This site is being phased out.
Trigonometric functions
The radian measure is usually introduced in a trigonometry course. It is often overlooked that this definition relies on the notion of $\pi$, which is a number defined as a limit (from a list of many). But limits only appear for the first time in a calculus course that comes after trigonometry! A vicious cycle? This is how we can build the radian measure from scratch.
We first develop calculus, without trig functions. We then create trigonometry for all possible measures of angles, as follows. Let's place a point $p$ arbitrarily on the unit circle; we call this angle $1$. We now create a function $g_p$ from the reals to the circle starting with this: $g_p(0)=(1,0)$ and $g_p(1)=p$. We expand this function
- by additivity, to the integers; then,
- by bisecting the angles, to all numbers that are expressible as fractions base $2$; and finally,
- by using limits, to the rest of the angles.
We then follow the usual method -- projection $P$ of the points on the circle to the axes -- to define trig functions from the reals to the reals as compositions: $$\sin_p (x)=P g_p(x),\ \cos_p (x)=P g_p(x),$$ These functions have been constructed for each $p$. These functions are proven, just as always, to be continuous and then differentiable (the derivatives are the same with some extra coefficients dependent on $p$ coming from the Chain Rule). The special limits at $x=0$, and then the derivatives, are also worked out with $p$ again present. Finally, we pick $p$ such that $(\sin_p(x))'=1$ at $x=0$. This is similar to how base $e$ is picked from all the bases of all exponential functions...
A direct route is as follows. Start with: $$\sin \pi=0,\ \cos \pi =-1$$ and then use the average identity to produce more and more values. The identity comes from the tangent of the average and sine and cosine expressed via tangents: $$\begin{array}{ll} \tan\left( \frac{\alpha+\beta}{2} \right) = \frac{\sin\alpha + \sin\beta}{\cos\alpha + \cos\beta}= -\,\frac{\cos\alpha - \cos\beta}{\sin\alpha - \sin\beta},\\ \sin \theta =\pm\frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}},\\ \cos \theta =\pm\frac{1}{\sqrt{1 + \tan^2 \theta}}. \end{array}$$ Then take a limit.