Gilbert Strang’s Calculus: Derivative of sin x and cos x

Calculus

Trigonometric Functions

Derivatives

Limits

Author

Chao Ma

Published

February 19, 2026

Goal

We want to prove:

$\frac{d}{dx}(\sin x)=\cos x$
$\frac{d}{dx}(\cos x)=-\sin x$

These formulas are true when angles are measured in radians.

Three facts we use

Pythagorean identity: \[ \sin^2\theta+\cos^2\theta=1. \]
Angle addition for sine: \[ \sin(x+h)=\sin x\cos h+\cos x\sin h. \]
Angle addition for cosine: \[ \cos(x+h)=\cos x\cos h-\sin x\sin h. \]

Circular motion picture

On the unit circle, the point at angle $t$ is \[ (\cos t,\sin t). \] So proving derivatives of $\sin t$ and $\cos t$ is equivalent to understanding velocity components in circular motion.

Why radians matter

\[ 2\pi\ \text{radians}=360^\circ. \]

Only in radian measure do we get the clean limit \[ \lim_{h\to 0}\frac{\sin h}{h}=1, \] which drives both derivative proofs.

Geometric inequalities

Two properties we need are:

$\sin\theta<\theta$ for $\theta>0$
$\tan\theta>\theta$ for $0<\theta<\frac\pi2$

For $\sin\theta<\theta$: in the unit circle, the chord subtended by angle $\theta$ is shorter than the arc, and that arc length is exactly $\theta$ (in radians). The vertical leg of the inscribed right triangle has length $\sin\theta$, and this leg is shorter than the chord. Therefore \[ \sin\theta < \text{chord} < \theta. \] So $\sin\theta<\theta$.

For $\tan\theta>\theta$: compare areas. The sector area is \[ A_{\text{sector}}=\frac12\theta \] on the unit circle, while the outer right triangle formed by the radius and tangent line has area \[ A_{\triangle}=\frac12\tan\theta. \] Because the sector lies strictly inside that tangent triangle, \[ \frac12\theta<\frac12\tan\theta \;\Rightarrow\;\theta<\tan\theta. \]

From $\theta<\tan\theta=\frac{\sin\theta}{\cos\theta}$, we get \[ \cos\theta<\frac{\sin\theta}{\theta}<1. \] By squeeze theorem, \[ \lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1. \]

The missing limit: (cos h - 1)/h -> 0

Yes, this part should be proved explicitly.

Use \[ \cos h-1=-2\sin^2\left(\frac h2\right). \] Then \[ \frac{\cos h-1}{h} =-\sin\left(\frac h2\right)\cdot \frac{\sin(h/2)}{h/2}. \] As $h\to 0$, \[ \sin\left(\frac h2\right)\to 0, \qquad \frac{\sin(h/2)}{h/2}\to 1, \] so \[ \lim_{h\to 0}\frac{\cos h-1}{h}=0. \]

Prove d(sin x)/dx = cos x

Using angle addition: \[ \frac{\sin(x+h)-\sin x}{h} =\frac{\sin x(\cos h-1)+\cos x\sin h}{h} =\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}. \] Take $h\to 0$: \[ \frac{d}{dx}(\sin x) =\sin x\cdot 0+\cos x\cdot 1 =\cos x. \]

Prove d(cos x)/dx = -sin x

Again from angle addition: \[ \frac{\cos(x+h)-\cos x}{h} =\frac{\cos x(\cos h-1)-\sin x\sin h}{h} =\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}. \] Take $h\to 0$: \[ \frac{d}{dx}(\cos x) =\cos x\cdot 0-\sin x\cdot 1 =-\sin x. \]

Takeaway. The two derivative formulas rest on two core limits, $\lim_{h\to 0}\frac{\sin h}{h}=1$ and $\lim_{h\to 0}\frac{\cos h-1}{h}=0$, which come from unit-circle geometry and radian measure.

--- title: "Gilbert Strang's Calculus: Derivative of sin x and cos x" author: "Chao Ma" date: "2026-02-19" categories: [Calculus, Trigonometric Functions, Derivatives, Limits] --- {{< video https://www.youtube.com/watch?v=FtQl1gAo12E&list=PLFW_V3qDH5jRyfpD9uiq6aKVTWIxpbIm3&index=7 >}} # Goal We want to prove: - $\frac{d}{dx}(\sin x)=\cos x$ - $\frac{d}{dx}(\cos x)=-\sin x$ These formulas are true when angles are measured in **radians**. # Three facts we use 1. Pythagorean identity: $$ \sin^2\theta+\cos^2\theta=1. $$ 2. Angle addition for sine: $$ \sin(x+h)=\sin x\cos h+\cos x\sin h. $$ 3. Angle addition for cosine: $$ \cos(x+h)=\cos x\cos h-\sin x\sin h. $$ # Circular motion picture On the unit circle, the point at angle $t$ is $$ (\cos t,\sin t). $$ So proving derivatives of $\sin t$ and $\cos t$ is equivalent to understanding velocity components in circular motion. ![Circular motion intuition](derivative-sin-cos-5.png) # Why radians matter $$ 2\pi\ \text{radians}=360^\circ. $$ Only in radian measure do we get the clean limit $$ \lim_{h\to 0}\frac{\sin h}{h}=1, $$ which drives both derivative proofs. # Geometric inequalities Two properties we need are: - $\sin\theta<\theta$ for $\theta>0$ - $\tan\theta>\theta$ for $0<\theta<\frac\pi2$ For $\sin\theta<\theta$: in the unit circle, the chord subtended by angle $\theta$ is shorter than the arc, and that arc length is exactly $\theta$ (in radians). The vertical leg of the inscribed right triangle has length $\sin\theta$, and this leg is shorter than the chord. Therefore $$ \sin\theta < \text{chord} < \theta. $$ So $\sin\theta<\theta$. ![Inequality for sin(theta)/theta](derivative-sin-cos-4.png) For $\tan\theta>\theta$: compare areas. The sector area is $$ A_{\text{sector}}=\frac12\theta $$ on the unit circle, while the outer right triangle formed by the radius and tangent line has area $$ A_{\triangle}=\frac12\tan\theta. $$ Because the sector lies strictly inside that tangent triangle, $$ \frac12\theta<\frac12\tan\theta \;\Rightarrow\;\theta<\tan\theta. $$ ![Sector and triangle comparison](derivative-sin-cos-3.png) From $\theta<\tan\theta=\frac{\sin\theta}{\cos\theta}$, we get $$ \cos\theta<\frac{\sin\theta}{\theta}<1. $$ By squeeze theorem, $$ \lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1. $$ # The missing limit: (cos h - 1)/h -> 0 Yes, this part should be proved explicitly. Use $$ \cos h-1=-2\sin^2\left(\frac h2\right). $$ Then $$ \frac{\cos h-1}{h} =-\sin\left(\frac h2\right)\cdot \frac{\sin(h/2)}{h/2}. $$ As $h\to 0$, $$ \sin\left(\frac h2\right)\to 0, \qquad \frac{\sin(h/2)}{h/2}\to 1, $$ so $$ \lim_{h\to 0}\frac{\cos h-1}{h}=0. $$ # Prove d(sin x)/dx = cos x Using angle addition: $$ \frac{\sin(x+h)-\sin x}{h} =\frac{\sin x(\cos h-1)+\cos x\sin h}{h} =\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}. $$ Take $h\to 0$: $$ \frac{d}{dx}(\sin x) =\sin x\cdot 0+\cos x\cdot 1 =\cos x. $$ # Prove d(cos x)/dx = -sin x Again from angle addition: $$ \frac{\cos(x+h)-\cos x}{h} =\frac{\cos x(\cos h-1)-\sin x\sin h}{h} =\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}. $$ Take $h\to 0$: $$ \frac{d}{dx}(\cos x) =\cos x\cdot 0-\sin x\cdot 1 =-\sin x. $$ --- **Takeaway.** The two derivative formulas rest on two core limits, $\lim_{h\to 0}\frac{\sin h}{h}=1$ and $\lim_{h\to 0}\frac{\cos h-1}{h}=0$, which come from unit-circle geometry and radian measure.