Precalculus · Unit 3: Trigonometric and Polar Functions · 14 min read · Updated 2026-05-11

Trigonometric identities (Pythagorean, sum/difference, double-angle) — AP Precalculus

AP Precalculus · Unit 3: Trigonometric and Polar Functions · 14 min read

1. Pythagorean Identities ★★☆☆☆ ⏱ 4 min

Pythagorean identities are the most fundamental trigonometric identities, derived directly from the Pythagorean theorem applied to the unit circle. They are used to convert between trigonometric functions, find unknown values when quadrant is known, and simplify expressions.

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

Dividing the core identity by $\cos^2 \theta$ (where $\cos\theta \neq 0$) and $\sin^2 \theta$ (where $\sin\theta \neq 0$) gives two additional related identities:

\tan^2 \theta + 1 = \sec^2 \theta

1 + \cot^2 \theta = \csc^2 \theta

📐 Worked Example

Given that $\sin \theta = \frac{3}{5}$ and $\frac{\pi}{2} < \theta < \pi$, find the exact value of $\sec \theta$.

Use the core Pythagorean identity to solve for $\cos^2 \theta$:
$\sin^2 \theta + \cos^2 \theta = 1 \implies \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$
Determine the sign of $\cos \theta$ from the quadrant: $\theta$ lies in the second quadrant, where cosine is negative, so:
$\cos \theta = -\frac{4}{5}$
Use the reciprocal identity for secant to get the final result:
$\sec \theta = \frac{1}{\cos \theta} = -\frac{5}{4}$
Verify with the tangent-secant Pythagorean identity to confirm:
$\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{3}{4} \implies \tan^2 \theta + 1 = \frac{9}{16} + 1 = \frac{25}{16} = \sec^2 \theta$

Exam tip: Always use the given quadrant information to assign the correct sign to your trig function—AP exam questions almost always include quadrant context specifically to test this sign check.

2. Sum and Difference Identities ★★★☆☆ ⏱ 4 min

Sum and difference identities let you find the sine, cosine, or tangent of a combined angle $\alpha \pm \beta$ when you know the trig values of $\alpha$ and $\beta$ individually. They are also used to simplify expressions and verify more complex identities.

\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta

\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}, \quad \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}

📐 Worked Example

Evaluate $\cos(15^\circ)$ exactly using sum/difference identities, using the fact that $15^\circ = 45^\circ - 30^\circ$.

Write the angle as a difference of two known special angles:
$\cos(15^\circ) = \cos(45^\circ - 30^\circ)$
Apply the cosine difference identity:
$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \implies \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$
Substitute the known exact special angle values: $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 30^\circ = \frac{1}{2}$
Multiply and combine terms to get the exact value:
$\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$

Exam tip: If you forget the identity sign, confirm with a known angle: $\cos(90^\circ - 30^\circ) = \sin 30^\circ = 0.5$, which will only match if the sign is positive for cosine of a difference, reinforcing the flip rule.

3. Double-Angle Identities ★★★☆☆ ⏱ 3 min

Double-angle identities are a special case of the sum identities, where $\alpha = \beta = \theta$, so you can calculate the trigonometric value of $2\theta$ from values for $\theta$. They are widely used to simplify products of trig functions, reduce power of squared terms, rewrite functions for graphing, and solve equations. The alternative forms for cosine are most often used for power reduction, a common AP exam question.

\sin(2\theta) = 2 \sin \theta \cos \theta

\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta

\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}

📐 Worked Example

Given $\cos \theta = \frac{1}{3}$ and $0 < \theta < \frac{\pi}{2}$, find the exact value of $\sin(2\theta)$.

The double-angle identity for sine requires both $\sin \theta$ and $\cos \theta$. We know $\cos \theta = \frac{1}{3}$, so first solve for $\sin \theta$.
Use the core Pythagorean identity, and assign the correct sign for first quadrant $\theta$:
$\sin^2 \theta + \left(\frac{1}{3}\right)^2 = 1 \implies \sin^2 \theta = \frac{8}{9} \implies \sin \theta = \frac{2\sqrt{2}}{3}$
Substitute into the double-angle identity for sine:
$\sin(2\theta) = 2 \sin \theta \cos \theta = 2 \left(\frac{2\sqrt{2}}{3}\right)\left(\frac{1}{3}\right) = \frac{4\sqrt{2}}{9}$
Verify with the Pythagorean identity for $2\theta$ to confirm:
$\cos(2\theta) = 2\left(\frac{1}{3}\right)^2 - 1 = -\frac{7}{9} \implies \sin^2(2\theta) + \cos^2(2\theta) = \frac{32}{81} + \frac{49}{81} = 1$

Exam tip: When finding a trig value for $2\theta$, always confirm the quadrant of $2\theta$ to check your sign—for example, if $\theta$ is between $\frac{\pi}{4}$ and $\frac{\pi}{2}$, $2\theta$ is between $\frac{\pi}{2}$ and $\pi$, where cosine is negative.

4. Mixed AP-Style Applications ★★★★☆ ⏱ 3 min

AP Precalculus exam questions often require combining multiple identities to simplify expressions, rewrite functions, or solve real-world problems. Below are common exam-style worked examples.

📐 Worked Example

Let $f(x) = \cos^2 x - \sin^2 x + 2 \sin x \cos x$. (a) Rewrite $f(x)$ as a sum of single-power trigonometric functions of $2x$. (b) Find the exact value of $f\left(\frac{5\pi}{12}\right)$. (c) Verify that $f(x) = \sqrt{2} \sin\left(2x + \frac{\pi}{4}\right)$.

Part (a): Apply double-angle identities to rewrite each term:
$\cos^2 x - \sin^2 x = \cos(2x), \quad 2 \sin x \cos x = \sin(2x) \implies f(x) = \cos(2x) + \sin(2x)$
Part (b): Substitute $x = \frac{5\pi}{12}$, so $2x = \frac{5\pi}{6}$:
$f\left(\frac{5\pi}{12}\right) = \cos\left(\frac{5\pi}{6}\right) + \sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 - \sqrt{3}}{2}$
Part (c): Expand the right-hand side using the sine sum identity:
$\sqrt{2} \sin\left(2x + \frac{\pi}{4}\right) = \sqrt{2}\left[\sin(2x)\cos\left(\frac{\pi}{4}\right) + \cos(2x)\sin\left(\frac{\pi}{4}\right)\right]$
$= \sqrt{2}\left[\sin(2x) \cdot \frac{\sqrt{2}}{2} + \cos(2x) \cdot \frac{\sqrt{2}}{2}\right] = \sin 2x + \cos 2x = f(x)$

Common Pitfalls

Why: Confusion between standard trig exponent notation ($\sin^2 \theta = (\sin \theta)^2$) and argument squaring ($\sin(\theta^2)$ is a completely different function)

Why: Students often copy the same sign rule as sine, leading to calculation errors

Why: Confusion between linear function properties and non-linear trigonometric functions

Why: Students remember the Pythagorean identity but ignore given quadrant information when finding the unknown trig value

Why: Students use identities to simplify and forget that division is only valid for non-zero divisors

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Trigonometric identities (Pythagorean, sum/difference, double-angle) — AP Precalculus

1. Pythagorean Identities ★★☆☆☆ ⏱ 4 min

2. Sum and Difference Identities ★★★☆☆ ⏱ 4 min

3. Double-Angle Identities ★★★☆☆ ⏱ 3 min

4. Mixed AP-Style Applications ★★★★☆ ⏱ 3 min

Common Pitfalls

Quick Reference Cheatsheet

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