The Identities Calculus Assumes — Cheat sheet

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.