Math Core

An identity is an equation that holds for every input where both sides are defined. In the last unit you evaluated trig functions one angle at a time. Identities let you work with whole expressions: rewrite one in a friendlier form, compute every trig value from one known value, and turn equations and integrals you'll meet later into ones you can actually solve.

The fundamental identities

Every basic identity comes from the unit circle. If the terminal side of $\theta$ meets the unit circle at $(x, y)$ , then $\cos\theta = x$ and $\sin\theta = y$ , and the other four functions are defined from these two.

Fundamental identities

Reciprocal and quotient

\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}

Pythagorean

\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta

Even and odd

\cos(-\theta) = \cos\theta, \quad \sec(-\theta) = \sec\theta; \qquad \sin(-\theta) = -\sin\theta, \quad \tan(-\theta) = -\tan\theta

(csc and cot are odd too.)

Cofunction

\sin\left(\tfrac{\pi}{2} - \theta\right) = \cos\theta, \qquad \tan\left(\tfrac{\pi}{2} - \theta\right) = \cot\theta, \qquad \sec\left(\tfrac{\pi}{2} - \theta\right) = \csc\theta

The first Pythagorean identity is just $x^2 + y^2 = 1$ , the equation of the unit circle. Divide it by $\cos^2\theta$ to get the second and by $\sin^2\theta$ to get the third, so you only need to memorize one. The even and odd identities come from reflecting across the $x$ -axis: $(x, y)$ becomes $(x, -y)$ , so cosine keeps its value and sine switches sign.

The Pythagorean identities also come in rearranged forms that you'll use all the time: $1 - \sin^2\theta = \cos^2\theta$ , $1 - \cos^2\theta = \sin^2\theta$ , $\sec^2\theta - 1 = \tan^2\theta$ and $\csc^2\theta - 1 = \cot^2\theta$ . Whenever you spot one of these left-hand sides, think about replacing it.

Finding all six values from one

If you know one trig value and the quadrant, the Pythagorean identities give you the rest. The identity tells you the size and the quadrant tells you the sign.

Worked example: Five values from one

Given $\sin\theta = -\dfrac{2}{3}$ with $\pi < \theta < \dfrac{3\pi}{2}$ , find the other five trig values.

Solution. From $\sin^2\theta + \cos^2\theta = 1$ ,

\cos^2\theta = 1 - \frac{4}{9} = \frac{5}{9}, \qquad \cos\theta = \pm\frac{\sqrt5}{3}.

$\theta$ is in Quadrant III, where cosine is negative, so $\cos\theta = -\dfrac{\sqrt5}{3}$ . Then

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-2/3}{-\sqrt5/3} = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5},

and the reciprocals are $\csc\theta = -\dfrac{3}{2}$ , $\sec\theta = -\dfrac{3}{\sqrt5} = -\dfrac{3\sqrt5}{5}$ , $\cot\theta = \dfrac{\sqrt5}{2}$ .

Simplifying expressions

To simplify a trig expression, you usually want fewer functions and fewer fractions. These moves handle most problems:

Rewrite in sines and cosines. This is the default when nothing else suggests itself.
Look for a Pythagorean pattern such as $1 - \cos^2 x$ or $\sec^2 x - 1$ .
Use algebra: factor, combine fractions over a common denominator, or cancel.
Multiply by a conjugate when you see $1 \pm \sin x$ or $1 \pm \cos x$ in a denominator. The product $(1 - \sin x)(1 + \sin x) = \cos^2 x$ becomes a single term.

Worked example: Simplify

Simplify (a) $\dfrac{\tan x \cos x}{\sin^2 x}$ and (b) $\dfrac{\sec^2 x - 1}{\sec^2 x}$ .

Solution.

(a) Write $\tan x$ as $\dfrac{\sin x}{\cos x}$ :

\frac{\tan x \cos x}{\sin^2 x} = \frac{\dfrac{\sin x}{\cos x}\cdot\cos x}{\sin^2 x} = \frac{\sin x}{\sin^2 x} = \frac{1}{\sin x} = \csc x.

(b) The numerator is $\tan^2 x$ . Then switch to sines and cosines:

\frac{\tan^2 x}{\sec^2 x} = \frac{\sin^2 x}{\cos^2 x}\cdot\cos^2 x = \sin^2 x.

Verifying identities

To verify an identity means to prove that the two sides are equal for every allowed input. The rule of the game is different from solving an equation: you may not treat the statement as true and do the same thing to both sides, because that assumes what you're trying to prove. Instead, start with one side and transform it, one justified step at a time, until it becomes the other side.

Some strategy:

Start with the more complicated side. It's easier to simplify than to "unsimplify."
Keep an eye on the target. If the other side is in terms of $\sec x$ , aim for $\cos x$ in a denominator.
If you get stuck, you may simplify each side separately to the same expression. The chain still connects the two sides.

Worked example: Verify an identity

Verify $\dfrac{\cos x}{1 - \sin x} = \sec x + \tan x$ .

Solution. Work on the left side and multiply by the conjugate of the denominator:

\begin{aligned} \frac{\cos x}{1 - \sin x}\cdot\frac{1 + \sin x}{1 + \sin x} &= \frac{\cos x\,(1 + \sin x)}{1 - \sin^2 x} \\ &= \frac{\cos x\,(1 + \sin x)}{\cos^2 x} \\ &= \frac{1 + \sin x}{\cos x} \\ &= \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x. \end{aligned}

The left side has become the right side, so the identity holds wherever both sides are defined.

Common mistake

Plugging in a few values can disprove an identity (one counterexample is enough) but it can never prove one. For instance, $\sin 2x = 2\sin x$ is true at $x = 0$ and $x = \pi$ , yet false at $x = \dfrac{\pi}{2}$ , where the left side is $0$ and the right side is $2$ .

Tip

A quick graph is a good sanity check before you try to prove something. If $y = \text{left side}$ and $y = \text{right side}$ produce the same curve, you're probably holding a true identity. If the curves separate anywhere, stop and look for a typo.

Both sides of the identity from the example trace the same curve (the second is dashed on top of the first).Open in grapher →

Using identities with algebra

Identities are also tools for computing values without ever finding the angle. The standard trick is to square a sum and let $\sin^2\theta + \cos^2\theta = 1$ collapse two of the terms.

Worked example: No angle needed

Suppose $\sin\theta - \cos\theta = \dfrac{1}{2}$ . Find $\sin\theta\cos\theta$ .

Solution. Square both sides. (Squaring is fine here, since you're deriving a consequence rather than verifying an identity.)

\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = \frac{1}{4} \quad\Longrightarrow\quad 1 - 2\sin\theta\cos\theta = \frac{1}{4}.

So $2\sin\theta\cos\theta = \dfrac{3}{4}$ and $\sin\theta\cos\theta = \dfrac{3}{8}$ .

In the next lesson you'll see that $2\sin\theta\cos\theta$ has its own name, $\sin 2\theta$ , which is one of several new identities built on the ones here.

Practice

Practice 1

If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II, find $\cos\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Simplify $\sin\theta\cot\theta$ .

Practice 3

If $\tan\theta = -2$ and $\theta$ is in Quadrant IV, find $\sec\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

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

Practice 5

Simplify $\dfrac{\sec x - \cos x}{\sin x}$ .

Practice 6

Which equation is not an identity?

Practice 7

If $\sin\theta + \cos\theta = \dfrac{7}{5}$ , find $\sin\theta\cos\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

If $\cot\theta = \dfrac{3}{4}$ and $\sin\theta < 0$ , find $\csc\theta + \sec\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.