Differentiating Sine Without Doing Any Work or Knowing Anything

If you do a google search for how to derive the facts that \sin'(x)=\cos(x) and \cos'(x)=-\sin(x), most of the derivations you’ll find rely on knowing something like the double angle formula and go through a direct, non-trivial evaluation of \lim_{\theta \to 0}\sin(\theta)/\theta.

It’s actually possible to compute these derivatives this without really remembering any specific facts about trigonometry (besides the fact that x^2+y^2= 1 for points (x,y) on a circle) or computing any limit directly. I think this deserves to be better known, so I thought I’d record it in a blog post.

I’ll state the theorem generically, without specifically referencing \sin or \cos, and using t instead of \theta to try to avoid allowing any hidden assumptions about angles to creep in.

Theorem: Suppose s and c are differentiable functions satisfying:

  • s(t)^2 + c(t)^2 = 1 for all t
  • The arclength of the path \gamma(t) = (c(t), s(t)) from t_0 to t_1 is t_1 - t_0 for any t_0 \leq t_1

Then one of the following two conditions holds:

  • s'(t)=c(t) and c'(t)=-s(t) for all t
  • s'(t)=-c(t) and c'(t)=s(t) for all t

Proof: We’ll use the two conditions to get two equations in our two unknowns and then solve. Differentiating s(t)^2 + c(t)^2 = 1 with respect to t (and dividing by 2) gives s(t)s'(t) + c(t)c'(t) = 0. That’s Equation 1.

The arclength condition is: \int_{t_0}^{t_1} \sqrt{s'(t)^2 + c'(t)^2}\,dt = t_1 - t_0. Differentiating that with respect to t_1 gives \sqrt{s'(t)^2 + c'(t)^2} = 1, and thus s'(t)^2 + c'(t)^2 = 1. That’s Equation 2.

These are our two equations. Solving for s'(t) in Equation 1 gives s'(t) = -c(t)c'(t)/s(t). Plugging that in to Equation 2 and simplifying (using s(t)^2 + c(t)^2 = 1) gives c'(t)^2 = s(t)^2. Thus, either c'(t) = s(t) or c'(t) = -s(t), which lead to the two possibilities respectively. QED.

From that generic result, s(t) and c(t) can then be characterized completely by assuming the initial conditions that s(0)=0 and c(0)=1 and by choosing s'(t) = c(t) (choosing the other alternative of s'(t) = -c(t) would correspond to going clockwise instead of counterclockwise around the circle).