Is it possible to deduce the derivatives of sine and cosine without geometry and “cheating?”

alternative-proofderivativesfunctional-equationsfunctionstrigonometry

I am asking if it's possible to find the derivatives without geometry, circular thinking, and definitions that don't make sense to introduce based on the intuitive meaning of sine and cosine (i.e that assume you already know the answer.)

Note: I take the Pythagorean identity, the zeroth and first derivatives at zero, and that sine is odd and cosine is even to be used as the definitions for sine and cosine.

Edit: I will also assume they are bounded by $1$ and $-1$ and never constant on a continuous interval.

I tried using the Pythagorean theorem and that sine is odd and cosine is even and reached that $$\cos\theta+i\sin\theta=e^{g(\theta)}$$ where $g$ is an odd function. However, I couldn't prove that $g''=0$ (equivilant to $g''$ is even) which we can use (along with $\sin'(0)=1$ and $\cos'(0)=0$) to deduce $g=i\theta$.

So, I thought maybe it's much easier. If we differentiate the Pythagorean identity, we get:
$$\sin\theta\sin'\theta=-\cos\theta\cos'\theta\tag{$\star$}\label{1}$$
Squaring both sides to avoid multiple cases for the signs, we get:
$$(\sin\theta\sin'\theta)^2=(\cos\theta\cos'\theta)^2$$
What are the possibilities here?

We certainly know $\sin^2\theta≠\cos^2\theta$ and $(\sin'\theta)^2≠(\cos\theta\cos'\theta)^2$, so it seems that the only possibility is:
$$(\sin'\theta)^2=(\cos\theta)^2\\
(\cos'\theta)^2=(\sin\theta)^2$$
My question is: Is this logic true, can we prove such a thing? If so, we can proceed as follows, differentiate the equations:
$$\sin'\theta\sin''\theta=\cos\theta\cos'\theta\\
\cos'\theta\cos''\theta=\sin\theta\sin'\theta$$
Now, substitute from $\eqref{1}$:
$$\sin''\theta=-\sin\theta\\
\cos''\theta=-\cos\theta$$
Afterwards, we can use ansatz $e^{\lambda\theta}$ and the zeroth and first derivatives at $0$ to solve for the exponential forms of sine and cosine. As a result, we can deduce the derivatives and the angle-sum formulas too.

Best Answer

Note: I take the Pythagorean identity, the zeroth and first derivatives at zero, and that sine is odd and cosine is even to be used as the definitions for sine and cosine.

This is not enough to uniquely specify sine and cosine. These conditions are satisfied, for example, by the real and imaginary parts of $e^{i (\theta + \theta^3)}$, namely $c(\theta) = \cos(\theta + \theta^3)$ and $s(\theta) = \sin(\theta + \theta^3)$.

However, the following conditions are similar to yours and do uniquely specify sine and cosine: they are the unique pair of differentiable functions $c(\theta), s(\theta)$ satisfying

  • $c(0) = 1, s(0) = 0$,
  • $c'(0) = 0, s'(0) = 1$,
  • $c(\theta)^2 + s(\theta)^2 = 1$, and
  • $c'(\theta)^2 + s'(\theta)^2 = 1$.

These conditions say that $\theta \mapsto (c(\theta), s(\theta))$ is a unit speed parameterization of the unit circle starting at $(1, 0)$ and moving counter-clockwise. Writing $v(\theta) = (c(\theta), s(\theta))$, differentiating the third condition $\| v \|^2 = 1$ gives $\langle v, v' \rangle = 0$, which together with the fourth condition $\| v' \|^2 = 1$ gives that $v'$ always differs from $v$ by a $90^{\circ}$ rotation (which the initial conditions imply is counter-clockwise). This gives $c'(\theta) = s(\theta), s'(\theta) = - c(\theta)$ as desired. From here we can get to Euler's formula, etc.

Of course one still has to prove that such a pair of functions exists.

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