THE WHOLE UNIT · ONE REFERENCE

Trigonometry
Cheat sheet.

The key rules, formulas and reminders from all 8 topics, gathered into reference cards.

Key formulas, conditions and traps · Read down each column.

The three ratios

Function	Ratio
$\sin\theta$	opposite / hypotenuse
$\cos\theta$	adjacent / hypotenuse
$\tan\theta$	opposite / adjacent

The other three are these upside down: $\cot\theta = 1/\tan\theta$ , $\sec\theta = 1/\cos\theta$ , $\csc\theta = 1/\sin\theta$ .

Traps

Trap	Rule
"Opposite" moves with the angle	name the sides from the angle you are using, not from the picture
Impossible values	$\sin\theta$ and $\cos\theta$ never exceed $1$

$\sin\theta = \cos(90^\circ - \theta)$ - the other acute angle sees the same legs swapped.

Values at 30, 45 and 60 Degrees

The values

	$30^\circ = \dfrac{\pi}{6}$	$45^\circ = \dfrac{\pi}{4}$	$60^\circ = \dfrac{\pi}{3}$
$\sin$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Cot, sec and csc: flip tan, cos and sin. The $30^\circ$ and $60^\circ$ columns are each other reversed.

Safety net

Triangle	Sides
30-60-90	$x$ , $x\sqrt{3}$ , $2x$
45-45-90	$x$ , $x$ , $x\sqrt{2}$

The points

Angle	Radians	$(\cos, \sin)$	$\tan$	$\cot$
$0^\circ$	$0$	$(1, 0)$	$0$	undefined
$90^\circ$	$\dfrac{\pi}{2}$	$(0, 1)$	undefined	$0$
$180^\circ$	$\pi$	$(-1, 0)$	$0$	undefined
$270^\circ$	$\dfrac{3\pi}{2}$	$(0, -1)$	undefined	$0$
$360^\circ$	$2\pi$	$(1, 0)$	$0$	undefined

The rule behind the table

Cosine is the first coordinate, sine the second. A zero on the bottom of a quotient means undefined: $\tan$ breaks where $\cos = 0$ ( $90^\circ$ , $270^\circ$ ), $\cot$ breaks where $\sin = 0$ ( $0^\circ$ , $180^\circ$ , $360^\circ$ ).

The Identities Calculus Assumes

The one to know

$\sin^2 x + \cos^2 x = 1$

Sign trap when rearranging: $\sin^2 x - 1 = -\cos^2 x$ , not $\cos^2 x$ .

Definitions

$\tan x = \dfrac{\sin x}{\cos x}$ , $\cot x = \dfrac{\cos x}{\sin x}$ , $\sec x = \dfrac{1}{\cos x}$ , $\csc x = \dfrac{1}{\sin x}$

Divided forms

$1 + \tan^2 x = \sec^2 x \qquad 1 + \cot^2 x = \csc^2 x$

Double angles

$\sin 2x = 2\sin x\cos x$

Form of $\cos 2x$	Use when the problem holds
$\cos^2 x - \sin^2 x$	both functions
$2\cos^2 x - 1$	only cosine
$1 - 2\sin^2 x$	only sine

For integration

$\cos^2 x = \dfrac{1 + \cos 2x}{2} \qquad \sin^2 x = \dfrac{1 - \cos 2x}{2}$

Half a turn

$\cos(\pi + x) = -\cos x \qquad \sin(\pi + x) = -\sin x$

Stuck?

Rewrite everything in $\sin$ and $\cos$ .

Inverse Trig: Choosing One Angle

The four windows

Function	Domain	Range	Negative input
$\arcsin$	$[-1, 1]$	$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$	negative angle
$\arccos$	$[-1, 1]$	$[0, \pi]$	obtuse angle
$\arctan$	$\mathbb{R}$	$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$	negative angle
$\text{arccot}$	$\mathbb{R}$	$(0, \pi)$	obtuse angle

Traps

Trap	Rule
$\sin^{-1} x$	means $\arcsin x$ , never $\dfrac{1}{\sin x}$
$\arcsin 2$	undefined - sine never reaches $2$
$\arcsin(\sin x)$	equals $x$ only when $x$ is already inside the window
$\sin(\arcsin x)$	always $x$ , for $x$ in $[-1, 1]$

Arcsin

Definition

$y = \arcsin x \iff \sin y = x$ , with $D = [-1, 1]$ and $R = \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (closed - the ends are reached).

Values

$x$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\arcsin x$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$

Negative inputs: $\arcsin(-x) = -\arcsin x$ - arcsin is odd.

Method

Job	Move
Free $x$ from $\sin(\ldots) = k$	apply $\arcsin$ to both sides
Free $x$ from $\arcsin(\ldots) = k$	apply $\sin$ to both sides
$\cos\left(\arcsin\dfrac{3}{5}\right)$ and friends	draw the triangle, Pythagoras gives the third side

Special sum

$x^2 + y^2 = 1 \Rightarrow \arcsin x + \arcsin y = \dfrac{\pi}{2}$ (check the condition first)

Arccos

Definition

$y = \arccos x \iff \cos y = x$ , with $D = [-1, 1]$ and $R = [0, \pi]$ . Never returns a negative angle.

The rule that is different

$\arccos(-x) = \pi - \arccos x$ - reflects to an obtuse angle. Not odd. Do not borrow arcsin's sign flip.

Values

$x$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\arccos x$	$\dfrac{\pi}{2}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{6}$	$0$

Negative inputs: reflect. $\arccos\left(-\dfrac{1}{2}\right) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$ .

Identities

$\arcsin x + \arccos x = \dfrac{\pi}{2}$ for every $x$ in $[-1, 1]$ .

Domain questions

Whatever sits inside must satisfy $-1 \le \text{inside} \le 1$ .

Arctan and Arccot

Definitions

$y = \arctan x \iff \tan y = x$ , with $R = \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ . $y = \text{arccot}\,x \iff \cot y = x$ , with $R = (0, \pi)$ . Both accept every real number; both ranges are open.

Traps

Trap	Rule
Arctan is odd	$\arctan(-x) = -\arctan x$
Arccot is not	$\text{arccot}(-x) = \pi - \text{arccot}\,x$ - reflects, like arccos
At zero	$\text{arccot}\,0 = \dfrac{\pi}{2}$ , while $\arctan 0 = 0$

Values

$x$	$0$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\arctan x$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$
$\text{arccot}\,x$	$\dfrac{\pi}{2}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{6}$

Negative inputs: arctan flips sign, arccot reflects.

Behaviour at infinity

$\arctan x \to \pm\dfrac{\pi}{2}$ ; $\text{arccot}\,x \to 0$ or $\pi$ . Horizontal asymptotes, never reached.

The conversion

$\arctan x + \text{arccot}\,x = \dfrac{\pi}{2}$ for every real $x$ - the fastest and safest way to turn one into the other.

Ratios in a Right Triangle

3 reference blocks

Read lesson ↗

The three ratios

Function	Ratio
$\sin\theta$	opposite / hypotenuse
$\cos\theta$	adjacent / hypotenuse
$\tan\theta$	opposite / adjacent

The other three

Function	Ratio	Also equals
$\cot\theta$	adjacent / opposite	$1/\tan\theta$
$\sec\theta$	hypotenuse / adjacent	$1/\cos\theta$
$\csc\theta$	hypotenuse / opposite	$1/\sin\theta$

Checks

Point	Note
Size does not matter	similar triangles share every ratio
Bound	$\sin\theta$ and $\cos\theta$ never exceed $1$
Complements	$\sin\theta = \cos(90^\circ - \theta)$
"Opposite" moves	the names are relative to the angle, not the page
The other angle	move to it and the two legs trade names

Values at 30, 45 and 60 Degrees

4 reference blocks

Read lesson ↗

All six, at all three angles

	$30^\circ$	$45^\circ$	$60^\circ$
$\sin$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\cot$	$\sqrt{3}$	$1$	$\dfrac{1}{\sqrt{3}}$
$\sec$	$\dfrac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$
$\csc$	$2$	$\sqrt{2}$	$\dfrac{2}{\sqrt{3}}$

In radians

Degrees	$30^\circ$	$45^\circ$	$60^\circ$
Radians	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$

Half a turn is $\pi = 180^\circ$ .

Rebuild it from two triangles

Triangle	From	Sides
30-60-90	half an equilateral triangle of side $2x$	$x$ , $x\sqrt{3}$ , $2x$
45-45-90	a square of side $x$ cut along its diagonal	$x$ , $x$ , $x\sqrt{2}$

The $x$ cancels in every ratio. That is why the size of the triangle never matters.

Patterns

Pattern	Why
The $30^\circ$ and $60^\circ$ columns are each other reversed	the two angles add to $90^\circ$
$\sin 45^\circ = \cos 45^\circ$	that triangle is isosceles
$\dfrac{1}{\sqrt{3}}$ is also written $\dfrac{\sqrt{3}}{3}$	some books clear the root from the bottom
Learn the radians too	calculus never asks in degrees

The Unit Circle

3 reference blocks

Read lesson ↗

Read them off the circle

Angle	Radians	Point	$\cos$	$\sin$	$\tan$	$\cot$
$0^\circ$	$0$	$(1, 0)$	$1$	$0$	$0$	undefined
$90^\circ$	$\dfrac{\pi}{2}$	$(0, 1)$	$0$	$1$	undefined	$0$
$180^\circ$	$\pi$	$(-1, 0)$	$-1$	$0$	$0$	undefined
$270^\circ$	$\dfrac{3\pi}{2}$	$(0, -1)$	$0$	$-1$	undefined	$0$
$360^\circ$	$2\pi$	$(1, 0)$	$1$	$0$	$0$	undefined

Tangent and cotangent break in opposite places

$\tan x = \dfrac{\sin x}{\cos x}$ divides by the cosine. $\cot x = \dfrac{\cos x}{\sin x}$ divides by the sine.

At	$\tan$	$\cot$
$0^\circ$ , $180^\circ$ , $360^\circ$	$0$	undefined
$90^\circ$ , $270^\circ$	undefined	$0$

A zero on the top gives $0$ . A zero on the bottom gives nothing at all.

Why it matters later

Fact	Consequence
$\cos x = 0$ at $90^\circ$ , $270^\circ$	tangent is undefined there - its asymptotes
$\sin x = 0$ at $0^\circ$ , $180^\circ$ , $360^\circ$	cotangent is undefined there instead
$\cos x$ is the first coordinate	no triangle needed on an axis
$360^\circ$ returns to $0^\circ$	periodicity: $\sin(x + 2\pi) = \sin x$

The Identities Calculus Assumes

8 reference blocks

Read lesson ↗

The one to know

$\sin^2 x + \cos^2 x = 1$ - Pythagoras on the unit circle. Everything else follows.

Rearrange it

When you see	Replace it with
$1 - \sin^2 x$	$\cos^2 x$
$1 - \cos^2 x$	$\sin^2 x$
$\sin^2 x - 1$	$-\cos^2 x$
$\cos^2 x - 1$	$-\sin^2 x$

Divide it

Divide by	Gives
$\cos^2 x$	$1 + \tan^2 x = \sec^2 x$
$\sin^2 x$	$1 + \cot^2 x = \csc^2 x$

Definitions

$\tan x = \dfrac{\sin x}{\cos x}$ , $\cot x = \dfrac{\cos x}{\sin x}$ , $\sec x = \dfrac{1}{\cos x}$ , $\csc x = \dfrac{1}{\sin x}$ .

And so $\tan x \cdot \cot x = 1$ .

Double angles

$\sin 2x = 2\sin x\cos x$

Form of $\cos 2x$	Use it when the problem holds
$\cos^2 x - \sin^2 x$	both sine and cosine
$2\cos^2 x - 1$	only cosine
$1 - 2\sin^2 x$	only sine

For integration

$\cos^2 x = \dfrac{1 + \cos 2x}{2}$ , $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ - rearrangements of the last two forms above.

Half a turn

$\cos(\pi + x) = -\cos x$ , $\sin(\pi + x) = -\sin x$ - adding half a turn carries the point to the opposite side of the circle, reversing both coordinates.

Fallback

Rewrite everything in $\sin$ and $\cos$ . Rarely elegant, almost always works.

Inverse Trig: Choosing One Angle

4 reference blocks

Read lesson ↗

The idea

Trig functions are not one-to-one, so each is restricted to a stretch where it is, and only that stretch is inverted. The restriction becomes the inverse's range.

The four windows

Function	Domain	Range
$\arcsin$	$[-1, 1]$	$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$\arccos$	$[-1, 1]$	$[0, \pi]$
$\arctan$	$\mathbb{R}$	$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
$\text{arccot}$	$\mathbb{R}$	$(0, \pi)$

Round trips

Expression	Value
$\sin(\arcsin x)$	$x$ , for every $x$ in $[-1, 1]$
$\arcsin(\sin x)$	only $x$ when $x$ is already inside the range

Trap

Domains are forced by the original function's reach: sine never leaves $[-1, 1]$ , so $\arcsin 2$ does not exist. Tangent reaches everything, so $\arctan$ accepts everything.

Arcsin

6 reference blocks

Read lesson ↗

Definition

$y = \arcsin x \iff \sin y = x$ with $y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . $D = [-1, 1]$ , $R = \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ .

Values worth knowing

$x$	$\arcsin x$
$-1$	$-\dfrac{\pi}{2}$
$-\dfrac{1}{2}$	$-\dfrac{\pi}{6}$
$0$	$0$
$\dfrac{1}{2}$	$\dfrac{\pi}{6}$
$\dfrac{\sqrt{2}}{2}$	$\dfrac{\pi}{4}$
$1$	$\dfrac{\pi}{2}$

Inverting

Trapped inside	Apply
$\sin(\ldots)$	$\arcsin$ to both sides
$\arcsin(\ldots)$	$\sin$ to both sides

Non-standard ratios

Let $\alpha$ be the inverse expression, draw the right triangle with that opposite side and hypotenuse, and use Pythagoras for the third side. Read any function of $\alpha$ off it.

The complementary sum

$x^2 + y^2 = 1 \Rightarrow \arcsin x + \arcsin y = \dfrac{\pi}{2}$ - the two legs of a unit-hypotenuse triangle, so the angles are complementary.

Properties

Property	Note
Odd	$\arcsin(-x) = -\arcsin x$
Increasing	across the whole domain
Closed range	$\pm\dfrac{\pi}{2}$ are reached
Outside $[-1,1]$	undefined

Arccos

6 reference blocks

Read lesson ↗

Definition

$y = \arccos x \iff \cos y = x$ , with $D = [-1, 1]$ and $R = [0, \pi]$ .

Arccos turns a number into an angle, and that angle is never negative.

Which quadrant

Input	Output lands in
positive	first quadrant, $0$ to $\dfrac{\pi}{2}$
negative	second quadrant, $\dfrac{\pi}{2}$ to $\pi$

$\arccos(-x) = \pi - \arccos x$ . Arccos reflects; it does not flip sign the way arcsin does.

Exact values

$x$	$-1$	$-\dfrac{\sqrt{3}}{2}$	$-\dfrac{\sqrt{2}}{2}$	$-\dfrac{1}{2}$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\arccos x$	$180^\circ$	$150^\circ$	$135^\circ$	$120^\circ$	$90^\circ$	$60^\circ$	$45^\circ$	$30^\circ$	$0^\circ$

As $x$ grows, $\arccos x$ falls. Arcsin climbs.

Identities

Identity	Condition
$\cos(\arccos x) = x$	any $x$ in $[-1, 1]$
$\arcsin x + \arccos x = \dfrac{\pi}{2}$	any $x$ in $[-1, 1]$
$\arccos x + \arccos y = \dfrac{\pi}{2}$	$x^2 + y^2 = 1$ , both positive

When there is no special angle

Name it $\alpha$ , so $\cos\alpha$ is the given ratio. That is adjacent over hypotenuse, so draw the triangle, let Pythagoras fill in the third side, and read off whatever is asked. The angle itself is never needed.

Domain of an arccos expression

Whatever sits inside must satisfy $-1 \le \text{inside} \le 1$ . Solve that compound inequality.

Arctan and Arccot

7 reference blocks

Read lesson ↗

Definitions

$y = \arctan x \iff \tan y = x$ , with $D = (-\infty, \infty)$ and $R = \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ .

$y = \text{arccot}\,x \iff \cot y = x$ , with $D = (-\infty, \infty)$ and $R = (0, \pi)$ .

Both accept every real number. Both ranges are open: the ends are approached, never reached.

Negative inputs

Function	Rule	Like
$\arctan(-x) = -\arctan x$	odd - flips sign	arcsin
$\text{arccot}(-x) = \pi - \text{arccot}\,x$	reflects - never negative	arccos

$-45^\circ$ and $315^\circ$ are the same point, but arctan reports $-45^\circ$ - the reading inside its range.

Exact values

$x$	$-\sqrt{3}$	$-1$	$-\dfrac{1}{\sqrt{3}}$	$0$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\arctan x$	$-60^\circ$	$-45^\circ$	$-30^\circ$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$
$\text{arccot}\,x$	$150^\circ$	$135^\circ$	$120^\circ$	$90^\circ$	$60^\circ$	$45^\circ$	$30^\circ$

$\text{arccot}\,0 = 90^\circ$ , while $\arctan 0 = 0$ . That difference catches people out.

Asymptotes

Arctan flattens towards $\pm\dfrac{\pi}{2}$ ; arccot towards $0$ and $\pi$ . Every real input, and the output never leaves a fixed interval - the standard example when limits at infinity are taught.

Identities

Identity	Condition
$\tan(\arctan x) = x$	any real $x$
$\cot(\text{arccot}\,x) = x$	any real $x$
$\arctan x + \text{arccot}\,x = \dfrac{\pi}{2}$	any real $x$ - the conversion between them

All four inverses, side by side

Function	Range	Quadrants	Negative input
$\arcsin$	$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$	first, fourth	negative angle
$\arctan$	$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$	first, fourth	negative angle
$\arccos$	$[0, \pi]$	first, second	obtuse angle
$\text{arccot}$	$(0, \pi)$	first, second	obtuse angle

When there is no special angle

Name it $\alpha$ . For arctan, $\tan\alpha$ is opposite over adjacent; for arccot, $\cot\alpha$ is adjacent over opposite. Draw the triangle, let Pythagoras give the hypotenuse, read off whatever is asked. The angle itself is never needed.