# 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).