Equivalence of Definitions of Sine and Cosine

Theorem
The unit circle and analytic definitions of cosine are equivalent on $\hointr 0 {2\pi}$.

The unit circle and analytic definitions of sine are also equivalent on the same interval.

Proof
Consider the vector-valued function $\map {\mathbf f} t = \paren {\cos t, \sin t}$

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 $\map {\mathbf f} t$ with the $x$-axis is $\theta$.

A line from $\paren {0, \sin \theta}$ to $\paren {\cos \theta, \sin \theta}$ is perpendicular to the $y$-axis, and its length is $\cos \theta$.

But that is the unit circle definition for cosine.

Similarly, a line from $\paren {\cos \theta, 0}$ to $\paren {\cos \theta, \sin \theta}$ is perpendicular to the $x$-axis, with length $\sin \theta$.

Again, that is the unit circle definition for sine.