Trig Derivatives - Can They Be Explained Intuitively but Non-Geometrically?

Question

Thanks for reading.

Every time I look for "intuitive" explanations of why trig derivatives and integrals are what they are, I always get graphics like the following:

However, Sines and Cosines show up in mathematical topics which aren't always geometric by nature...even if perhaps connections to triangles and circles can always be found.

Are there ways to explain, still intuitively (by intuitive I mean explanations that "click," I don't mean hand-wavy...I hope that makes sense), but non-geometrically, why the derivative of sine is cosine (and why the integral of cosine is sine, derivative of cosine is -sine, etc...)?

Thank you!

Answer 1

I always like to consider the graphical forms of the equations. For derivatives especially. For example, the derivative of $\cos{\theta}$ measures the slope of the function $\cos{\theta}$ itself. If you can imagine the graph of $\cos{\theta}$ you can immediately realise that it dips off at the start which implies a negative gradient. So, if the derivative describes a measure of the slope of a function, then if we were to graph the derivative of $\cos{\theta}$ against $\theta$, then the derivative function will have to start at negative values. As you may know, $\sin{\theta}$ and $\cos{\theta}$ are closely related in their derivatives. Normally, the equation of the form $\sin{\theta}$ starts with positive values, however, since the slope of the $\cos{\theta}$ graph is initially negative this means that the derivative of $\cos{\theta}$ must be $-\sin{\theta}$. Sorry that it's alot to take In, but this is what I do when investigating the derivatives of trigonometric functions. It might be worth while getting used to what each function looks like before using this thought process.

Answer 2

It comes down to how you choose to define sine and cosine. There are multiple possible definitions, which are not obviously equivalent, but all give rise to the same functions and therefore can all be used to intuit $\frac{d}{dx} \sin(x) = \cos(x)$ and $\frac{d}{dx} \cos(x) = -\sin(x)$. I'll go through the three most common definitions:

Geometric definition:

If you're working with the geometric definitions for sine and cosine, then the geometric proof for their derivatives is clearly the most intuitive. Whether or not it is intuitive to you is a function of your intuition for geometry and calculus, not the sine and cosine functions specifically.

Differential equations:

As I mentioned in the comments, you can define sine and cosine as solutions to the differential equation $$ f(x) + f''(x) = 0 $$ with the initial value conditions $\sin(0) = 0$, $\sin'(0) = 1$ and $\cos(0) = 1$, $\cos'(0) =0$. If you've got a good intuition for differential equations, it ought to click that this uniquely defines $\sin(x)$ and $\cos(x)$. It might not be clear it's equivalent to the geometric definition. But it's trivially in this case that $\sin'(x) = \cos(x)$ and $\cos'(x) = -\sin(x)$. We observe $\sin'(x)$ solves the same differential equation, because we can differentiate both sides of $\sin(x) + \sin''(x) = 0$ and see that $$ \sin'(x) + \sin'''(x) = 0 $$ and then we also observe that $\sin'(x)$ has the initial values $\sin'(0) = 1$ and $\sin''(0) = -\sin(0) = 0$, so $\sin'(x)$ satisfies the same initial value problem as $\cos(x)$ and hence they are equal. Similarly, you can convince yourself $\cos'(x) = -\sin(x)$.

Complex analysis:

I doubt there are many people for whom this is more intuitive than the other two definitions, but I'll include it for the sake of completeness. For real $x$, we can define sine and cosine to be the real-analytic functions such that Euler's identity $e^{i x} = \cos x + i\sin x$ is true, i.e.:$$ \sin(x) = \Im (e^{i x})\\ \cos(x) = \Re (e^{i x}) $$ Then \begin{eqnarray} \frac{d}{dx} e^{i x} &=& \frac{d}{dx}\left(\cos(x) + i\sin(x)\right)\\ i e^{i x} &=& \cos'(x) + i\sin'(x)\\ i(\cos(x) + i \sin(x)) &=& \cos'(x) + i\sin'(x)\\ -\sin(x) +i\cos(x) &=& \cos'(x) + i \sin'(x) \end{eqnarray} Equating real and imaginary parts yields the desired identity.