Derivatives of Trigonometric Functions

If you have ever watched a pendulum swing back and forth, or listened to the gentle hum of an electrical appliance, you have witnessed trigonometric functions in action. These periodic, wave-like functions describe so much of the rhythmic behavior we see in nature: the rise and fall of ocean tides, the oscillation of a guitar string, the alternating current that powers your devices. Now that you understand how to take derivatives, it is time to extend that skill to these essential functions.

The good news is that the derivatives of trigonometric functions follow elegant, memorable patterns. Once you learn that the derivative of sine is cosine, and the derivative of cosine is negative sine, you will have the foundation for everything else. The other four trig functions (tangent, cotangent, secant, and cosecant) have derivatives that can all be derived from these two basic facts.

By the end of this lesson, you will be able to differentiate any expression involving trigonometric functions, and you will understand why these derivatives take the forms they do. This knowledge opens the door to analyzing oscillating systems, understanding wave behavior, and solving a wide range of problems in physics and engineering.

Core Concepts

Review: The Unit Circle and Trigonometric Functions

Before diving into derivatives, let us briefly recall what sine and cosine actually measure. On the unit circle (a circle of radius 1 centered at the origin), if you start at the point $(1, 0)$ and rotate counterclockwise by an angle $\theta$, you land at the point $(\cos\theta, \sin\theta)$.

• $\cos\theta$ is the x-coordinate: the horizontal displacement from the origin
• $\sin\theta$ is the y-coordinate: the vertical displacement from the origin

This geometric interpretation will help you understand why the derivatives work out the way they do. As you rotate around the circle, the rate at which your vertical position changes depends on your current horizontal position, and vice versa.

The other four trigonometric functions are built from sine and cosine:

• $\tan\theta = \frac{\sin\theta}{\cos\theta}$
• $\cot\theta = \frac{\cos\theta}{\sin\theta}$
• $\sec\theta = \frac{1}{\cos\theta}$
• $\csc\theta = \frac{1}{\sin\theta}$

Why Radians Matter (Not Degrees!)

Before we establish any derivative formulas, there is a crucial point you must understand: these derivative formulas only work when angles are measured in radians. If you use degrees, the formulas will be wrong.

Why does this matter? Consider the limit:

$$\lim_{x \to 0} \frac{\sin x}{x}$$

When $x$ is in radians, this limit equals exactly 1. But if $x$ is in degrees, the limit would be $\frac{\pi}{180}$ (approximately 0.01745), which would make all our derivative formulas unnecessarily complicated.

The reason radians are “natural” is that a radian is defined by the arc length on the unit circle. An angle of 1 radian corresponds to an arc of length 1 on a unit circle. This direct connection between angle measure and arc length is what makes the calculus of trig functions work out so cleanly.

From now on, always assume angles are in radians unless explicitly stated otherwise. Your calculator should be in radian mode whenever you are doing calculus.

The Key Limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

The foundation of all trigonometric derivatives rests on one crucial limit:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

This is not obvious, and proving it rigorously requires some geometric arguments involving the unit circle (often called the “squeeze theorem” proof). The intuition is this: when $x$ is a small angle in radians, $\sin x$ is approximately equal to $x$ itself. Try it on your calculator: $\sin(0.1) \approx 0.0998$, which is very close to 0.1.

There is a companion limit that we also need:

$$\lim_{x \to 0} \frac{\cos x - 1}{x} = 0$$

This says that $\cos x - 1$ approaches zero faster than $x$ does as $x$ approaches 0. Again, this makes sense: $\cos(0.1) \approx 0.995$, so $\cos(0.1) - 1 \approx -0.005$, which is much smaller in magnitude than 0.1.

These two limits, combined with the limit definition of the derivative, give us our derivative formulas.

Derivative of Sine: $\frac{d}{dx}[\sin x] = \cos x$

Let us derive this using the definition of the derivative:

$$\frac{d}{dx}[\sin x] = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h}$$

Using the angle addition formula $\sin(x + h) = \sin x \cos h + \cos x \sin h$:

$$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$

$$= \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}$$

$$= \lim_{h \to 0} \left[ \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} \right]$$

$$= \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$

Therefore:

$$\boxed{\frac{d}{dx}[\sin x] = \cos x}$$

This result has a beautiful geometric interpretation. On the unit circle, as you move around, your vertical position ($\sin\theta$) changes fastest when you are at the leftmost or rightmost points (where $\cos\theta = \pm 1$). It changes slowest (momentarily not at all) when you are at the top or bottom of the circle (where $\cos\theta = 0$).

Derivative of Cosine: $\frac{d}{dx}[\cos x] = -\sin x$

A similar derivation using the angle addition formula $\cos(x + h) = \cos x \cos h - \sin x \sin h$ gives:

$$\frac{d}{dx}[\cos x] = \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h}$$

$$= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$$

$$= \lim_{h \to 0} \left[ \cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h} \right]$$

$$= \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$$

Therefore:

$$\boxed{\frac{d}{dx}[\cos x] = -\sin x}$$

Notice the negative sign. This is easy to forget, so pay attention to it. The derivative of cosine is negative sine, not positive sine.

Derivatives of the Other Trigonometric Functions

With the derivatives of sine and cosine established, we can derive the derivatives of the other four trig functions using the quotient rule.

Derivative of Tangent:

Since $\tan x = \frac{\sin x}{\cos x}$, we apply the quotient rule:

$$\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$

$$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$

$$\boxed{\frac{d}{dx}[\tan x] = \sec^2 x}$$

Derivative of Cotangent:

Since $\cot x = \frac{\cos x}{\sin x}$:

$$\frac{d}{dx}[\cot x] = \frac{\sin x \cdot (-\sin x) - \cos x \cdot \cos x}{\sin^2 x}$$

$$= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x$$

$$\boxed{\frac{d}{dx}[\cot x] = -\csc^2 x}$$

Derivative of Secant:

Since $\sec x = \frac{1}{\cos x}$, we can use the quotient rule (or think of it as $(\cos x)^{-1}$ and use the chain rule):

$$\frac{d}{dx}[\sec x] = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x$$

$$\boxed{\frac{d}{dx}[\sec x] = \sec x \tan x}$$

Derivative of Cosecant:

Since $\csc x = \frac{1}{\sin x}$:

$$\frac{d}{dx}[\csc x] = \frac{0 \cdot \sin x - 1 \cdot \cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x} = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\csc x \cot x$$

$$\boxed{\frac{d}{dx}[\csc x] = -\csc x \cot x}$$

Patterns to Remember

Notice the patterns in these derivatives:

1. Sine and Cosine are cyclical: The derivative of $\sin x$ is $\cos x$; the derivative of $\cos x$ is $-\sin x$; the derivative of $-\sin x$ is $-\cos x$; the derivative of $-\cos x$ is $\sin x$. If you keep differentiating, you cycle through these four functions.

2. “Co” functions have negatives: The derivatives of the “co” functions (cosine, cotangent, cosecant) all have negative signs. The non-“co” functions (sine, tangent, secant) have positive signs in their derivatives.

3. Tangent and secant are paired: $\frac{d}{dx}[\tan x] = \sec^2 x$ and $\frac{d}{dx}[\sec x] = \sec x \tan x$

4. Cotangent and cosecant are paired: $\frac{d}{dx}[\cot x] = -\csc^2 x$ and $\frac{d}{dx}[\csc x] = -\csc x \cot x$

Chain Rule with Trigonometric Functions

In most real applications, you will not just differentiate $\sin x$ or $\cos x$ alone. You will encounter composite functions like $\sin(3x)$, $\cos(x^2)$, or $\tan(e^x)$. For these, you need to combine the trigonometric derivative rules with the chain rule.

Remember the chain rule: if $y = f(g(x))$, then $\frac{dy}{dx} = f’(g(x)) \cdot g’(x)$.

For trigonometric functions, this becomes:

$$\frac{d}{dx}[\sin(u)] = \cos(u) \cdot \frac{du}{dx}$$

$$\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx}$$

$$\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot \frac{du}{dx}$$

And similarly for the other three functions. The key is to differentiate the outer function (the trig function) and then multiply by the derivative of the inner function (whatever is inside the trig function).

Higher-Order Derivatives

Because sine and cosine cycle through each other, their higher derivatives follow a predictable pattern:

$$\frac{d}{dx}[\sin x] = \cos x$$ $$\frac{d^2}{dx^2}[\sin x] = -\sin x$$ $$\frac{d^3}{dx^3}[\sin x] = -\cos x$$ $$\frac{d^4}{dx^4}[\sin x] = \sin x$$

After four derivatives, you are back where you started. This cyclical behavior is unique among common functions and reflects the periodic nature of trigonometric functions.

Notation and Terminology

Term	Meaning	Example
$\frac{d}{dx}[\sin x]$	Derivative of sine equals cosine	$\frac{d}{dx}[\sin x] = \cos x$
$\frac{d}{dx}[\cos x]$	Derivative of cosine equals negative sine	$\frac{d}{dx}[\cos x] = -\sin x$
$\frac{d}{dx}[\tan x]$	Derivative of tangent equals secant squared	$\frac{d}{dx}[\tan x] = \sec^2 x$
$\frac{d}{dx}[\cot x]$	Derivative of cotangent equals negative cosecant squared	$\frac{d}{dx}[\cot x] = -\csc^2 x$
$\frac{d}{dx}[\sec x]$	Derivative of secant	$\frac{d}{dx}[\sec x] = \sec x \tan x$
$\frac{d}{dx}[\csc x]$	Derivative of cosecant	$\frac{d}{dx}[\csc x] = -\csc x \cot x$
Radian	Natural unit for angles in calculus	$\pi$ radians $= 180°$
Chain rule with trig	Differentiate outer trig, multiply by derivative of inner	$\frac{d}{dx}[\sin(3x)] = 3\cos(3x)$

Examples

Key Properties and Rules

The Six Basic Trigonometric Derivatives

$$\frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x$$

$$\frac{d}{dx}[\tan x] = \sec^2 x \qquad \frac{d}{dx}[\cot x] = -\csc^2 x$$

$$\frac{d}{dx}[\sec x] = \sec x \tan x \qquad \frac{d}{dx}[\csc x] = -\csc x \cot x$$

Chain Rule Versions

When the argument is a function $u = g(x)$:

$$\frac{d}{dx}[\sin u] = \cos u \cdot \frac{du}{dx} \qquad \frac{d}{dx}[\cos u] = -\sin u \cdot \frac{du}{dx}$$

$$\frac{d}{dx}[\tan u] = \sec^2 u \cdot \frac{du}{dx} \qquad \frac{d}{dx}[\cot u] = -\csc^2 u \cdot \frac{du}{dx}$$

$$\frac{d}{dx}[\sec u] = \sec u \tan u \cdot \frac{du}{dx} \qquad \frac{d}{dx}[\csc u] = -\csc u \cot u \cdot \frac{du}{dx}$$

The Essential Limits

$$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{\cos x - 1}{x} = 0$$

$$\lim_{x \to 0} \frac{\tan x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$

Higher Derivatives of Sine and Cosine

$n$	$\frac{d^n}{dx^n}[\sin x]$	$\frac{d^n}{dx^n}[\cos x]$
1	$\cos x$	$-\sin x$
2	$-\sin x$	$-\cos x$
3	$-\cos x$	$\sin x$
4	$\sin x$	$\cos x$

The pattern repeats every four derivatives.

Memory Aids

“Co” means negative: The derivatives of cosine, cotangent, and cosecant all have negative signs.

Pairing: Tangent and secant appear together; cotangent and cosecant appear together.

Squares for tan and cot: $\frac{d}{dx}[\tan x] = \sec^2 x$ and $\frac{d}{dx}[\cot x] = -\csc^2 x$

Products for sec and csc: $\frac{d}{dx}[\sec x] = \sec x \tan x$ and $\frac{d}{dx}[\csc x] = -\csc x \cot x$

Real-World Applications

Simple Harmonic Motion (Pendulums and Springs)

When a pendulum swings or a spring oscillates, its position can be modeled by a function like:

$$x(t) = A\cos(\omega t + \phi)$$

where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase shift.

The velocity is the derivative of position: $$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$$

The acceleration is the derivative of velocity: $$a(t) = \frac{dv}{dt} = -A\omega^2\cos(\omega t + \phi)$$

Notice that $a(t) = -\omega^2 x(t)$. This relationship, where acceleration is proportional to position but opposite in sign, is the defining characteristic of simple harmonic motion.

Sound Waves and Music

Sound waves are pressure variations that can be modeled using sine and cosine functions. A pure tone at frequency $f$ can be represented as:

$$p(t) = A\sin(2\pi f t)$$

The rate of change of pressure determines how quickly the air molecules are accelerating, which affects how we perceive the sound. Understanding derivatives of trigonometric functions is essential for analyzing audio signals, designing speakers and microphones, and processing music digitally.

Alternating Current (AC) Circuits

The voltage in an AC electrical outlet follows a sinusoidal pattern:

$$V(t) = V_0 \sin(\omega t)$$

where $V_0$ is the peak voltage and $\omega$ is related to the frequency (60 Hz in the US, 50 Hz in Europe).

The current in a circuit depends on the derivative of voltage. In a capacitor, for instance:

$$I(t) = C\frac{dV}{dt} = CV_0\omega\cos(\omega t)$$

This 90-degree phase shift between voltage and current in capacitors and inductors is a fundamental concept in electrical engineering, and understanding it requires knowing the derivatives of sine and cosine.

Circular Motion

When an object moves in a circle of radius $r$ at constant angular velocity $\omega$, its position can be written as:

$$x(t) = r\cos(\omega t), \qquad y(t) = r\sin(\omega t)$$

The velocity components are: $$v_x = -r\omega\sin(\omega t), \qquad v_y = r\omega\cos(\omega t)$$

The acceleration components are: $$a_x = -r\omega^2\cos(\omega t), \qquad a_y = -r\omega^2\sin(\omega t)$$

This shows that the acceleration vector always points toward the center of the circle (centripetal acceleration), with magnitude $r\omega^2$. Understanding derivatives of trig functions reveals the physics of circular motion.

Modeling Daylight Hours

The number of daylight hours throughout the year can be modeled approximately by:

$$D(t) = 12 + A\sin\left(\frac{2\pi}{365}(t - 80)\right)$$

where $t$ is the day of the year and $A$ depends on latitude.

The derivative tells us how quickly daylight hours are changing:

$$D’(t) = A \cdot \frac{2\pi}{365} \cos\left(\frac{2\pi}{365}(t - 80)\right)$$

Near the equinoxes, when $D(t) = 12$, the cosine term is at its maximum, so daylight hours are changing most rapidly. Near the solstices, the cosine is near zero, so daylight hours change slowly from day to day.

Self-Test Problems

Summary

• The fundamental derivatives: $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$. These two formulas are the foundation for all trigonometric derivatives.

• The key limit that makes these derivatives work is $\lim_{x \to 0} \frac{\sin x}{x} = 1$, which only holds when $x$ is measured in radians.

• The other four derivatives follow from quotient rule applications:

  • $\frac{d}{dx}[\tan x] = \sec^2 x$
  • $\frac{d}{dx}[\cot x] = -\csc^2 x$
  • $\frac{d}{dx}[\sec x] = \sec x \tan x$
  • $\frac{d}{dx}[\csc x] = -\csc x \cot x$

• Memory aid: The “co” functions (cosine, cotangent, cosecant) all have negative signs in their derivatives.

• Chain rule is essential: Most applications involve composite functions like $\sin(3x)$ or $\cos(x^2)$. Always multiply by the derivative of the inner function.

• Sine and cosine cycle: Repeated differentiation of sine or cosine cycles through four functions: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$.

• Real-world applications include simple harmonic motion (pendulums and springs), sound and electromagnetic waves, AC circuits, and circular motion. These phenomena are fundamentally described by trigonometric functions, and their derivatives reveal velocity, acceleration, and rates of change.

• Always use radians in calculus. The derivative formulas are only correct when angles are measured in radians.