Mathematics · Calculus

Derivatives of Sine and Cosine

Q: Why must x be in radians when differentiating sin x?

The rule (sin x)′ = cos x rests on the limit $\frac{\sin{}(h)}{h} \to{} 1$ as h → 0, and this holds only in radians, where an angle is measured directly as arc length on the unit circle. If you work in degrees, the limit is $\frac{\pi{}}{180} \approx{} 0.01745$ instead, and every derivative acquires this factor: (sin x°)′ = (π/180)·cos x°. The elegant relations like (sin x)″ = −sin x would be lost. Practical consequence: always work in radians in calculus and set the calculator to RAD. A typical exam error is an extremum at "$x = 90$" instead of $x =\frac{ \pi{}}{2}$ because the calculator was in DEG mode.

Q: How are the extrema and inflection points of sine and cosine related?

Via the derivatives: extrema of sin x lie where $\cos{} x = 0$, that is at $x =\frac{ \pi{}}{2} + k\cdot{}\pi{}$; inflection points of sin x lie where the second derivative −sin x vanishes, that is exactly at the zeros $x = k\cdot{}\pi{}$ of the sine itself. So the curve inflects at every axis crossing and is steepest there with slope ±1. For the cosine everything is shifted by $\frac{\pi{}}{2}$: maxima and minima at $x = k\cdot{}\pi{}$, inflection points at $x =\frac{ \pi{}}{2} + k\cdot{}\pi{}$. This regularity makes trigonometric curve analysis very predictable: once you have one feature, you know all others by shifting quarter periods.

Q: What is the derivative of tan x and how does it follow from sin and cos?

The tangent is the quotient $\tan{} x =\frac{ \sin{} x}{\cos{}} x$. The quotient rule gives (tan x)′ = (cos x·cos x − sin x·(−sin x))/cos²x = (cos²x + sin²x)/cos²x. With the trigonometric Pythagoras $\sin{}^{2}x + \cos{}^{2}x = 1$ this simplifies to (tan x)′ = 1/cos²x, equivalently $1 + \tan{}^{2}x$. Since $\frac{1}{\cos{}^{2}}x$ is always positive, the tangent increases strictly on each of its branches; at the poles $x =\frac{ \pi{}}{2} + k\cdot{}\pi{}$, where $\cos{} x = 0$, it does not exist. The form $1 + \tan{}^{2}x$ is useful in physics and integration, for instance when derivatives are to be expressed through tan itself.

Sine and cosine differentiate into each other cyclically: sin becomes cos, cos becomes −sin, each in radians.

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Formula

(sin x)^{'} = cos x, (cos x)^{'} = - sin x

x in radians (rad) · function values dimensionless

Variables & units

Variables & units
Symbol	Meaning	Unit
sin x, cos x	Trigonometric functions on the unit circle	dimensionless
$x$	Angle in radians	rad

LaTeX

(\sin x)' = \cos x, \quad (\cos x)' = -\sin x

QUANTA STUDYquanta-study.de

Sine curve with three short tangents and a dashed cosine curve; at the same points cos x takes the values 1, 0 and −1. — The slope of sin x equals cos x at every point: slope 1 at x = 0, horizontal at π/2, slope −1 at π.

Derivation & background

The derivatives of the trigonometric functions form a cycle of four: sin → cos → −sin → −cos → sin. They hold in this simple form only in radians; in degrees the factor $\frac{\pi{}}{180}$ would appear. The foundation is the limit $\frac{\sin{}(h)}{h} \to{} 1$ as h → 0. Because $(\sin x)'' = -\sin x$ , sine and cosine solve the oscillation equation and describe harmonic oscillations.

Exam blueprint

Validity range

Holds for all real x, but only in radians. In degrees, (sin x°)′ = (π/180)·cos x° instead.

Derivation steps

Addition theorems in the difference quotient plus the limit $\frac{\sin{}(h)}{h} \to{} 1$ .

1 $\sin{}(x + h) - \sin{} x = \sin{} x\cdot{}(\cos{} h - 1) + \cos{} x\cdot{}\sin{} h$ .
2With $\frac{(\cos{} h - 1)}{h} \to{} 0$ and $\frac{\sin{}(h)}{h} \to{} 1$ , cos x remains; analogously for cos x.

Rearrangements

Inner factor

(\sin(kx))' = k \cos(kx)

The angular frequency k moves to the front as a factor.

Second derivative

(\sin x)'' = -\sin x

Sine solves the oscillation equation y″ = −y.

Antiderivatives

\int \sin x \, dx = -\cos x + C

Watch the sign: integrating sin gives minus cos, integrating cos gives plus sin.

Task variant

Differentiate $f(x) = 4\cdot{}\cos{}(3x)$ .

Outer derivative −sin, inner 3: f′ $(x) = 4\cdot{}(-\sin{}(3x))\cdot{}3 = -12\cdot{}\sin{}(3x)$ .
At $x = 0$ : f′ $(0) = 0$ , a maximum sits there.

Where does $f(x) = \sin{} x$ have horizontal tangents in [0; 2π]?

f′ $(x) = \cos{} x = 0$ at $x =\frac{ \pi{}}{2}$ and $x =\frac{ 3\pi{}}{2}$ , that is at the maximum $(\frac{\pi{}}{2}|1)$ and minimum $(\frac{3\pi{}}{2}|-1)$ .

Common mistakes

Writing (cos x)′ = sin x without the minus.

Correct: (cos x)′ = −sin x. Memory cycle sin → cos → −sin → −cos.

Working in degrees.

The differentiation rules hold in radians; set the calculator to RAD.

Forgetting the inner derivative 2 for sin(2x).

(sin(2x))′ = 2·cos(2x).

Exam context

Trigonometric curve analysis, oscillation problems and extrema of periodic functions.

Formula cluster

Trigonometric calculus

Derivative cycle, chain rule and oscillation models belong together.

(f∘g)' = f'(g(x))·g'(x)

Chain Rule of Differentiation

a/sin α = b/sin β = c/sin γ

Law of Sines

Worked example

$f(x) = 3\cdot{}\sin{}(2x)$ : f′ $(x) = 3\cdot{}\cos{}(2x)\cdot{}2 = 6\cdot{}\cos{}(2x)$ . At $x = 0$ : f′ $(0) = 6\cdot{}\cos{} 0 = 6$ , amplitude 3 times inner derivative 2 ✓.

Applications

Harmonic oscillations and waves, curve analysis of trigonometric functions, optimization with periodic quantities, AC circuit analysis

How do you calculate with Derivatives of Sine and Cosine?

Here is how to work through a typical Derivatives of Sine and Cosine ((sin x)' = cos x, (cos x)' = −sin x) task step by step:

1
Task
Differentiate $f(x) = 4\cdot{}\cos{}(3x)$ .
Solution path
1. Outer derivative −sin, inner 3: f′ $(x) = 4\cdot{}(-\sin{}(3x))\cdot{}3 = -12\cdot{}\sin{}(3x)$ .
2. At $x = 0$ : f′ $(0) = 0$ , a maximum sits there.
2
Task
Where does $f(x) = \sin{} x$ have horizontal tangents in [0; 2π]?
Solution path
f′ $(x) = \cos{} x = 0$ at $x =\frac{ \pi{}}{2}$ and $x =\frac{ 3\pi{}}{2}$ , that is at the maximum $(\frac{\pi{}}{2}|1)$ and minimum $(\frac{3\pi{}}{2}|-1)$ .

Frequently asked questions about Derivatives of Sine and Cosine

What are the derivatives of sine and cosine?+

We have (sin x)′ = cos x and (cos x)′ = −sin x, each for x in radians. Together with the second derivatives this forms a cycle of four: sin → cos → −sin → −cos and back to sin. After four differentiations you are back at the start. The minus appears exactly at the transition from cosine: the cosine starts at its maximum at $x = 0$ , then falls, so its slope must start negative, and −sin x does exactly that. Conversely the sine starts at 0 with maximal slope 1, and $\cos{} 0 = 1$ fits. Once you trace the cycle on the graphs, you mix up the signs far less often.

Why must x be in radians when differentiating sin x?+

The rule (sin x)′ = cos x rests on the limit $\frac{\sin{}(h)}{h} \to{} 1$ as h → 0, and this holds only in radians, where an angle is measured directly as arc length on the unit circle. If you work in degrees, the limit is $\frac{\pi{}}{180} \approx{} 0.01745$ instead, and every derivative acquires this factor: (sin x°)′ = (π/180)·cos x°. The elegant relations like (sin x)″ = −sin x would be lost. Practical consequence: always work in radians in calculus and set the calculator to RAD. A typical exam error is an extremum at " $x = 90$ " instead of $x =\frac{ \pi{}}{2}$ because the calculator was in DEG mode.

How do I differentiate

a\cdot{}\sin{}(bx + c)

Use the chain rule: the outer function sin becomes cos, the inner function bx + c has derivative b. So $(a\cdot{}\sin{}(bx + c))$ ′ = a·b·cos(bx + c). The amplitude a is kept as a factor, the angular frequency b enters multiplicatively, the phase shift c stays unchanged in the argument. Example: $f(x) = 3\cdot{}\sin{}(2x - \pi{})$ has f′ $(x) = 6\cdot{}\cos{}(2x - \pi{})$ . Physically the factor b means: a faster oscillation with the same amplitude has a proportionally larger maximum velocity. The standard mistake is forgetting b; when in doubt, check with a point, for example the initial slope at $x = 0$ .

How are the extrema and inflection points of sine and cosine related?+

Via the derivatives: extrema of sin x lie where $\cos{} x = 0$ , that is at $x =\frac{ \pi{}}{2} + k\cdot{}\pi{}$ ; inflection points of sin x lie where the second derivative −sin x vanishes, that is exactly at the zeros $x = k\cdot{}\pi{}$ of the sine itself. So the curve inflects at every axis crossing and is steepest there with slope ±1. For the cosine everything is shifted by $\frac{\pi{}}{2}$ : maxima and minima at $x = k\cdot{}\pi{}$ , inflection points at $x =\frac{ \pi{}}{2} + k\cdot{}\pi{}$ . This regularity makes trigonometric curve analysis very predictable: once you have one feature, you know all others by shifting quarter periods.

What is the derivative of tan x and how does it follow from sin and cos?+

The tangent is the quotient $\tan{} x =\frac{ \sin{} x}{\cos{}} x$ . The quotient rule gives (tan x)′ = (cos x·cos x − sin x·(−sin x))/cos²x = (cos²x + sin²x)/cos²x. With the trigonometric Pythagoras $\sin{}^{2}x + \cos{}^{2}x = 1$ this simplifies to (tan x)′ = 1/cos²x, equivalently $1 + \tan{}^{2}x$ . Since $\frac{1}{\cos{}^{2}}x$ is always positive, the tangent increases strictly on each of its branches; at the poles $x =\frac{ \pi{}}{2} + k\cdot{}\pi{}$ , where $\cos{} x = 0$ , it does not exist. The form $1 + \tan{}^{2}x$ is useful in physics and integration, for instance when derivatives are to be expressed through tan itself.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Calculus Volume 1, §3.5: Derivatives of Trigonometric Functions

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Derivative of the Natural Logarithm

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Derivatives of Sine and Cosine

Sine and cosine differentiate into each other cyclically: sin becomes cos, cos becomes −sin, each in radians.

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

sin x, cos x

Trigonometric functions on the unit circle

dimensionless

x

Angle in radians

rad

Derivation & background

Exam blueprint

Validity range

Holds for all real x, but only in radians. In degrees, (sin x°)′ = (π/180)·cos x° instead.

Derivation steps

Addition theorems in the difference quotient plus the limit $\frac{\sin{}(h)}{h} \to{} 1$ .

1 $\sin{}(x + h) - \sin{} x = \sin{} x\cdot{}(\cos{} h - 1) + \cos{} x\cdot{}\sin{} h$ .
2With $\frac{(\cos{} h - 1)}{h} \to{} 0$ and $\frac{\sin{}(h)}{h} \to{} 1$ , cos x remains; analogously for cos x.

How do you calculate with Derivatives of Sine and Cosine?

Here is how to work through a typical Derivatives of Sine and Cosine ((sin x)' = cos x, (cos x)' = −sin x) task step by step:

Task

Differentiate $f(x) = 4\cdot{}\cos{}(3x)$ .

Solution path

Outer derivative −sin, inner 3: f′ $(x) = 4\cdot{}(-\sin{}(3x))\cdot{}3 = -12\cdot{}\sin{}(3x)$ .
At $x = 0$ : f′ $(0) = 0$ , a maximum sits there.

Task

Where does $f(x) = \sin{} x$ have horizontal tangents in [0; 2π]?

Solution path

f′ $(x) = \cos{} x = 0$ at $x =\frac{ \pi{}}{2}$ and $x =\frac{ 3\pi{}}{2}$ , that is at the maximum $(\frac{\pi{}}{2}|1)$ and minimum $(\frac{3\pi{}}{2}|-1)$ .

Frequently asked questions about Derivatives of Sine and Cosine

What are the derivatives of sine and cosine?+

Why must x be in radians when differentiating sin x?+

How do I differentiate

a\cdot{}\sin{}(bx + c)

How are the extrema and inflection points of sine and cosine related?+

What is the derivative of tan x and how does it follow from sin and cos?+

Retain Derivatives of Sine and Cosine for exams

Create a curated FSRS exam set for (sin x)' = cos x, (cos x)' = −sin x: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential