Yes, you can do this without using trigonometric functions. As David points out, you can parametrise all points on the unit circle as
$$\left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$$
Given a starting point $(x,y)$ on the unit circle, a point $(x',y')$ on the unit circle that is $1^{\circ}$ away from it satisfies $(x,y).(x',y') = \cos(1^{\circ}) \approx 0.9998477$. So if
$$(x',y') = \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$$
we get
$$\frac{1-t^2}{1+t^2}x + \frac{2t}{1+t^2}y = \cos(1^{\circ})$$
Now, $x$ and $y$ (and $\cos(1^{\circ})$) are known, so we can multiply through by $1+t^2$ to get a quadratic equation in $t$. The two solutions represent the two points $1^{\circ}$either side of $(x,y)$.
I seem to have used a trigonometric function here, but $\cos(1^{\circ})$ is really just a constant...