = -\sin(w-\phi)\, r \ \delta.$$

(When measuring in degrees, multiply $\delta$ by $180/\pi \approx 57.3.$)

If that sine term weren't there, we would be looking at a traditional expression: the change $r\delta$ in $\hat Y$ is $r$ times the change in $w.$ This is modified by the sine term, though, but it's not difficult to understand: it is zero when $w=\phi,$ increases to $1$ when $w$ is 90 degrees clockwise of $\phi$ ($w=\phi+\pi/2$), returns to $0$ when $w$ is directly opposite $\phi$ ($w=\phi+\pi$), decreases to $-1$ when $w$ is 90 degrees counterclockwise of $\phi$ ($w=\phi + 3\pi/2$), and finally returns to $0.$

Thus, $\phi$ is a kind of origin of the angular effect: when $w=\pm \phi,$ a little change in $w$ doesn't appreciably change the response, whereas otherwise the change depends on how far away from the origin $w$ is, swinging back and forth between $-r$ and $r.$

We can summarize this simply:

A small change in the angle $w$ changes the fit $\hat Y$ by an amount between $-r$ and $r,$ where $r^2 = a_1^2 + a_2^2$ and the change is measured in radians. That amount depends on where $w$ is situated relative to the angle made by the vector $(a_1, a_2)$ in the plane counterclockwise of the vector $(1,0)$ (that is, "north of east" in geographers' terms).