If you do a google search for how to derive the facts that and , 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 .
It’s actually possible to compute these derivatives this without really remembering any specific facts about trigonometry (besides the fact that for points 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 or , and using instead of to try to avoid allowing any hidden assumptions about angles to creep in.
Theorem: Suppose and are differentiable functions satisfying:
- for all
- The arclength of the path from to is for any
Then one of the following two conditions holds:
- and for all
- and for all
Proof: We’ll use the two conditions to get two equations in our two unknowns and then solve. Differentiating with respect to (and dividing by 2) gives . That’s Equation 1.
The arclength condition is: . Differentiating that with respect to gives , and thus . That’s Equation 2.
These are our two equations. Solving for in Equation 1 gives . Plugging that in to Equation 2 and simplifying (using ) gives . Thus, either or , which lead to the two possibilities respectively. QED.
From that generic result, and can then be characterized completely by assuming the initial conditions that and and by choosing (choosing the other alternative of would correspond to going clockwise instead of counterclockwise around the circle).