TITLE: Using the IVP definition of $\cos$ and $\sin$, how can we show that $\cos^2(x)+\sin^2(x) = 1$ without any "magic"?
QUESTION [5 upvotes]: One way to define the exponential, hyperbolic and circular functions is to assert that they're the unique solutions to certain IVP systems:
The exponential function:
$$\exp'(x) = \exp(x), \qquad \exp(0) = 1$$
Hyperbolic functions:
$$\cosh'(x) = \sinh(x)$$
$$\sinh'(x) = \cosh(x)$$
$$\cosh(0) = 1$$
$$\sinh(0) = 0$$
The circular functions:
$$\cos'(x) = -\sin(x)$$
$$\sin'(x) = \cos(x)$$
$$\cos(0) = 1$$
$$\sin(0) = 0$$

Question. Using the IVP definition of $\cos$ and $\sin$, how can we show that $\cos^2(x)+\sin^2(x) = 1$?
I'm looking for a satisfying proof that I can present to young people, so:

no complex numbers, and
no power series, and, more generally
no "magic"!

In fact, the closer the proof is to being "just plain algebra", the better.

Motivation.
There's a geometric way of defining the circular functions: first, position yourself at $(1,0)$ on the unit circle. Then start walking anticlockwise at unit speed. It follows that if the time elapsed is $t$, your $x$-coordinate will be $\cos t$ and your $y$-coordinate will be $\sin t$.
The above information can be organized into four statements:

We walk on the unit circle centered at the origin.
We position ourselves at $(1,0)$ to begin with.
We walk at unit speed.
We walk anticlockwise.

These can be formalized as follows:

$\cos^2 t + \sin^2 t = 1$
$\cos(0) = 1, \sin(0) = 0$
$(\cos't)^2+(\sin't)^2 = 1$
$\sin'(0) > 0$

Now conditions 2 and 4 follow easily from the IVP definition. Furthermore, if we can show 1, then 3 follows easily. But, I haven't been able to show 1 in a "non-magic way" e.g. without complex numbers.

REPLY [5 votes]: Since $\frac{d}{dx}\sin(x)=\cos(x)$ and $\frac{d}{dx}\cos(x)=-\sin(x)$ we have
$$ \frac{d}{dx}\left(\sin^2(x)+\cos^2(x)\right) = \sin(x)\cos(x)-\sin(x)\cos(x) = 0 $$
hence $\sin^2(x)+\cos^2(x)$ is a constant function, equal to $\sin^2(0)+\cos^2(0)=1$.
This easily gives 1.,2. and 3. and 4..