Math Core

An identity is an equation that is true for every value of the variable where both sides are defined, like $2(x + 1) = 2x + 2$ . Trigonometry has a small set of identities that connect sine, cosine and tangent to each other. With them you can find every trig value of an angle from just one, and rewrite messy expressions in simpler form.

Three reciprocal functions

Before the identities, meet the last three trigonometric functions. Each is the reciprocal of one you already know.

Definition

Reciprocal functions

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

These are cosecant, secant and cotangent. Each is undefined where its denominator is $0$ .

For example, $\sec\dfrac{\pi}{3} = \dfrac{1}{\cos(\pi/3)} = \dfrac{1}{1/2} = 2$ , and $\csc\dfrac{\pi}{2} = \dfrac{1}{1} = 1$ . A reciprocal always has the same sign as the original function, so the quadrant sign rules carry over.

Common mistake

The pairs are not the ones the names suggest. Secant goes with cosine, and cosecant goes with sine. One way to remember it: each pair has exactly one "co-" in it (cosine with secant, sine with cosecant, tangent with cotangent).

The quotient identities

You already know the first one from the tangent lesson:

\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad\qquad \cot\theta = \frac{\cos\theta}{\sin\theta}.

The Pythagorean identity

The point at angle $\theta$ on the unit circle is $(\cos\theta, \sin\theta)$ , and every point on that circle satisfies $x^2 + y^2 = 1$ . Substituting gives the most important identity in trigonometry.

The Pythagorean identities

\sin^2\theta + \cos^2\theta = 1

Dividing every term by $\cos^2\theta$ or by $\sin^2\theta$ gives two more versions:

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

The notation $\sin^2\theta$ means $(\sin\theta)^2$ , the square of the sine. It does not mean $\sin(\theta^2)$ .

Here is where the second version comes from:

\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \quad\Longrightarrow\quad \tan^2\theta + 1 = \sec^2\theta.

Finding all values from one

If you know one trig value and the quadrant, the Pythagorean identity gives the others. The identity finds the size of the missing value, and the quadrant decides its sign.

Worked example: From sine to everything

Suppose $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II. Find $\cos\theta$ , $\tan\theta$ and $\sec\theta$ .

Use $\sin^2\theta + \cos^2\theta = 1$ :

\begin{aligned} \left(\tfrac{3}{5}\right)^2 + \cos^2\theta &= 1 \\ \cos^2\theta &= 1 - \tfrac{9}{25} = \tfrac{16}{25} \\ \cos\theta &= \pm\tfrac{4}{5} \end{aligned}

Cosine is negative in Quadrant II, so $\cos\theta = -\dfrac{4}{5}$ . Then

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}, \qquad \sec\theta = \frac{1}{\cos\theta} = -\frac{5}{4}.

Worked example: Starting from tangent

Suppose $\tan\theta = -2$ and $\theta$ is in Quadrant IV. Find $\cos\theta$ .

Use $\tan^2\theta + 1 = \sec^2\theta$ : $\sec^2\theta = 4 + 1 = 5$ , so $\sec\theta = \pm\sqrt{5}$ . Cosine (and so secant) is positive in Quadrant IV, so $\sec\theta = \sqrt{5}$ and

\cos\theta = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}.

Tip

You can also sketch a right triangle for the reference angle. For $\sin\theta = \dfrac{3}{5}$ , draw legs $3$ and $4$ and hypotenuse $5$ , read off the ratios, and then attach signs from the quadrant. It's the same computation in picture form.

The negative-angle identities

From the symmetry of the graphs (and the reflection across the $x$ -axis on the unit circle):

\sin(-\theta) = -\sin\theta \qquad \cos(-\theta) = \cos\theta \qquad \tan(-\theta) = -\tan\theta

Cosine is even, while sine and tangent are odd.

Simplifying expressions

To simplify a trig expression, a reliable plan is:

Rewrite everything in terms of sine and cosine.
Combine fractions and cancel common factors.
Look for $\sin^2\theta + \cos^2\theta$ , or a rearrangement like $1 - \cos^2\theta = \sin^2\theta$ , to replace.

Worked example: Simplifying

Simplify each expression.

$\sin\theta \cot\theta$
$\dfrac{1 - \sin^2\theta}{\cos\theta}$
$\sec\theta - \sin\theta\tan\theta$

Solutions.

$\sin\theta \cdot \dfrac{\cos\theta}{\sin\theta} = \cos\theta$ .
By the Pythagorean identity, $1 - \sin^2\theta = \cos^2\theta$ , so the expression is $\dfrac{\cos^2\theta}{\cos\theta} = \cos\theta$ .
Write both terms over $\cos\theta$ :

\frac{1}{\cos\theta} - \frac{\sin^2\theta}{\cos\theta} = \frac{1 - \sin^2\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta.

To verify an identity, start with the more complicated side and transform it, step by step, until it matches the other side. Don't move terms across the equals sign, because you'd be assuming the very thing you're trying to show.