Trigonometry is inherently geometric, yet, students are often taught to memorize the derivatives of various trigonometric functions through formulas and tables. Let’s put the geometry back into trigonometry with some simple visual proofs.

The derivatives of sine and cosine

Sine and cosine are the coordinates of a point on the unit circle, $x = \cos\theta$ and $y = \sin\theta$. To find their derivative, we study how those coordinates change as the point slides around the circle by a small angle $\delta\theta$.

The legs of the small similar triangle are precisely the change in the coordinates of the point on the unit circle. As $\delta\theta$ vanishes, the length of the hypotenuse of the small triangle approaches the arc length $\delta\theta$. Therefore,

$$\delta y = \cos\theta \, \delta\theta, \qquad \delta x = -\sin\theta \, \delta\theta,$$

the minus sign is included because counterclockwise motion carries the point to the left. Dividing by $\delta\theta$ yields the derivatives,

$$\frac{d}{d\theta}\sin\theta = \cos\theta, \qquad \frac{d}{d\theta}\cos\theta = -\sin\theta.$$

The derivative of tangent and secant

The tangent is given by the vertical leg of the triangle drawn below — we just extended the hypotenuse of the triangle determined by $\theta$ past the unit circle. We denote the length of this leg by $t = \tan\theta$. Since the base of the triangle has length $1$, that leg stands on the line tangent to the circle, which is where the function gets its name from.

Nudging the angle by $\delta\theta$ extends the vertical leg by an amount $\delta t$. This allows us to create a small triangle that is almost similar to the large triangle, the only difference being the angle $\alpha = \theta + \delta\theta$. In the limit, $\alpha \to \theta$, so we can read from the small triangle

$$\cos\theta = \frac{\ell\,\delta\theta}{\delta t} \implies \frac{\delta t}{\delta \theta} = \ell^2$$

Where we have used the fact that $1/\cos\theta = \ell$ (reading from the large triangle). Therefore, the derivative of tangent is,

$$\frac{d}{d\theta} \tan\theta = 1 + \tan^2\theta = \sec^2\theta.$$

Looking at the small triangle, we can read off $\frac{\delta \ell}{\delta \theta} = \ell\tan\theta$, which implies the derivative of secant,

$$\frac{d}{d \theta} \sec\theta = \sec\theta\tan\theta$$

The derivatives of cotangent and cosecant

Taking our triangle construction from above and resting its hypotenuse on the $x$-axis lets us extend one of its legs until it collides with the $y$-axis. A few lines of trig shows us that the $y$-intercept is $\tilde{\ell} = \csc\theta$ and that the segment from the point on the unit circle point to the intercept has length $\tilde{t} = \cot\theta$. This is where the prefixes “co” come from: the contact point splits the line into $t$ and $\tilde{t}$, and its two intercepts sit at distances $\ell$ and $\tilde{\ell}$ from the origin.

Nudging the angle drops the intercept by $\delta\tilde{\ell}$ and swings the line by a distance of $\tilde{t}\,\delta\theta$. The magnified triangle carries the angle $\theta$ between the $y$-axis and the line. Taking the sine of this angle yields,

$$\frac{d}{d\theta}\csc\theta = \frac{\delta \tilde{\ell}}{\delta \theta} = - \frac{\tilde{t}}{\sin \theta} = - \csc\theta\cot\theta$$

with the sign because the intercept falls as $\theta$ grows.

The cotangent leg $\tilde{t}$ loses length at both ends. At the intercept it gives up $\delta\tilde{\ell}\cos\theta$ from the small triangle, and at the contact point with the unit circle it gives up $\delta\theta$. Together $\delta \tilde{t} = \delta \tilde{\ell} \cos\theta - \delta\theta$, so

$$\begin{align*} \frac{d}{d\theta}\cot\theta = \frac{\delta \tilde{t}}{\delta \theta} &= -1 + \frac{\delta \tilde{\ell}}{\delta\theta}\cos\theta \\ &= - (1 + \cot^2\theta) \\ &= -\csc^2\theta \end{align*}$$

The two ends are exactly the two terms of $1 + \cot^2\theta = \csc^2\theta$.