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

AP Precalculus· AP Precalculus CED — 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.

📘 Definition

Pythagorean Trigonometric Identities

A set of core identities derived from the unit circle Pythagorean theorem that relate the squares of basic trigonometric functions.

Example:

The core identity $sin^{2} θ + cos^{2} θ = 1$ holds for all real $θ$ .

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

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

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

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

📐 Worked Example

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

1
Use the core Pythagorean identity to solve for $cos^{2} θ$ :
$\sin^2 \theta + \cos^2 \theta = 1 \implies \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$
2
Determine the sign of $cos θ$ from the quadrant: $θ$ lies in the second quadrant, where cosine is negative, so:
$\cos \theta = -\frac{4}{5}$
3
Use the reciprocal identity for secant to get the final result:
$\sec \theta = \frac{1}{\cos \theta} = -\frac{5}{4}$
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 $α \pm β$ when you know the trig values of $α$ and $β$ 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 $cos (1 5^{\circ})$ exactly using sum/difference identities, using the fact that $1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ .

1
Write the angle as a difference of two known special angles:
$\cos(15^\circ) = \cos(45^\circ - 30^\circ)$
2
Apply the cosine difference identity:
$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \implies \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$
3
Substitute the known exact special angle values: $cos 4 5^{\circ} = \frac{2}{2}$ , $cos 3 0^{\circ} = \frac{3}{2}$ , $sin 4 5^{\circ} = \frac{2}{2}$ , $sin 3 0^{\circ} = \frac{1}{2}$
4
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: $cos (9 0^{\circ} - 3 0^{\circ}) = sin 3 0^{\circ} = 0.5$ , 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 $α = β = θ$ , so you can calculate the trigonometric value of $2 θ$ from values for $θ$ . 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 $cos θ = \frac{1}{3}$ and $0 < θ < \frac{π}{2}$ , find the exact value of $sin (2 θ)$ .

1
The double-angle identity for sine requires both $sin θ$ and $cos θ$ . We know $cos θ = \frac{1}{3}$ , so first solve for $sin θ$ .
2
Use the core Pythagorean identity, and assign the correct sign for first quadrant $θ$ :
$\sin^2 \theta + \left(\frac{1}{3}\right)^2 = 1 \implies \sin^2 \theta = \frac{8}{9} \implies \sin \theta = \frac{2\sqrt{2}}{3}$
3
Substitute into the double-angle identity for sine:
$\sin(2\theta) = 2 \sin \theta \cos \theta = 2 \left(\frac{2\sqrt{2}}{3}\right)\left(\frac{1}{3}\right) = \frac{4\sqrt{2}}{9}$
4
Verify with the Pythagorean identity for $2 θ$ 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 $2 θ$ , always confirm the quadrant of $2 θ$ to check your sign—for example, if $θ$ is between $\frac{π}{4}$ and $\frac{π}{2}$ , $2 θ$ is between $\frac{π}{2}$ and $π$ , 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

If $sin x = \frac{2}{3}$ and $\frac{π}{2} < x < π$ , what is the value of $tan (2 x)$ ?
Options: A) $- \frac{4 5}{9}$ , B) $- 45$ , C) $\frac{4 5}{5}$ , D) $45$

1
Find $cos x$ using the Pythagorean identity. Since $x$ is in the second quadrant, $cos x$ is negative:
$\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \frac{4}{9}} = -\frac{\sqrt{5}}{3}$
2
Calculate $tan x$ :
$\tan x = \frac{\sin x}{\cos x} = \frac{2/3}{-\sqrt{5}/3} = -\frac{2}{\sqrt{5}}$
3
Substitute into the double-angle identity for tangent:
$\tan(2x) = \frac{2\tan x}{1 - \tan^2 x} = \frac{2(-2/\sqrt{5})}{1 - 4/5} = \frac{-4/\sqrt{5}}{1/5} = -4\sqrt{5}$
4
The correct answer is Option B.

📐 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 $2 x$ .
(b) Find the exact value of $f (\frac{5 π}{12})$ .
(c) Verify that $f (x) = 2 sin (2 x + \frac{π}{4})$ .

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

5. Common Pitfalls

Wrong move:

Writing $sin^{2} θ = sin (θ^{2})$ (interpreting the exponent notation incorrectly)

Why:

Confusion between standard trig exponent notation ( $sin^{2} θ = (sin θ)^{2}$ ) and argument squaring ( $sin (θ^{2})$ is a completely different function)

Correct move:

Always remember that the exponent is applied to the entire trig function, not the angle argument. Write $(sin θ)^{2}$ if you are unsure about notation.

Wrong move:

Writing $cos (α + β) = cos α cos β + sin α sin β$ with the wrong sign on the middle term

Why:

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

Correct move:

Use the "sine keeps, cosine flips" mnemonic every time you expand a sum/difference identity to confirm the middle term sign.

Wrong move:

Factoring a constant out of the trig argument to write $sin (2 θ) = 2 sin θ$

Why:

Confusion between linear function properties and non-linear trigonometric functions

Correct move:

Never factor a constant out of a trigonometric argument—always use the appropriate double-angle identity for doubling an angle.

Wrong move:

Forgetting to change the sign of an unknown trig function for $θ$ in Q2/Q3/Q4, leading to incorrect final values

Why:

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

Correct move:

Always assign the sign of the unknown trig function based on the given quadrant before substituting into any identity.

Wrong move:

Dividing both sides of a trigonometric equation by $sin θ$ (or another trig function) when simplifying, leading to missing solutions

Why:

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

Correct move:

Always move all terms to one side of the equation and factor out the common trig term instead of dividing.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Core Pythagorean	$sin^{2} θ + cos^{2} θ = 1$	Holds for all real $θ$
Pythagorean (tangent-secant)	$1 + tan^{2} θ = sec^{2} θ$	Valid for $θ \neq = \frac{π}{2} + k π$
Pythagorean (cotangent-cosecant)	$1 + cot^{2} θ = csc^{2} θ$	Valid for $θ \neq = k π$
Sine Sum/Difference	$sin (α \pm β) = sin α cos β \pm cos α sin β$	Middle sign matches argument sign
Cosine Sum/Difference	$cos (α \pm β) = cos α cos β \mp sin α sin β$	Middle sign flips from argument sign
Tangent Sum/Difference	$tan (α \pm β) = \frac{t a n α \pm t a n β}{1 \mp t a n α t a n β}$	Denominator sign also flips
Double-Angle Sine	$sin (2 θ) = 2 sin θ cos θ$	Works for all real $θ$
Double-Angle Cosine	$cos (2 θ) = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$	Use alternative forms for power reduction
Double-Angle Tangent	$tan (2 θ) = \frac{2 t a n θ}{1 - t a n ^{2} θ}$	Undefined when $1 - tan^{2} θ = 0$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
Find exact value using double angle
2023 · AP Precalculus
Simplify expression with identities

What's Next

This topic is the foundation for all advanced trigonometric topics in AP Precalculus Unit 3. Immediately after mastering these identities, you will use them to simplify inverse trigonometric compositions, solve complex trigonometric equations, and rewrite polar and parametric trigonometric functions for graphing and analysis. Without being able to quickly and correctly apply these identities, you will struggle with harmonic form conversion, power reduction, and the calculus preview topics that appear on the AP exam. This topic also provides the core trigonometric tools needed to convert between rectangular and polar coordinates and analyze the graphs of polar functions, a major weighted component of Unit 3.