Pairs of trig functions: Pythagorean identities and derivatives

Question

Or, "Why do the six trig functions split up into three pairs of best friends?"

There are three Pythagorean identities:

$$ \begin{align*} \sin^2 x + \cos^2 x &= 1\\ \tan^2 x + 1 &= \sec^2 x \\ \cot^2 x + 1 &= \csc^2 x \end{align*} $$

And the derivatives of our trig functions are:

\begin{array}{c | c} f(x) & f'(x) \\ \hline \sin x & \cos x \\ \tan x & \sec^2 x \\ \sec x & \sec x \tan x \end{array}

(where the derivatives of cofunctions are given by $\frac{d}{dx}[f(\frac{\pi}{2} - x)] = - f'(\frac{\pi}{2} - x)$ via the chain rule, giving e.g. $\frac{d}{dx}[\cot x] = -\csc^2 x$).

I am curious: Is there any reason that sine and cosine, secant and tangent, and cosecant and cotangent like to hang out in those particular pairs so much?

Or, put another way: Given the Pythagorean identities, is it possible to see that the same pairs of trig functions will hang out together when it comes to differentiating, as well? I guess starting from the derivatives of trig functions would be OK too.

Answer 1

If $f$, $g$ satisfy the "Pythagorean" identity $f(x)^2+g(x)^2=1$, then $f\,f'+g\,g'=0$. This gives a relation between $f$, $g$ and its derivatives, but not necessarily like the ones for the trigonometric functions. For instance, if $f(x)=x$ and $g(x)=\sqrt{1-x^2}$, we have $$f'(x)=1,\quad g'(x)=-\frac{x}{\sqrt{1-x^2}}=-\frac{f(x)}{g(x)}.$$