Equivalence of Definitions of Sine and Cosine

Theorem
In the following, $\theta$ understood to take values in $\hointr 0 {2 \pi}$.

Proof
Consider the following vector-valued function $\mathbf f : \R \to {\closedint {-1} 1}^2$:
 * $\map {\mathbf f} t = \tuple {\cos t, \sin t}$

where $\cos t$ and $\sin t$ are defined analytically.

Then, for any $t$ the distance to the origin is:

Therefore, $\map {\mathbf f} t$ always lies on the unit circle.

For arbitrary $\theta \in \hointr 0 {2 \pi}$, the arc length on $\closedint 0 \theta$ is:

Thus, by the definition of radians, the angle made by $\map {\mathbf f} \theta$ with the $x$-axis is $\theta$.

A straight line segment from $\tuple {0, \sin \theta}$ to $\tuple {\cos \theta, \sin \theta}$ is perpendicular to the $y$-axis.

Its length is $\size {\cos \theta}$, and it is on the side of the $y$-axis corresponding to the sign of $\cos \theta$.

But that is the unit circle definition for cosine.

Similarly, a straight line segment from $\tuple {\cos \theta, 0}$ to $\tuple {\cos \theta, \sin \theta}$ is perpendicular to the $x$-axis, with length $\size {\sin \theta}$ and appropriate direction.

Again, that is the unit circle definition for sine.