Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For the sake of this proof, the trigonometric functions $\cos$ and $\sin$ are defined as the coordinates of a point on the unit circle, rather than any of the modern analytic definitions.

Let $\vec x(\theta)$ be a function giving the position of a point on the unit circle as a function of the angle $\theta$, that is, $\vec x(\theta) = \left( \begin{array}{c} \cos(\theta)\\ \sin(\theta)\\ \end{array} \right)$. Decompose $\vec x'(\theta)$ into a colinear and an orthogonal component with $\vec x(\theta)$. If the colinear component is non-null, then on the assumption that $\vec x'(\theta)$ is continuous (could this be demonstrated?), it would be non-null over some small interval (right?), and then by taking the integral over that interval, we would have that $\vec x(\theta)$ either moves away from or towards the center of the origin, which is contrary to our hypotheses. Thus the colinear component is always null, and $\vec x'(\theta)$ is always tangent to the circle.

But for any vector $\left( \begin{array}{c} a\\ b\\ \end{array} \right)$, a vector is orthogonal to it if and only if it is colinear with $\left( \begin{array}{c} -b\\ a\\ \end{array} \right)$. Therefore $\vec x'(\theta) = k\left( \begin{array}{c} -\sin(\theta)\\ \cos(\theta)\\ \end{array} \right)$ and we have $\cos' = -k\sin$, $\sin' = k\cos$, for some $k$.

There are gaps in this proof (in italics), but are there ways of filling them in while remaining within the geometric definitions of the trigonometric functions? It would be pleasing to be able to prove the derivatives of the trig functions purely on the basis of the fairly obvious geometrical nature of circular motion.

share|cite|improve this question
A rather difficult and subtle point with this kind of definition of trig functions is to define angles. – Baby Dragon Apr 2 '13 at 20:56
@BabyDragon length of arc divided by length of arm? Although I realize the integrals used to compute those might themselves require the trig functions... – Jack M Apr 2 '13 at 21:07
@JackM you don't run into the risk of circular definitions since the part of the unit circle in the upper half plane can be given as the graph of the function $[-1,1] \owns x \mapsto \sqrt{1-x^2} \in \mathbb R$. You don't actually need to compute the integrals to get a definition of $\cos$ and $\sin$ (that would indeed be circular), you just need to prove convergence. But compared to the power series definition this becomes very very ugly if one tries to do all the details. In fact giving a proper definition of arc length without mentioning $d$-dimensional Hausdorff measure is quite tricky. – kahen Apr 5 '13 at 15:56

A hint: In order to determine $k$ use the fact that $|\vec x(\theta)|^2\equiv1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.