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

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Pythagorean identities, angle sum and difference identities, double-angle identities, simplifying trigonometric expressions, verifying identities, evaluating trigonometric functions, and solving trigonometric equations aligned to Unit 3 learning objectives.

You should already know: Unit circle definitions of sine, cosine, and tangent. Basic algebraic manipulation of rational and polynomial expressions. Special angle trigonometric values.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Trigonometric identities (Pythagorean, sum/difference, double-angle)?

Trigonometric identities are equalities that hold for all input values where both sides of the equation are defined, used to rewrite trigonometric expressions in simpler or more useful forms for problem-solving. This topic is core to AP Precalculus Unit 3: Trigonometric and Polar Functions, accounting for approximately 4-6% of the total exam weight per the official College Board CED. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often as a required intermediate step in multi-step problems involving graphing periodic functions, solving equations, or modeling real-world periodic phenomena. Unlike conditional trigonometric equations that are only true for specific inputs, identities can be used freely to substitute and rewrite expressions. This topic groups three core, frequently tested families of identities: Pythagorean identities from the unit circle, sum/difference identities for combining angles, and double-angle identities as a special case of sum identities for doubling an angle.

2. Pythagorean Identities

The Pythagorean identities are the most fundamental and frequently used trigonometric identities, derived directly from the Pythagorean theorem applied to the unit circle. For any angle $\theta$, the point on the unit circle corresponding to $\theta$ has coordinates $(\cos \theta ,\sin \theta )$, so by the Pythagorean theorem $x^2+y^2=1$, giving the core identity: $${\sin }^2\theta +{\cos }^2\theta =1$$ Two additional related identities are derived by dividing both sides of the core identity by ${\cos }^2\theta$ or ${\sin }^2\theta$ respectively. Dividing by ${\cos }^2\theta$ (for all $\theta$ where cosine is non-zero) gives: $${\tan }^2\theta +1={\sec }^2\theta$$ Dividing by ${\sin }^2\theta$ (for all $\theta$ where sine is non-zero) gives: $$1+{\cot }^2\theta ={\csc }^2\theta$$ These identities are used to convert between sine and cosine, eliminate one trig function from an expression, or find the value of one trig function given another when the quadrant of the angle is known.

Worked Example

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

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

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.

3. Sum and Difference Identities

Sum and difference identities allow you to find the sine, cosine, or tangent of the sum or difference of two angles ($\alpha \pm \beta$) when you already know the trigonometric values of $\alpha$ and $\beta$ individually. They are also used to simplify expressions, verify more complex identities, and solve trigonometric equations. The core identities for sine and cosine are: $$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$$ $$\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$$ $$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$$ $$\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$$ For tangent, dividing the sine sum identity by the cosine sum identity and simplifying gives: $$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta },\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$$ A simple mnemonic to remember the signs: "sine keeps the sign, cosine flips the sign"—the sign of the middle term in the expanded identity matches the sign in the angle argument for sine, and is flipped for cosine.

Worked Example

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

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: $\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$, so $\cos (45^{\circ }-30^{\circ })=\cos 45^{\circ }\cos 30^{\circ }+\sin 45^{\circ }\sin 30^{\circ }$.
3. Substitute the known exact special angle values: $\cos 45^{\circ }=\frac{\sqrt{2}}{2}$, $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$, $\sin 45^{\circ }=\frac{\sqrt{2}}{2}$, $\sin 30^{\circ }=\frac{1}{2}$.
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: $\cos (90^{\circ }-30^{\circ })=\sin 30^{\circ }=0.5$, which will only match if the sign is positive for cosine of a difference, reinforcing the flip rule.

4. Double-Angle Identities

Double-angle identities are a special case of the sum identities, where $\alpha =\beta =\theta$, so we calculate the trigonometric value of $2\theta$ from the values for $\theta$. They are used extensively for simplifying products of trig functions, reducing the power of squared trig terms, rewriting functions for graphing, and solving trigonometric equations. Starting from the sine sum identity with $\alpha =\beta =\theta$: $$\sin (\theta +\theta )=\sin \theta \cos \theta +\cos \theta \sin \theta =2\sin \theta \cos \theta$$ giving the double-angle identity for sine: $$\sin (2\theta )=2\sin \theta \cos \theta$$ For cosine, starting from the cosine sum identity: $$\cos (\theta +\theta )=\cos \theta \cos \theta -\sin \theta \sin \theta$$ so the base form is: $$\cos (2\theta )={\cos }^2\theta -{\sin }^2\theta$$ Using the Pythagorean identity to substitute for ${\sin }^2\theta$ or ${\cos }^2\theta$, we get two alternative extremely useful forms: $$\cos (2\theta )=2{\cos }^2\theta -1\text{and}\cos (2\theta )=1-2{\sin }^2\theta$$ For tangent, we get: $$\tan (2\theta )=\frac{2\tan \theta }{1-{\tan }^2\theta }$$ The alternative forms for cosine are most often used for power reduction, rewriting squared sine or cosine in terms of first-power cosine of a double angle, a common AP exam question.

Worked Example

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

1. The double-angle identity for sine requires both $\sin \theta$ and $\cos \theta$; we know $\cos \theta =\frac{1}{3}$, so we first find $\sin \theta$.
2. Use the core Pythagorean identity to solve for $\sin \theta$: ${\sin }^2\theta +{\left(\frac{1}{3}\right)}^2=1⟹{\sin }^2\theta =\frac{8}{9}$. Since $\theta$ is in the first quadrant, $\sin \theta$ is positive, so $\sin \theta =\frac{2\sqrt{2}}{3}$.
3. Substitute into the double-angle identity: $\sin (2\theta )=2\sin \theta \cos \theta =2\left(\frac{2\sqrt{2}}{3}\right)\left(\frac{1}{3}\right)=\frac{4\sqrt{2}}{9}$.
4. Verify with the Pythagorean identity for $2\theta$: $\cos (2\theta )=2{\left(\frac{1}{3}\right)}^2-1=-\frac{7}{9}$, so ${\sin }^2(2\theta )+{\cos }^2(2\theta )=\frac{32}{81}+\frac{49}{81}=1$, which confirms the result is correct.

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, so always check this before finalizing your answer.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing ${\sin }^2\theta =\sin ({\theta }^2)$ (interpreting the exponent notation incorrectly). Why: Confusion between standard trig exponent notation (${\sin }^2\theta =(\sin \theta {)}^2$) and argument squaring ($\sin ({\theta }^2)$ is a completely different function). Correct move: Always remember that the exponent is applied to the entire trig function, not the angle argument, and write $(\sin \theta {)}^2$ if you are unsure about notation.
• Wrong move: Writing $\cos (\alpha +\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ with the wrong sign on the middle term. Why: Students often copy the same sign rule as sine, leading to errors. Correct move: Use the "sine keeps, cosine flips" mnemonic every time you expand a sum/difference identity to confirm the sign.
• Wrong move: Factoring a constant out of the trig argument to write $\sin (2\theta )=2\sin \theta$. 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 $\sin \theta$ for $\theta$ in Q2, leading to a positive $\sin (2\theta )$ when it should be negative. Why: Students remember the Pythagorean identity but ignore 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 $\sin \theta$ when simplifying, leading to missing solutions where $\sin \theta =0$. 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. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

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

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

Question 2 (Free Response)

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 $2x$, with no squared terms or products. (b) Find the exact value of $f\left(\frac{5\pi }{12}\right)$ using your rewritten expression. (c) Verify that $f(x)=\sqrt{2}\sin \left(2x+\frac{\pi }{4}\right)$ using a sum identity.

Worked Solution: (a) Using double-angle identities, ${\cos }^{2}x-{\sin }^{2}x=\cos (2x)$ and $2\sin x\cos x=\sin (2x)$. So $f(x)=\cos (2x)+\sin (2x)$. (b) For $x=\frac{5\pi }{12}$, $2x=\frac{5\pi }{6}$. Substitute: $f\left(\frac{5\pi }{12}\right)=\cos \left(\frac{5\pi }{6}\right)+\sin \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{1-\sqrt{3}}{2}$. (c) Expand the right-hand side with the sine sum identity: $$\begin{align*} \sqrt{2} \sin\left(2x + \frac{\pi}{4}\right) &= \sqrt{2}\left[\sin(2x)\cos\left(\frac{\pi}{4}\right) + \cos(2x)\sin\left(\frac{\pi}{4}\right)\right] \ &= \sqrt{2}\left[\sin(2x) \cdot \frac{\sqrt{2}}{2} + \cos(2x) \cdot \frac{\sqrt{2}}{2}\right] \ &= \sqrt{2} \cdot \frac{\sqrt{2}}{2}\left(\sin 2x + \cos 2x\right) = \sin 2x + \cos 2x = f(x) \end{align*}$$ The identity is verified.

Question 3 (Application / Real-World Style)

The pressure variation of two overlapping identical sound waves is given by $P(t)=24\sin (250\pi t)\cos (250\pi t)$, where $P$ is pressure in pascals and $t$ is time in seconds. (a) Rewrite $P(t)$ as a single sine function with no product of trigonometric functions. (b) Find the frequency of the combined wave in cycles per second (for a wave $A\sin (\omega t)$, frequency $f=\frac{\omega }{2\pi }$).

Worked Solution: (a) Use the double-angle identity for sine: $\sin (2\theta )=2\sin \theta \cos \theta$, so $\sin \theta \cos \theta =\frac{1}{2}\sin (2\theta )$. Substitute $\theta =250\pi t$: $$P(t)=24\cdot \frac{1}{2}\sin \left(2\cdot 250\pi t\right)=12\sin (500\pi t)$$ (b) Calculate frequency: $f=\frac{\omega }{2\pi }=\frac{500\pi }{2\pi }=250$ cycles per second. Interpretation: The combined wave from two overlapping 125 Hz sound waves has a frequency of 250 Hz, twice the frequency of the original individual waves.

7. Quick Reference Cheatsheet

8. 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. Follow-on topics to study next:

• Inverse Trigonometric Functions
• Solving Trigonometric Equations
• Polar Coordinates and Graphs
• Parametric Trigonometric Curves