I literally saw the sin one yesterday, so I'll rewrite it!!

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)

"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..."

Isn't that from the definition of cos and sin, even geometrically?

The version of rigorous infinitesimals I've seen didn't let you have these identities, because e^2 != 0 (because it still maintains the field axiom xy = 0 => x=0 V y=0); but the argument still goes through by carrying the whole power series in place of these simplifications – you can write the higher order terms as e*f(x, e) and drop them when you evaluate st(...).

You can get the relationship:

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.