Trigonometric Identities — AP® Precalculus Notes

Two expressions can look nothing alike and still have the same graph. $1 - 2\sin^2 x$ and $\cos(2x)$ are one example. An equation that is true for every input where both sides are defined is an identity. On the exam, identities do three jobs: they turn an angle like $\frac{\pi}{12}$ into exact values you know, they rewrite an expression as a single trigonometric function, and they turn a hard equation into one you can solve.

The Pythagorean identity

The point $(\cos x, \sin x)$ is on the unit circle $X^2 + Y^2 = 1$ , so

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

for every $x$ . (Here $\sin^2 x$ means $(\sin x)^2$ .) Dividing every term by $\cos^2 x$ , and then by $\sin^2 x$ , gives two more forms:

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

Use them to find one value from another. If $\sin x = \frac{3}{5}$ and $x$ is in Quadrant II, then $\cos^2 x$ $= 1 - \frac{9}{25}$ $= \frac{16}{25}$ , so $\cos x = -\frac{4}{5}$ , negative because of the quadrant.

Sum and difference identities

$\sin(a \pm b) = \sin a\cos b \pm \cos a\sin b$

$\cos(a \pm b) = \cos a\cos b \mp \sin a\sin b$

Watch the signs in the cosine identity: a sum gives a minus, a difference gives a plus. And notice what they say is false: $\sin(a + b)$ is not $\sin a + \sin b$ .

Worked example 1. Find $\sin\frac{5\pi}{12}$ exactly.

Write $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$ , two angles you know.

$\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right)$ $= \sin\frac{\pi}{4}\cos\frac{\pi}{6}$ $+ \cos\frac{\pi}{4}\sin\frac{\pi}{6}$ $= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}$ $+ \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$ $= \frac{\sqrt{6} + \sqrt{2}}{4}$ .

The difference identities also explain the phase shift between the graphs: $\cos\left(x - \frac{\pi}{2}\right)$ $= \cos x\cdot 0$ $+ \sin x\cdot 1$ $= \sin x$ .

Double-angle identities

Put $b = a$ in the sum identities:

$\sin(2a) = 2\sin a\cos a$

$\cos(2a) = \cos^2 a - \sin^2 a = 2\cos^2 a - 1 = 1 - 2\sin^2 a$

The last two forms come from replacing $\sin^2 a$ or $\cos^2 a$ with the Pythagorean identity. Figure 1 shows the first form as heights on a graph.

Graphs of cos squared x and sin squared x (dashed) and of cos(2x) (solid) for x from 0 to 2 pi. At every x the solid height equals the first dashed height minus the second. — **Figure 1.** $\cos(2x)$ (solid) is the difference of $\cos^2 x$ and $\sin^2 x$ (dashed). LevlPrep original diagram.

Family	Identities	Typical use
Pythagorean	$\sin^2 x + \cos^2 x = 1$ ; $\tan^2 x + 1 = \sec^2 x$ ; $1 + \cot^2 x = \csc^2 x$	replace a square; find one ratio from another
Sum and difference	$\sin(a \pm b)$ , $\cos(a \pm b)$	exact values at angles like $\frac{\pi}{12}$ ; shifts
Double angle	$\sin(2a) = 2\sin a\cos a$ ; three forms of $\cos(2a)$	an equation with $2x$ and $x$ together

Rewriting as a single function

When a question says "as a single trigonometric function", the answer is one function like $\tan x$ , not a quotient like $\frac{\sin x}{\cos x}$ . A routine that works:

Write every function in terms of $\sin$ and $\cos$ , and expand any $2x$ with a double-angle identity.
Use a Pythagorean identity to replace sums of squares.
Cancel common factors.
Name the one function that is left.

Worked example 2. Rewrite $\frac{1 - \cos(2x)}{\sin(2x)}$ as a single trigonometric function.

Choose the form of $\cos(2x)$ that cancels the $1$ : $1 - \cos(2x)$ $= 1 - (1 - 2\sin^2 x)$ $= 2\sin^2 x$ .

Expand the bottom: $\sin(2x) = 2\sin x\cos x$ .

Cancel: $\frac{2\sin^2 x}{2\sin x\cos x}$ $= \frac{\sin x}{\cos x}$ $= \tan x$ , wherever the original expression is defined.

Solving equations with identities

An equation that mixes $\sin(2x)$ and $\sin x$ cannot be solved until both use the same angle.

Worked example 3. Solve $\cos(2x) = \cos x$ on $[0, 2\pi)$ .

Use the form with only cosines: $2\cos^2 x - 1 = \cos x$ , so $2\cos^2 x - \cos x - 1 = 0$ .

Factor: $(2\cos x + 1)(\cos x - 1)$ $= 0$ . So $\cos x = -\frac{1}{2}$ ( $x = \frac{2\pi}{3}, \frac{4\pi}{3}$ ) or $\cos x = 1$ ( $x = 0$ ).

Answer: $x = 0$ , $\frac{2\pi}{3}$ , $\frac{4\pi}{3}$ .

Common slips

Splitting a function over a sum: $\cos(a + b) \ne \cos a + \cos b$ .
Pulling the $2$ out of $\sin(2x)$ .
Stopping at $\frac{\sin x}{\cos x}$ when one function was asked for.
Dividing an equation by $\sin x$ and losing the solutions where $\sin x = 0$ ; factor instead.