Let e be an infinitesimal.
We write st(x) for the function dropping the infinitesimal part of a number.
Then f'(x) = st(1/e (f(x+e)-f(x)))
Now use the angle sum identity and cos(e) = 1 - e^2, sin(e) = e. I don't know how to justify these values other than the power series identities for sin and cos...
So st(1/e (sin(x+e)-sin(x))) = st(1/e (sin(x)cos(e)+cos(x)sin(e)-sin(x)) = st(1/e (sin(x)(1-e^2)+cos(x)e - sin(x))) = st(1/e (sin(x) - sin(x) - e^2 sin(x) + cos(x)e)) = st(e sin(x) + cos(x)) = cos(x)
Isn't that from the definition of cos and sin, even geometrically?
1 - x^2 < cos(x)^2 < 1 / (1 + x^2)
from ordinary trigonometry, although you can't just do this with nilsquare infinitesimals; you need a more sophisticated setup.