I found this geometric reasoning (or proof) of a derivative of a Tangent function in Tristan Needham’s book on Visual Differential Geometry and it struck me as something really intuitive, simple and beautiful.
I wanted to do similar derivations for other simple trigonometric functions, namely, Sine and Cosine of an angle. Although I was able to get the answers, it was not that succint or elegant as the proof of the Tan function.
So here goes…
Consider a triangle inscribed in a circle of radius L, center A, with angle \( \theta \), opposite side of length T and hypoteneuse of length L as shown. We then increment the angle by \(\delta \theta \) and extend side T (BC) by \( \delta T \) so that the ray AP meets the extended side BC at G.
Figure 1: Right angled triangle in a circle of radius L
We then form the triangle \( \Delta CGP \) with opposite side of length \( \delta L \), and the hypoteneuse with length \( \delta T \), with angle \(\angle FCG = \theta\).
As \( \delta \theta \to 0 \), i.e., gets infinitesimally small, angle \( \angle GFC \to 90^{0}\).
We then have, \( \tan \theta = T \), \( \delta S = L\delta \theta \).
Derivative of the tangent (from Needham)
Triangle \( \Delta ABC \) and \( \Delta CFG \) are similar in the limit and also \( FC \to \delta S \).
Hence,
\( \frac{\delta T}{L} = \frac{\delta S}{1} \)
\( \delta T = L\delta S = L.L\delta \theta = L^{2}\delta \theta \)
\( \frac{\delta T}{\delta \theta} = L^{2} = 1 + T^{2} \)
Derivative of the sine
For the sine (as well as the cosine), it didn’t seem so straighforward but still doable.
\begin{equation} \frac{\delta \sin \theta}{\delta \theta} = \lim_{\delta \theta \to 0}\frac{\sin (\theta + \delta \theta) - \sin (\theta)}{\delta \theta} \end{equation}
From triangles \( \Delta ABG \) and \( \Delta ABC \), we get:
\( \sin (\theta + \delta \theta) = \frac{T + \delta T}{L + \delta L} \), and \( \sin (\theta) = \frac{T}{L} \).
Using \( \delta \theta = \frac{\delta S}{L} \) along with the values for \( \sin (\theta + \delta \theta) \) and \( \sin (\theta) \) and simplifying:
\begin{equation} \frac{\sin (\theta + \delta \theta) - \sin (\theta)}{\delta \theta} = \frac{L\delta T - T\delta L}{\delta S (L + \delta L)} \end{equation}
Comparing the similar triangles:
\( \frac{\delta T}{\delta S} = \frac{L}{1} \) and \( \frac{\delta L}{\delta S} = T \) and plugging this back into \( (2) \), we get:
\begin{equation} \frac{\sin (\theta + \delta \theta) - \sin (\theta)}{\delta \theta} = \frac{L^{2} - T^{2}}{L + \delta L} = \frac{1}{L + \delta L} \end{equation}
From triangle \( \Delta ABG \), \( \cos (\theta + \delta\theta) = \frac{1}{L + \delta L} \).
And from \( (1) \),
\( \lim_{\delta \theta \to 0}\frac{\sin (\theta + \delta \theta) - \sin (\theta)}{\delta \theta} = \lim_{\delta \theta \to 0}\cos (\theta + \delta \theta) = \cos (\theta) \), which is the derivative of \( \sin (\theta) \).
Derivative of the cosine
Proceeding similarly, we have:
\begin{equation} \frac{\delta \cos \theta}{\delta \theta} = \lim_{\delta \theta \to 0}\frac{\cos (\theta + \delta \theta) - \cos (\theta)}{\delta \theta} \end{equation}
From the two triangles, we have \( \cos (\theta + \delta \theta) = \frac{1}{L + \delta L} \), and \( \cos \theta = \frac{1}{L} \) and \( \delta \theta = \frac{\delta S}{L} \). Simplifying, we get:
\begin{equation} \frac{\cos (\theta + \delta \theta) - \cos (\theta)}{\delta \theta} = \frac{-\delta L}{\delta S(L + \delta L)} \end{equation}
Putting \( \frac{\delta L}{\delta S} = \tan \theta \) and \( \frac{1}{L + \delta L} = \cos (\theta + \delta \theta) \) into \( (5) \), we get:
\( \frac{\cos (\theta + \delta \theta) - \cos (\theta)}{\delta \theta} = (-\tan \theta)(\cos (\theta + \delta \theta)) \)
And from \( (4) \),
\( \lim_{\delta \theta \to 0}\frac{\cos (\theta + \delta \theta) - \cos (\theta)}{\delta \theta} = \lim_{\delta \theta \to 0} (-\tan \theta)(\cos (\theta + \delta \theta)) = -\sin (\theta) \), which is the derivative of \( \cos (\theta) \).