How many ways are there to define sine and cosine?

I like to define them by the differential equation:

$$y''=-y$$

and then, choosing the initial conditions we get these two functions. This also give rise nice definition for $\pi$ being the fundamental period for the solutions of these equations.

(This is by no means a comprehensive list.)

Right-triangle definition: the sine (cosine) of an acute angle is the ratio of the lengths of the leg opposite (adjacent to) the given angle to the hypotenuse of the triangle.

Bizarre Geometric Definition: the cosine of an angle in a triangle is $\frac{a^2+b^2-c^2}{2ab}$, where $a$ and $b$ are the lengths of the sides adjacent to the angle and $c$ is the length of the side opposite the angle; the sine of an acute angle is the cosine of its complement; the sine of an obtuse angle is the cosine of its supplement's complement. (edit: this one might be even worse than I'd originally thought as a definition, so perhaps just ignore it.)

Rotation-transformation definition: the sine (cosine) of a magnitude of rotation is the vertical (horizontal) coordinate of the image of the point (1,0) under a rotation of the given magnitude centered at the origin.

Unit circle definition: the sine (cosine) of a directed angle with vertex at the origin and initial ray on the positive x-axis is the y-coordinate (x-coordinate) of the point of intersection of the terminal ray of the angle with the unit circle centered at the origin.

Power series definition: $$\begin{align} \sin x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \\\\ \cos x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \end{align}$$

Exponential definition: $$\sin x=\frac{e^{ix}-e^{-ix}}{2i};\quad\quad\cos x=\frac{e^{ix}+e^{-ix}}{2}$$

An interesting construction is given by Michael Spivak in his book Calculus, chapter 15. The steps are basically the following:

$1.$ We define what a directed angle is. $2.$ We define a unit circle by $x^2+y^2=1$, and show that every angle between the $x$-axis and a line origined from $(0,0)$ defines a point $(x,y)$ in that circle.

$3.$ We define $x = \cos \theta$ and $y = \sin \theta$.

$4.$ We note that the area of the circular sector is always $x/2$, so maybe we can define this functions explicitly with this fact:

$5.$ We define $\pi$ as the area of the unit circle, this is:

$$\pi = 2 \int_{-1}^1 \sqrt{1-x^2} dx$$

$6.$ We give an explicit formula for the area of the circular sector, namely:

$$A(x) = \frac{x\sqrt{1-x^2}}{2}+\int_x^1 \sqrt{1-t^2}dx$$

and show that it is continuous, and takes all values from $0$ to $\pi/2$. We may also plot it, since we can show that $2A(x)$ is actually the inverse of $\cos x$.

$7.$ We define $\cos x$ as the only number in $[-1,1]$ such that

$$A(\cos x) = \frac{x}{2}$$

and thus define

$$\sin x = \sqrt{1-\cos^2x}$$

$8.$ We show that for $0<x<\pi$

$$\cos(x)' = - \sin(x)$$

$$\sin(x)' = \cos(x)$$

$9.$ We finally extend the functions to all real values by showing that for $\pi \leq x \leq 2\pi$,

$$-\sin(2\pi-x) = \sin x$$

$$\cos(2\pi-x) = \cos x$$

and then that

$$\cos(2\pi k+x) = \cos x$$

$$\sin(2\pi k+x) = \sin x$$