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

Question

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.

Answer 1

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.

Answer 2

There are Multiple Issues here :

(1) When you take these Criteria :
"(1A) Pythagorean identity, (1B) the zeroth at zero (1C) first derivatives at zero, & (1D) sine is odd & cosine is even"
Then there are Multiple Solutions which will work & we will not get $\sin$ & $\cos$.

We might take the Example of $\sin{\theta^{3}}$ & $\cos{\theta^{3}}$ which also will satisfy all the Criteria.
Solution is not unique.
[[ Correction based on Comment by user "Gerry Myerson" ]]

(2) When you say "We certainly know [squares are not Equal]" how can we say that without "cheating" or without knowing geometry + trigonometry ?
[[ We know this only @ $\theta=0$ , but it might be equal elsewhere ]]

(3) When you say "We certainly know [Derivative with squares are not Equal]" how can we say that without "cheating" or without knowing Derivative in advance ?
[[ We know this only @ $\theta=0$ , but it might be equal elsewhere ]]

(4) When we Square it to avoid negative sign, we will get more solutions into the picture.
[[ OP has mentioned that we can go back to original Equation to eliminate the unwanted solutions , which is correct & hence this Point (4) is then not valid ]]

(5) When we have $(AB)^{2}=(CD)^{2}$ & somehow know that $(A)^{2}$ is not $(C)^{2}$ & $(B)^{2}$ is not $(CD)^{2}$ , we can not conclude that $(A)^{2}$ is $(D)^{2}$ & $(B)^{2}$ is $(C)^{2}$

We can take the Example of the Integer Case where $(A,B,C,D)$ is $(2,4,1,8)$
[[ In Current Case, OP wants to show that what we have is actually $(AB)^{2}=(BA)^{2}$ which requires more analysis ]]

In short, this current approach specifically will not work & we have to use geometry + trigonometry to get the Derivative.
More-over, there can be no theoretical approach in general which can avoid geometry when using this Assumption or Definition : Sine & Cosine are the ratios of the right angle triangle.
That is because , no matter what calculations we use , we have to eventually show that it is equal to the ratios of the right angle triangle.

Proof By Contradiction :
Assume that we can deduce Sine & Cosine without geometry. Assume that we can then show that it is Exactly Equal to trigonometry Sine & Cosine.
Then this means we have shown the validity of Euclidean Geometry without using geometry , which means Parallel Postulate is absolutely true & all other geometries (Riemann & Lobachevski) are not true.
Naturally, we have to go back on our Assumption & conclude that Sine & Cosine are inherently geometric & are Euclidean Geometry Products.

This Answer is contrary to the Accepted Answer [[ which I have upvoted , seeing the interesting idea & the nice approach ]] which is actually using geometry in Disguise even though OP wants to avoid that.