The Identities Calculus Assumes

Idea

Calculus will not ask you to prove these. It expects you to know them and use them fast.

Nearly all of them come from one fact: Pythagoras, on the unit circle. The radius to the point $(\cos x, \sin x)$ is the hypotenuse of a right triangle whose legs are the two coordinates, and that hypotenuse is $1$ .

Rule

The one everything else comes from.

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

Tangent and cotangent are quotients. Each is the other upside down.

$\tan x = \dfrac{\sin x}{\cos x}$

$\cot x = \dfrac{\cos x}{\sin x}$

$\tan x \cdot \cot x = 1$

Secant and cosecant are reciprocals.

$\sec x = \dfrac{1}{\cos x}$

$\csc x = \dfrac{1}{\sin x}$

Double angles. The three cosine forms are one identity, written three ways.

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

$\cos 2x = \cos^2 x - \sin^2 x$

$\cos 2x = 2\cos^2 x - 1$

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

An identity is a claim about every angle, so any angle you know can test it. At $30^\circ$ : $\sin^2 30^\circ + \cos^2 30^\circ = \dfrac{1}{4} + \dfrac{3}{4} = 1$ . If a rearrangement of yours fails a check like that, the rearrangement is wrong.

How it is used

One fact, four shapes

Move a term across and it looks different. It is still the same identity.

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

↳ $1 - \sin^2 x = \cos^2 x$

↳ $1 - \cos^2 x = \sin^2 x$

↳ $\sin^2 x - 1 = -\cos^2 x$

↳ $\cos^2 x - 1 = -\sin^2 x$

Watch the minus sign on the last two. It is the usual slip.

Two more, by dividing

Divide $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ , then by $\sin^2 x$ .

$1 + \tan^2 x = \sec^2 x$

$1 + \cot^2 x = \csc^2 x$

Do not memorise these. You can rebuild both in one line.

Choosing a form of cos 2x

All three are equal, so none is wrong. Pick the one that keeps the problem in a single function.

If the problem holds	Use
both sine and cosine	$\cos 2x = \cos^2 x - \sin^2 x$
only cosine	$\cos 2x = 2\cos^2 x - 1$
only sine	$\cos 2x = 1 - 2\sin^2 x$

The last two come from the first. Swap $\sin^2 x$ for $1 - \cos^2 x$ , or $\cos^2 x$ for $1 - \sin^2 x$ .

The forms integration needs

Rearrange the last two for the squared term. A squared sine or cosine cannot be integrated as it stands. A $\cos 2x$ can.

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

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

Half a turn reverses both coordinates

Adding $180^\circ$ to an angle carries its point to the opposite side of the circle, so both coordinates change sign.

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

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

When you are stuck

Rewrite everything in $\sin$ and $\cos$ . Rarely the shortest route, almost always a route.

Worked example

Simplify $\dfrac{\sin^2 x}{1 - \cos x}$ .

The numerator is the part to attack, because the Pythagorean identity can rewrite it. Replace $\sin^2 x$ with $1 - \cos^2 x$ , which is a difference of two squares:

\frac{1 - \cos^2 x}{1 - \cos x} = \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x}

The common factor cancels, provided $\cos x \ne 1$ :

1 + \cos x

Explore

Pick an angle and square the two readouts in your head. Now drag anywhere and watch the running total under the readouts: the two squares trade against each other, and the total never leaves $1$ .

Every identity in this lesson is that one fact rearranged.

The unit circle

Try this. Try 0°, 90°, 180° and 270°. The horizontal projection is cosine; the vertical projection is sine. Track their signs between axes.

Angle (degrees)45

θ = 45° = 0.25π rad · cos θ = 0.707 · sin θ = 0.707 · sin² θ + cos² θ = 1