MGU Mathematics Series — Trigonometry and Precalculus

Part II: Trigonometric Foundations

Identities · Verification · Equations · Applications

Chapter 6 of 14

MGU Mathematics Series · Trigonometry and Precalculus · Library Textbook Edition

Chapter 6:
Trigonometric Identities
and Equations

Reciprocal, quotient, Pythagorean, cofunction, sum, difference, double-angle, and half-angle identities — as tools for rewriting, verifying, and solving

■ Part II — Trigonometric Foundations ■

■ 14 Sections • Chapter 6 of 14 ■

Follows Chapter 5: Graphs of Trigonometric Functions · Prepares students for Chapter 7: Triangle Trigonometry and Applications

Chapter 6 — Front Matter

Purpose and Overview

In the chapters before this one, you learned what trigonometric functions are, how they are defined on the unit circle, and how their graphs behave. You can now evaluate sin(θ), cos(θ), and tan(θ) at special angles, interpret amplitude and period, and graph periodic functions. This chapter asks a deeper question: what happens when two different-looking trigonometric expressions turn out to be equal for every allowed angle?

A trigonometric identity is an equation between trigonometric expressions that is true for every value of the variable where both sides are defined. Identities are not equations to solve — they are tools for rewriting. They let you replace a complicated expression with a simpler one, convert between different trigonometric functions, simplify expressions before applying calculus, and solve equations that would otherwise be intractable.

This chapter builds the identity toolkit in layers. It begins with the most fundamental identities — those that follow directly from the unit circle definitions — and builds toward sum and difference formulas, double-angle and half-angle identities, and their applications in solving equations. Throughout, the emphasis is on strategy and meaning, not on memorizing a list of formulas. You should understand why each identity is true and how to choose which identity to apply.

Opening Question

A physics instructor writes the expression (1 − cos²(θ)) / sin(θ) on the board and asks students to simplify it as much as possible. One student stares at it, unsure where to begin. Another student recognizes that 1 − cos²(θ) equals sin²(θ) by a Pythagorean identity, rewrites the expression as sin²(θ) / sin(θ), and simplifies to sin(θ) in two steps. The ability to recognize and apply identities is what separates a fluent problem-solver from one who is stuck. By the end of this chapter, you will have that fluency.

14Sections

13Objectives

18Key Terms

20+Worked Examples

Learning Objectives

1Explain what a trigonometric identity is and distinguish it from a conditional equation.
2State and apply the reciprocal identities.
3State and apply the quotient identities.
4State and apply the three Pythagorean identities and their alternate forms.
5State and apply the even and odd identities for sine, cosine, and tangent.
6State and apply the cofunction identities.
7Simplify trigonometric expressions using combinations of identities.
8Verify trigonometric identities by transforming one side into the other.
9State and apply the sum and difference identities for sine, cosine, and tangent.
10State and apply the double-angle identities.
11State and apply the half-angle identities.
12Solve basic trigonometric equations using inverse functions.
13Solve trigonometric equations on a specified interval, including equations requiring identity substitution.

Sections at a Glance

6.1What a Trigonometric Identity IsIdentity vs. conditional equation; domain of validity; checking by substitution; why identities are tools, not problems to solve
6.2Reciprocal Identitiescsc, sec, cot as reciprocals; deriving from unit circle definitions; using reciprocals to simplify
6.3Quotient Identitiestan = sin/cos; cot = cos/sin; connection to unit circle; why these follow from definitions
6.4Pythagorean IdentitiesThree Pythagorean identities; deriving from x² + y² = 1; alternate forms; strategic rewriting
6.5Even and Odd IdentitiesEven functions: cos(−θ) = cos(θ); odd functions: sin(−θ) = −sin(θ), tan(−θ) = −tan(θ); graph symmetry connection
6.6Cofunction IdentitiesComplementary angles; sin and cos, tan and cot, sec and csc cofunctions; geometric origin
6.7Simplifying Trigonometric ExpressionsStrategy: convert to sin/cos, factor, combine fractions, apply Pythagorean identities; worked simplifications
6.8Verifying Trigonometric IdentitiesRules of verification; working one side only; choosing the more complex side; strategy overview
6.9Sum and Difference Identitiessin(A ± B), cos(A ± B), tan(A ± B); deriving exact values at non-special angles; applications
6.10Double-Angle Identitiessin(2θ), cos(2θ) in three forms, tan(2θ); deriving from sum identities; applications in calculus and physics
6.11Half-Angle Identitiessin(θ/2), cos(θ/2), tan(θ/2); sign determined by quadrant; using half-angle identities for exact values
6.12Solving Basic Trigonometric EquationsInverse functions; principal values; all solutions; using periodicity; isolating the trig function first
6.13Solving Trigonometric Equations on IntervalsRestricting to [0, 2π); using identities to reduce; quadratic-type equations; multiple angle equations
6.14Common Identity and Equation MistakesSix recurring errors with corrections; prevention strategies; checking answers

Section 6.1

What a Trigonometric Identity Is

An equation asserts that two expressions are equal for some values of the variable. A trigonometric identity asserts that two expressions are equal for every value of the variable in the domain where both expressions are defined. This distinction is not minor. When you solve an equation, you are finding specific values. When you work with an identity, you are establishing a universal truth.

For example, consider the equation sin(θ) = 0. This is true only when θ = 0, π, 2π, and so on — specific values. Now consider the equation sin²(θ) + cos²(θ) = 1. This is true for every real number θ. No matter what angle you choose, the sum of the squares of its sine and cosine equals 1. That is an identity.

Identities are useful precisely because they allow you to rewrite an expression in a different but equivalent form. In calculus, this lets you simplify integrands before integrating. In physics, it converts between representations of waves. In problem-solving, it replaces a form you cannot work with for one that you can.

Trigonometric Identity — Definition

A trigonometric identity is an equation involving trigonometric functions that is true for all values of the variable for which both sides are defined.

A conditional equation is an equation that is true only for certain specific values. Solving a conditional equation means finding those values.

When working with an identity, your goal is not to find θ — it is to show that two expressions are always equal, or to rewrite one expression in terms of another.

Example 6.1.1

Distinguishing an identity from a conditional equation

ProblemDetermine whether each equation is an identity or a conditional equation. (a) 2 sin(θ) = 1 (b) sin²(θ) + cos²(θ) = 1
ThinkTest whether each equation holds for all values or only some values.
Analyze (a)Try θ = π/6: sin(π/6) = 1/2, so 2(1/2) = 1. True. Try θ = π/4: sin(π/4) = √2/2, so 2(√2/2) = √2 ≠ 1. False for θ = π/4. This is a conditional equation — it is only true for certain θ.
Analyze (b)Try θ = 0: 0 + 1 = 1. Try θ = π/4: (√2/2)² + (√2/2)² = 1/2 + 1/2 = 1. Try θ = π/3: (√3/2)² + (1/2)² = 3/4 + 1/4 = 1. The sum is always 1 — this follows from the unit circle definition and is an identity.

Answer: (a) is a conditional equation. (b) is an identity.

Checking an Identity

To test whether an equation might be an identity, substitute a specific value. If the equation fails at even one point in the domain, it is not an identity. If it holds at every test value, it may be an identity — but you still need to prove it algebraically, since finitely many test cases cannot confirm universal truth.

Practice 6.1Concept Check

In your own words, explain the difference between a trigonometric identity and a conditional trigonometric equation. Give one example of each.
For each equation, substitute θ = π/6 and θ = π/4 to determine whether it could be an identity. Do not prove it formally.

(a) cos(θ) = sin(θ) (b) tan(θ) · cos(θ) = sin(θ) (c) sin(θ) + cos(θ) = 1
Why is substituting one specific value not enough to prove an identity is true?

Section 6.2

Reciprocal Identities

The six trigonometric functions are not six independent objects. Three of them — cosecant, secant, and cotangent — are simply the reciprocals of the other three. This is not an accident or a convention; it follows directly from how all six functions are defined on the unit circle.

Recall from Chapter 4: at an angle θ on the unit circle, the point on the circle has coordinates (cos θ, sin θ). The six functions were defined as: sin θ = y-coordinate, cos θ = x-coordinate, tan θ = y/x (when x ≠ 0), csc θ = 1/y (when y ≠ 0), sec θ = 1/x (when x ≠ 0), and cot θ = x/y (when y ≠ 0).

The reciprocal relationships follow immediately:

Reciprocal Identities

csc(θ) = 1 / sin(θ) (valid when sin(θ) ≠ 0)

sec(θ) = 1 / cos(θ) (valid when cos(θ) ≠ 0)

cot(θ) = 1 / tan(θ) (valid when tan(θ) ≠ 0)

Equivalently: sin(θ) · csc(θ) = 1 cos(θ) · sec(θ) = 1 tan(θ) · cot(θ) = 1

Example 6.2.1

Applying reciprocal identities to simplify

ProblemSimplify: sin(θ) · sec(θ)
ThinkReplace sec(θ) with its reciprocal identity: sec(θ) = 1/cos(θ).
Solve sin(θ) · sec(θ) = sin(θ) · (1/cos(θ)) = sin(θ)/cos(θ) = tan(θ)
CheckAt θ = π/4: sin(π/4)·sec(π/4) = (√2/2)·√2 = 1 = tan(π/4). ✓

Answer: tan(θ)

Example 6.2.2

Finding the value of a reciprocal function

ProblemIf sin(θ) = 3/5, find csc(θ).
Solve csc(θ) = 1/sin(θ) = 1/(3/5) = 5/3

Answer: csc(θ) = 5/3

Practice 6.2Reciprocal Identities

If cos(θ) = 4/5, find sec(θ).
If tan(θ) = 2, find cot(θ).
Simplify: cos(θ) · csc(θ)
Simplify: tan(θ) · csc(θ)
Simplify: cot(θ) · sin(θ)
Simplify: sec(θ) · sin(θ)

Section 6.3

Quotient Identities

On the unit circle, sin(θ) is the y-coordinate and cos(θ) is the x-coordinate of the terminal point. Tangent was defined as the ratio y/x, which means tan(θ) = sin(θ)/cos(θ). Cotangent, the reciprocal of tangent, equals x/y = cos(θ)/sin(θ). These are not new definitions — they are consequences of definitions already established in Chapter 4.

Quotient Identities

tan(θ) = sin(θ) / cos(θ) (valid when cos(θ) ≠ 0)

cot(θ) = cos(θ) / sin(θ) (valid when sin(θ) ≠ 0)

Example 6.3.1

Applying the quotient identity

ProblemSimplify: (sin(θ) / cos(θ)) + cos(θ)/sin(θ)
ThinkRecognize that sin/cos = tan and cos/sin = cot. Alternatively, combine the fractions over a common denominator.
Solve sin(θ)/cos(θ) + cos(θ)/sin(θ) = (sin²(θ) + cos²(θ)) / (sin(θ)cos(θ)) = 1 / (sin(θ)cos(θ)) [Pythagorean identity — see 6.4]
NoteThis example previews how quotient and Pythagorean identities are often used together.

Answer: 1 / (sin(θ)cos(θ))

Practice 6.3Quotient Identities

Express tan(θ) using only sin and cos.
Express cot(θ) using only sin and cos.
Simplify: cos(θ) · tan(θ)
Simplify: sin(θ) · cot(θ)
If sin(θ) = 5/13 and cos(θ) = 12/13, find tan(θ) and cot(θ).
Simplify: (sin(θ) · cot(θ)) / cos(θ)

Section 6.4

Pythagorean Identities

The Pythagorean identities are arguably the most important identities in all of trigonometry. They express the fundamental geometric constraint that every point on the unit circle satisfies: the sum of the squares of the x- and y-coordinates always equals 1. Everything else follows from there.

Recall the unit circle: for any angle θ, the terminal point is (cos θ, sin θ), and since this point lies on a circle of radius 1, the Pythagorean Theorem gives:

The Three Pythagorean Identities

Identity 1: sin²(θ) + cos²(θ) = 1

Identity 2: 1 + tan²(θ) = sec²(θ) (divide Identity 1 by cos²(θ))

Identity 3: 1 + cot²(θ) = csc²(θ) (divide Identity 1 by sin²(θ))

Each Pythagorean identity has useful alternate forms obtained by rearranging. Memorizing these alternate forms speeds up simplification considerably:

From Identity 1:sin²(θ) = 1 − cos²(θ)

From Identity 1:cos²(θ) = 1 − sin²(θ)

From Identity 2:tan²(θ) = sec²(θ) − 1

From Identity 3:cot²(θ) = csc²(θ) − 1

Example 6.4.1

Deriving the second Pythagorean identity

ProblemShow that 1 + tan²(θ) = sec²(θ) follows from sin²(θ) + cos²(θ) = 1.
ThinkDivide every term of sin²(θ) + cos²(θ) = 1 by cos²(θ), assuming cos(θ) ≠ 0.
Solve sin²(θ)/cos²(θ) + cos²(θ)/cos²(θ) = 1/cos²(θ) tan²(θ) + 1 = sec²(θ) ∴ 1 + tan²(θ) = sec²(θ) ✓

Answer: The second Pythagorean identity follows directly from dividing the first by cos²(θ).

Example 6.4.2

Using a Pythagorean identity to simplify

ProblemSimplify: (1 − cos²(θ)) / sin(θ)
ThinkRecognize that 1 − cos²(θ) = sin²(θ) by the alternate form of Identity 1.
Solve (1 − cos²(θ)) / sin(θ) = sin²(θ) / sin(θ) = sin(θ)
CheckAt θ = π/6: (1 − 3/4)/(1/2) = (1/4)/(1/2) = 1/2 = sin(π/6). ✓

Answer: sin(θ)

Example 6.4.3

Finding a missing function value using a Pythagorean identity

ProblemIf sin(θ) = −3/5 and θ is in Quadrant III, find cos(θ) and tan(θ).
ThinkUse sin²(θ) + cos²(θ) = 1. In Quadrant III, cos(θ) is negative.
Solve (−3/5)² + cos²(θ) = 1 9/25 + cos²(θ) = 1 cos²(θ) = 16/25 cos(θ) = −4/5 (negative in QIII) tan(θ) = sin(θ)/cos(θ) = (−3/5)/(−4/5) = 3/4

Answer: cos(θ) = −4/5, tan(θ) = 3/4

Student Warning — Pythagorean Identity Common Error

Students sometimes write sin²(θ) + cos²(θ) = 1 but then misapply it. The identity says the sum of the squares equals 1 — it does not say sin(θ) + cos(θ) = 1. Those two expressions are completely different. At θ = π/4, sin(θ) + cos(θ) = √2 ≈ 1.414, not 1.

Practice 6.4Pythagorean Identities

Simplify: 1 − sin²(θ)
Simplify: sec²(θ) − 1
Simplify: cos²(θ)(1 + tan²(θ))
Simplify: sin²(θ) + cos²(θ) + tan²(θ)
If cos(θ) = 5/13, θ in QI, find sin(θ) and tan(θ).
If tan(θ) = −2, θ in QII, find sec(θ).
Simplify: (sec²(θ) − 1) · cos²(θ)
Simplify: sin⁴(θ) − cos⁴(θ) [Hint: factor as a difference of squares first.]

Section 6.5

Even and Odd Identities

In Chapter 2, you learned that an even function satisfies f(−x) = f(x) — its graph is symmetric about the y-axis. An odd function satisfies f(−x) = −f(x) — its graph has rotational symmetry about the origin. These properties apply to trigonometric functions as well, and they give us useful identities for negative angles.

Look at the unit circle. The point corresponding to angle θ has coordinates (cos θ, sin θ). The point corresponding to angle −θ is the reflection of this point across the x-axis. Its coordinates are (cos θ, −sin θ). This means cos(−θ) = cos(θ) — cosine is even — and sin(−θ) = −sin(θ) — sine is odd. Tangent, as the ratio of sine to cosine, is also odd.

Even and Odd Identities

Cosine is even: cos(−θ) = cos(θ)

Sine is odd: sin(−θ) = −sin(θ)

Tangent is odd: tan(−θ) = −tan(θ)

The reciprocal functions follow: sec is even, csc is odd, cot is odd.

Example 6.5.1

Evaluating trigonometric functions of negative angles

ProblemFind exact values of: (a) sin(−π/6) (b) cos(−π/3) (c) tan(−π/4)
Solve (a) sin(−π/6) = −sin(π/6) = −1/2 (b) cos(−π/3) = cos(π/3) = 1/2 (c) tan(−π/4) = −tan(π/4) = −1

Answer: (a) −1/2 (b) 1/2 (c) −1

Example 6.5.2

Simplifying with even/odd identities

ProblemSimplify: sin(−θ) · sec(−θ)
ThinkApply even/odd identities: sin(−θ) = −sin(θ) and sec(−θ) = sec(θ) (since sec is even). Then simplify using reciprocal identities.
Solve sin(−θ) · sec(−θ) = (−sin(θ)) · sec(θ) = −sin(θ) · (1/cos(θ)) = −sin(θ)/cos(θ) = −tan(θ)

Answer: −tan(θ)

Practice 6.5Even and Odd Identities

sin(−π/4)
cos(−2π/3)
tan(−π/3)
csc(−π/6)
Simplify: cos(−θ) · tan(−θ)
Simplify: sin(−θ) + sin(θ)
Simplify: cos(−θ) + cos(θ)
If sin(θ) = 0.6, find sin(−θ) and cos(−θ) given cos(θ) = 0.8.

Section 6.6

Cofunction Identities

The word "cofunction" contains a clue: cosine is the cofunction of sine, cotangent is the cofunction of tangent, and cosecant is the cofunction of secant. Two functions are cofunctions if their values at complementary angles are equal. Complementary angles sum to 90° or π/2 radians.

This relationship has geometric roots. In a right triangle with angles α and β = (π/2 − α), the side opposite α is the same side that is adjacent to β. So sin(α) — which involves the side opposite α — equals cos(β) = cos(π/2 − α). This is where the name "cosine" comes from: cosine of angle α means sine of the complementary angle.

Cofunction Identities

sin(θ) = cos(π/2 − θ) and cos(θ) = sin(π/2 − θ)

tan(θ) = cot(π/2 − θ) and cot(θ) = tan(π/2 − θ)

sec(θ) = csc(π/2 − θ) and csc(θ) = sec(π/2 − θ)

In degrees: replace π/2 with 90°.

Example 6.6.1

Using cofunction identities

ProblemFind the value: (a) sin(π/3) using cosine (b) cos(20°) using sine
Solve (a) sin(π/3) = cos(π/2 − π/3) = cos(π/6) = √3/2
Solve (b) cos(20°) = sin(90° − 20°) = sin(70°)

Answer: (a) √3/2 (b) sin(70°)

Cofunction Identities in Calculus

These identities appear frequently in calculus when derivatives and integrals of trigonometric functions are computed. Knowing that sin and cos are cofunctions, and tan and cot are cofunctions, helps you recognize transformed expressions quickly.

Practice 6.6Cofunction Identities

Write sin(π/4) in terms of cosine.
Write cos(π/6) in terms of sine.
Write tan(40°) in terms of cotangent.
Write csc(π/5) in terms of secant.
Simplify: sin(π/2 − θ) · sec(θ)
Simplify: cos(π/2 − θ) · csc(θ)
If sin(25°) = 0.423, find cos(65°).
Simplify: tan(π/2 − θ) · tan(θ)

Section 6.7

Simplifying Trigonometric Expressions

Simplifying a trigonometric expression means rewriting it in an equivalent form that is shorter, more uniform, or more useful. Unlike verification, where you are asked to prove that two specific expressions are equal, simplification is open-ended: you are looking for the "simplest" form, which usually means an expression involving fewer functions, fewer fractions, or only sines and cosines.

There is no single algorithm for simplification, but the following strategies are broadly effective:

Simplification Strategy

Strategy 1: Convert everything to sin and cos using reciprocal and quotient identities. This unifies mixed expressions.

Strategy 2: Apply a Pythagorean identity when you see sin²(θ) + cos²(θ), 1 − cos²(θ), 1 − sin²(θ), sec²(θ) − 1, or csc²(θ) − 1.

Strategy 3: Factor when the expression contains two terms that share a factor, or when you can group into a product.

Strategy 4: Combine fractions over a common denominator when the expression contains addition or subtraction of fractions.

Strategy 5: Multiply by a conjugate when you see 1 ± sin(θ), 1 ± cos(θ), or similar expressions in a denominator.

Strategy 6: After each step, look for new patterns — a Pythagorean identity that now applies, or a cancellation that is now visible.

Example 6.7.1

Convert to sin and cos, then simplify

ProblemSimplify: cot(θ) · sin(θ) + cos(θ)
ThinkReplace cot(θ) with cos(θ)/sin(θ). Simplify each term.
Solve cot(θ) · sin(θ) + cos(θ) = (cos(θ)/sin(θ)) · sin(θ) + cos(θ) = cos(θ) + cos(θ) = 2cos(θ)
CheckAt θ = π/4: cot(π/4)·sin(π/4) + cos(π/4) = 1·(√2/2) + √2/2 = √2 = 2cos(π/4). ✓

Answer: 2cos(θ)

Example 6.7.2

Apply a Pythagorean identity after factoring

ProblemSimplify: sin²(θ)·sec²(θ) − sin²(θ)
ThinkFactor out sin²(θ), then apply a Pythagorean identity.
Solve sin²(θ)·sec²(θ) − sin²(θ) = sin²(θ)(sec²(θ) − 1) = sin²(θ) · tan²(θ) [since sec²(θ)−1 = tan²(θ)] = sin²(θ) · sin²(θ)/cos²(θ) = sin⁴(θ)/cos²(θ) = sin⁴(θ) · sec²(θ)

Answer: sin⁴(θ)·sec²(θ) or equivalently sin⁴(θ)/cos²(θ)

Example 6.7.3

Combine fractions over a common denominator

ProblemSimplify: 1/(1 − sin(θ)) − 1/(1 + sin(θ))
ThinkSubtract fractions using the common denominator (1 − sin(θ))(1 + sin(θ)) = 1 − sin²(θ) = cos²(θ).
Solve 1/(1−sin(θ)) − 1/(1+sin(θ)) = [(1+sin(θ)) − (1−sin(θ))] / [(1−sin(θ))(1+sin(θ))] = [2sin(θ)] / [1 − sin²(θ)] = 2sin(θ) / cos²(θ) = 2 · (sin(θ)/cos(θ)) · (1/cos(θ)) = 2tan(θ)sec(θ)

Answer: 2tan(θ)sec(θ)

Practice 6.7Simplifying Expressions

Simplify each expression as much as possible. Show all steps.

tan(θ) · cos(θ)
sec(θ) · cos²(θ)
sin(θ) · csc(θ) − cos²(θ)
(1 − cos²(θ)) · csc(θ)
tan²(θ) · csc²(θ)
sec²(θ) − sec²(θ)·sin²(θ)
sin(θ)/cos(θ) + cos(θ)/sin(θ)
(cos(θ) − sin(θ))(cos(θ) + sin(θ))
1/(cos(θ)) − cos(θ)
(1 − cos²(θ))/tan²(θ)

Section 6.8

Verifying Trigonometric Identities

Verifying a trigonometric identity means proving that two expressions are equal for all values in their domain. This is different from simplifying — here, you are given both sides and asked to show they match. The central rule of verification is this: work on one side only. Do not move terms across the equals sign, do not add the same thing to both sides, and do not multiply both sides by the same factor. Any of these operations assumes what you are trying to prove, which makes the argument circular.

Choose the side that appears more complex — usually the one with more terms, fractions, or mixed functions — and transform it until it looks exactly like the other side.

Rules for Verifying Identities

Rule 1: Work on one side only. Choose the more complex side.

Rule 2: Use known identities — reciprocal, quotient, Pythagorean, even/odd, cofunction — to rewrite the working side.

Rule 3: Algebraic tools — factoring, combining fractions, canceling common factors — are valid throughout.

Rule 4: Do not cross the equals sign. Do not add, subtract, multiply, or divide both sides.

Rule 5: Show every step clearly. The goal is a transparent proof, not just a correct answer.

Example 6.8.1

Verify a basic identity

ProblemVerify: tan(θ) · cos(θ) = sin(θ)
ThinkWork on the left side. Replace tan(θ) using the quotient identity.
Left Side tan(θ) · cos(θ) = (sin(θ)/cos(θ)) · cos(θ) = sin(θ) ✓

The left side simplifies to sin(θ), which equals the right side. Identity verified.

Example 6.8.2

Verify using Pythagorean identity and factoring

ProblemVerify: (sin²(θ) − 1) / sin(θ) = −cos(θ) · cot(θ)
ThinkWork on the left side. The numerator has the form sin²(θ) − 1 = −(1 − sin²(θ)) = −cos²(θ).
Left Side (sin²(θ) − 1) / sin(θ) = −cos²(θ) / sin(θ) = −cos(θ) · (cos(θ)/sin(θ)) = −cos(θ) · cot(θ) ✓

Identity verified.

Example 6.8.3

Verify using fraction manipulation

ProblemVerify: (1 + tan²(θ)) / tan²(θ) = csc²(θ)
ThinkWork on the left side. Separate the fraction or recognize 1 + tan²(θ) = sec²(θ).
Left Side (1 + tan²(θ)) / tan²(θ) = sec²(θ) / tan²(θ) = (1/cos²(θ)) / (sin²(θ)/cos²(θ)) = (1/cos²(θ)) · (cos²(θ)/sin²(θ)) = 1/sin²(θ) = csc²(θ) ✓

Identity verified.

Student Warning — What Not to Do When Verifying

A common mistake is to "verify" by writing the equation and then doing algebra on both sides simultaneously — adding to both sides, multiplying both sides, until both sides look the same. This is invalid because it assumes the conclusion. You must transform only one side using identities and algebra until it looks exactly like the other side.

Practice 6.8Verifying Identities

Verify each identity. Work on the more complex side. Show all steps.

cos(θ) · tan(θ) = sin(θ)
sin(θ) · sec(θ) = tan(θ)
csc(θ) · cos(θ) = cot(θ)
sin²(θ) · sec²(θ) + sin²(θ) · csc²(θ) = sec²(θ)
(1 − sin²(θ)) / cos(θ) = cos(θ)
sec²(θ) − sin²(θ) · sec²(θ) = 1
(csc(θ) − 1)(csc(θ) + 1) = cot²(θ)
(sin(θ) + cos(θ))² = 1 + 2sin(θ)cos(θ)
tan(θ) + cot(θ) = sec(θ)csc(θ)
cos⁴(θ) − sin⁴(θ) = cos²(θ) − sin²(θ)

Section 6.9

Sum and Difference Identities

One of the most natural questions in trigonometry is: if I know sin(A) and sin(B), can I find sin(A + B)? The answer is yes, but the formula is not sin(A) + sin(B). This is a critical point — the sine function is not additive. sin(A + B) is something more complex, involving both sine and cosine of A and B together.

The sum and difference identities are among the most powerful tools in the subject. They allow you to compute exact values at angles that are not standard unit-circle angles — for example, sin(75°) = sin(45° + 30°). They also serve as the foundation for deriving all double-angle and half-angle identities.

Sum and Difference Identities

Cosine:

cos(A + B) = cos(A)cos(B) − sin(A)sin(B)

cos(A − B) = cos(A)cos(B) + sin(A)sin(B)

Sine:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

sin(A − B) = sin(A)cos(B) − cos(A)sin(B)

Tangent:

tan(A + B) = (tan(A) + tan(B)) / (1 − tan(A)tan(B))

tan(A − B) = (tan(A) − tan(B)) / (1 + tan(A)tan(B))

Memory Aid for Sum and Difference Identities

For cosine, the sign between the terms flips: cos(A + B) uses minus, cos(A − B) uses plus. For sine, the sign matches the sum or difference. A common memory phrase: "cosine switches, sine matches."

Example 6.9.1

Exact value of sin(75°)

ProblemFind the exact value of sin(75°).
Think75° = 45° + 30°. Both are special angles. Use the sine sum identity.
Solve sin(75°) = sin(45° + 30°) = sin(45°)cos(30°) + cos(45°)sin(30°) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4

Answer: (√6 + √2)/4

Example 6.9.2

Exact value of cos(π/12)

ProblemFind the exact value of cos(π/12).
Thinkπ/12 = 15° = 45° − 30°. Use the cosine difference identity.
Solve cos(π/12) = cos(π/4 − π/6) = cos(π/4)cos(π/6) + sin(π/4)sin(π/6) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4

Answer: (√6 + √2)/4

Example 6.9.3

Verifying an identity using sum formulas

ProblemVerify: sin(θ + π/2) = cos(θ)
ThinkExpand the left side using the sine sum identity.
Left Side sin(θ + π/2) = sin(θ)cos(π/2) + cos(θ)sin(π/2) = sin(θ)(0) + cos(θ)(1) = cos(θ) ✓

Identity verified. This also explains the phase-shift relationship between sine and cosine graphs.

Example 6.9.4

Using a sum identity given function values

ProblemIf sin(A) = 3/5 (A in QI) and cos(B) = −5/13 (B in QII), find sin(A + B).
ThinkFind cos(A) and sin(B) using Pythagorean identities. Then apply the sine sum identity.
Find cos(A) A in QI → cos(A) > 0 cos(A) = √(1 − 9/25) = √(16/25) = 4/5
Find sin(B) B in QII → sin(B) > 0 sin(B) = √(1 − 25/169) = √(144/169) = 12/13
Solve sin(A + B) = sin(A)cos(B) + cos(A)sin(B) = (3/5)(−5/13) + (4/5)(12/13) = −15/65 + 48/65 = 33/65

Answer: sin(A + B) = 33/65

Practice 6.9Sum and Difference Identities

Find exact values or verify as directed. Show all steps.

Find sin(105°)
Find cos(15°)
Find tan(π/12)
Find sin(7π/12)
Find cos(195°)
Find tan(105°)
Verify: cos(π − θ) = −cos(θ)
Verify: sin(π − θ) = sin(θ)

For 9–10: use given values to find the requested quantity.

sin(A) = 5/13, A in QI; cos(B) = 3/5, B in QI. Find cos(A − B).
cos(A) = −4/5, A in QII; sin(B) = −12/13, B in QIII. Find sin(A + B).

Section 6.10

Double-Angle Identities

The double-angle identities express sin(2θ), cos(2θ), and tan(2θ) in terms of sin(θ) and cos(θ). They are not new identities discovered independently — they are direct consequences of the sum identities when A = B = θ.

Set A = B = θ in the sine sum identity: sin(θ + θ) = sin(θ)cos(θ) + cos(θ)sin(θ) = 2sin(θ)cos(θ). That is the sine double-angle identity. For cosine, the same substitution in the cosine sum identity yields three equivalent forms, each useful in different contexts.

Double-Angle Identities

Sine: sin(2θ) = 2sin(θ)cos(θ)

Cosine — three equivalent forms:

cos(2θ) = cos²(θ) − sin²(θ) (direct from sum identity)

cos(2θ) = 2cos²(θ) − 1 (substitute sin²(θ) = 1 − cos²(θ))

cos(2θ) = 1 − 2sin²(θ) (substitute cos²(θ) = 1 − sin²(θ))

Tangent: tan(2θ) = 2tan(θ) / (1 − tan²(θ))

Example 6.10.1

Deriving cos(2θ) = 2cos²(θ) − 1 from the first form

Start cos(2θ) = cos²(θ) − sin²(θ)
SubstituteReplace sin²(θ) with 1 − cos²(θ): = cos²(θ) − (1 − cos²(θ)) = cos²(θ) − 1 + cos²(θ) = 2cos²(θ) − 1 ✓

The second form follows from the first by a single substitution. Similarly, replacing cos²(θ) with 1 − sin²(θ) gives the third form: 1 − 2sin²(θ).

Example 6.10.2

Finding sin(2θ) and cos(2θ) given sin(θ)

ProblemIf sin(θ) = 3/5 and θ is in Quadrant I, find sin(2θ), cos(2θ), and tan(2θ).
Find cos(θ) cos(θ) = 4/5 (QI, positive)
sin(2θ) sin(2θ) = 2sin(θ)cos(θ) = 2(3/5)(4/5) = 24/25
cos(2θ) cos(2θ) = cos²(θ) − sin²(θ) = 16/25 − 9/25 = 7/25
tan(2θ) tan(2θ) = sin(2θ)/cos(2θ) = (24/25)/(7/25) = 24/7

Answer: sin(2θ) = 24/25, cos(2θ) = 7/25, tan(2θ) = 24/7

Example 6.10.3

Using cos(2θ) to simplify an expression

ProblemSimplify: 2cos²(θ) − 1
ThinkRecognize the double-angle form: 2cos²(θ) − 1 = cos(2θ).
AnswerThis expression equals cos(2θ) by the second form of the double-angle identity. This recognition is essential in calculus for simplifying integrands such as cos²(θ) = (1 + cos(2θ))/2.

Answer: cos(2θ)

Double-Angle Identities in Calculus

The forms cos(2θ) = 1 − 2sin²(θ) and cos(2θ) = 2cos²(θ) − 1 rearrange to give the power-reducing formulas sin²(θ) = (1 − cos(2θ))/2 and cos²(θ) = (1 + cos(2θ))/2. These are used constantly in Calculus II when integrating even powers of sine and cosine. Knowing the double-angle identities fluently will save you significant work in your next course.

Practice 6.10Double-Angle Identities

sin(2θ) if sin(θ) = 5/13, QI
cos(2θ) if cos(θ) = −4/5, QII
tan(2θ) if tan(θ) = 3, QI
sin(2θ) if cos(θ) = 1/2, QI
Simplify: 1 − 2sin²(θ)
Simplify: 2sin(θ)cos(θ)
Simplify: cos²(θ) − sin²(θ)
Verify: sin(2θ) / (2cos(θ)) = sin(θ)
Find cos(2θ) if sin(θ) = −2/3, QIII.
Simplify: cos²(θ) = (1 + cos(2θ))/2. [Verify this power-reducing form from the double-angle identity.]

Section 6.11

Half-Angle Identities

The half-angle identities express sin(θ/2), cos(θ/2), and tan(θ/2) in terms of cos(θ). They are derived by solving for the single-angle terms in the double-angle identities. Specifically, start from cos(2α) = 1 − 2sin²(α) and solve for sin(α) — then replace α with θ/2 so that 2α = θ. The same process gives the cosine half-angle identity from cos(2α) = 2cos²(α) − 1.

An important feature of these identities is the ± sign: the formula gives the magnitude, and you determine the sign based on which quadrant θ/2 falls in.

Half-Angle Identities

sin(θ/2) = ± √((1 − cos(θ))/2)

cos(θ/2) = ± √((1 + cos(θ))/2)

tan(θ/2) = ± √((1 − cos(θ))/(1 + cos(θ)))

Equivalent forms for tan(θ/2):

tan(θ/2) = sin(θ) / (1 + cos(θ)) and tan(θ/2) = (1 − cos(θ)) / sin(θ)

The sign (+ or −) is determined by the quadrant of θ/2, not θ. Determine the quadrant of θ/2 first.

Example 6.11.1

Deriving the sine half-angle identity

StartBegin with: cos(2α) = 1 − 2sin²(α)
Solve for sin(α) 2sin²(α) = 1 − cos(2α) sin²(α) = (1 − cos(2α))/2 sin(α) = ± √((1 − cos(2α))/2)
SubstituteReplace α with θ/2, so 2α = θ: sin(θ/2) = ± √((1 − cos(θ))/2)

The half-angle identity for sine follows directly from the double-angle identity for cosine.

Example 6.11.2

Exact value using a half-angle identity

ProblemFind the exact value of sin(22.5°).
Think22.5° = 45°/2. Use the sine half-angle identity with θ = 45°. Since 22.5° is in QI, sin(22.5°) > 0, so we take the positive root.
Solve sin(22.5°) = sin(45°/2) = + √((1 − cos(45°))/2) = √((1 − √2/2)/2) = √((2 − √2)/4) = √(2 − √2) / 2

Answer: sin(22.5°) = √(2 − √2) / 2

Example 6.11.3

Half-angle identity with a given cosine value

ProblemIf cos(θ) = 3/4 and 0 < θ < π/2, find sin(θ/2) and cos(θ/2).
Determine quadrant of θ/20 < θ < π/2 implies 0 < θ/2 < π/4, so θ/2 is in QI. Both sin(θ/2) and cos(θ/2) are positive.
Solve sin(θ/2) = √((1 − 3/4)/2) = √(1/8) = 1/(2√2) = √2/4 cos(θ/2) = √((1 + 3/4)/2) = √(7/8) = √7/(2√2) = √14/4

Answer: sin(θ/2) = √2/4, cos(θ/2) = √14/4

Student Warning — The ± Sign

The ± in half-angle identities is not optional. You must determine the sign by finding the quadrant of θ/2. If θ = 300°, then θ/2 = 150°, which is in Quadrant II — where sine is positive but cosine is negative. Ignoring the sign will give wrong values.

Practice 6.11Half-Angle Identities

Find cos(π/8) using the half-angle identity.
Find sin(π/12) using the half-angle identity.
Find tan(22.5°) using the half-angle identity.
Find sin(157.5°) using the half-angle identity.
cos(θ) = 1/3, θ in QI. Find sin(θ/2) and cos(θ/2).
cos(θ) = −5/7, θ in QII. Find cos(θ/2). Determine the sign first.
Verify: cos(θ) = 1 − 2sin²(θ/2). [Rearrange from the half-angle identity.]
Simplify: ± √((1 − cos(60°))/2)

Section 6.12

Solving Basic Trigonometric Equations

A trigonometric equation is a conditional equation — it is true only for certain values of the variable. Unlike an identity, which holds everywhere in its domain, a trigonometric equation poses a specific problem: for which values of θ is the equation true?

The key challenge is periodicity. Because trigonometric functions repeat their values in cycles, every basic trigonometric equation has infinitely many solutions. The equation sin(θ) = 1/2 is satisfied by θ = π/6, but also by θ = 5π/6, and by θ = π/6 + 2πk and θ = 5π/6 + 2πk for every integer k. Learning to write the complete solution set — or the solutions restricted to a given interval — is the central skill of this section.

The general approach is: isolate the trigonometric function, use an inverse trigonometric function to find the principal value, then use the symmetry and periodicity of the unit circle to find all other solutions.

General Strategy for Solving Trigonometric Equations

Step 1: Isolate the trigonometric function. If the equation has more than one function, use identities to convert to a single function.

Step 2: Find the principal value using an inverse trigonometric function. This gives the reference angle.

Step 3: Use the unit circle to find all solutions in [0, 2π) by considering which quadrants give the correct sign.

Step 4: Write the general solution by adding the period (2π for sin and cos, π for tan) times an integer k.

Step 5: Check each solution in the original equation.

Example 6.12.1

Solving a basic sine equation

ProblemSolve sin(θ) = 1/2 for all real θ.
ThinkFind where sin(θ) = 1/2 on the unit circle. The reference angle is π/6. Sine is positive in QI and QII.
Solve QI solution: θ = π/6 QII solution: θ = π − π/6 = 5π/6 General solution: θ = π/6 + 2πk or θ = 5π/6 + 2πk, k ∈ ℤ
Checksin(π/6) = 1/2 ✓ sin(5π/6) = 1/2 ✓

Answer: θ = π/6 + 2πk or θ = 5π/6 + 2πk, for any integer k.

Example 6.12.2

Solving a cosine equation

ProblemSolve cos(θ) = −√3/2 for all real θ.
ThinkReference angle: cos⁻¹(√3/2) = π/6. But cos(θ) is negative, so θ must be in QII or QIII.
Solve QII solution: θ = π − π/6 = 5π/6 QIII solution: θ = π + π/6 = 7π/6 General solution: θ = 5π/6 + 2πk or θ = 7π/6 + 2πk, k ∈ ℤ

Answer: θ = 5π/6 + 2πk or θ = 7π/6 + 2πk, for any integer k.

Example 6.12.3

Solving after isolating the trig function

ProblemSolve 2sin(θ) − 1 = 0 for all real θ.
Isolate 2sin(θ) = 1 sin(θ) = 1/2
SolveSame as Example 6.12.1: θ = π/6 + 2πk or θ = 5π/6 + 2πk, k ∈ ℤ

Answer: θ = π/6 + 2πk or θ = 5π/6 + 2πk.

Example 6.12.4

Solving a tangent equation

ProblemSolve tan(θ) = −1 for all real θ.
Thinktan has period π. Reference angle: π/4. Tangent is negative in QII and QIV.
Solve QII: θ = π − π/4 = 3π/4 General solution (period π): θ = 3π/4 + πk, k ∈ ℤ
Checktan(3π/4) = −1 ✓ tan(3π/4 + π) = tan(7π/4) = −1 ✓

Answer: θ = 3π/4 + πk, for any integer k.

Practice 6.12Solving Basic Equations

Solve for all real solutions. Write the general solution.

sin(θ) = √3/2
cos(θ) = 1/2
tan(θ) = √3
2cos(θ) + √2 = 0
sin(θ) = −1
sec(θ) = 2
csc(θ) = −√2
√3 tan(θ) + 1 = 0

Section 6.13

Solving Trigonometric Equations on Intervals

In many applied problems — and in calculus — you need solutions not over all real numbers but on a specific interval such as [0, 2π) or [0°, 360°). On such an interval, there are finitely many solutions. The technique is the same: isolate the function, find the reference angle, apply quadrant reasoning, and list only those solutions that fall within the given interval.

This section also covers equations that are not immediately in standard form. Some require factoring, some require using a Pythagorean identity to convert to a single function, and some involve multiple angles. Each situation calls for a slightly different approach, but the underlying strategy remains consistent.

Example 6.13.1

Solving on [0, 2π): basic

ProblemSolve cos(θ) = √2/2 on [0, 2π).
ThinkReference angle: cos⁻¹(√2/2) = π/4. Cosine is positive in QI and QIV.
Solve QI: θ = π/4 QIV: θ = 2π − π/4 = 7π/4

Answer: θ = π/4 and θ = 7π/4

Example 6.13.2

Solving a quadratic-type trigonometric equation

ProblemSolve 2sin²(θ) − sin(θ) − 1 = 0 on [0, 2π).
ThinkThis is quadratic in sin(θ). Let u = sin(θ) and factor: 2u² − u − 1 = (2u + 1)(u − 1) = 0.
Solve for u 2u + 1 = 0 → sin(θ) = −1/2 u − 1 = 0 → sin(θ) = 1
Solve for θ sin(θ) = −1/2: θ = 7π/6, 11π/6 sin(θ) = 1: θ = π/2
CheckSubstitute each solution into 2sin²(θ) − sin(θ) − 1 = 0 to confirm. ✓

Answer: θ = π/2, 7π/6, 11π/6

Example 6.13.3

Solving using a Pythagorean identity substitution

ProblemSolve 2cos²(θ) + sin(θ) − 1 = 0 on [0, 2π).
ThinkThe equation has both cos² and sin. Replace cos²(θ) with 1 − sin²(θ) to get a single-variable equation.
Substitute 2(1 − sin²(θ)) + sin(θ) − 1 = 0 2 − 2sin²(θ) + sin(θ) − 1 = 0 −2sin²(θ) + sin(θ) + 1 = 0 2sin²(θ) − sin(θ) − 1 = 0
Factor (2sin(θ) + 1)(sin(θ) − 1) = 0 sin(θ) = −1/2 or sin(θ) = 1
Solve sin(θ) = −1/2: θ = 7π/6, 11π/6 sin(θ) = 1: θ = π/2

Answer: θ = π/2, 7π/6, 11π/6

Example 6.13.4

Solving an equation involving a double angle

ProblemSolve sin(2θ) = sin(θ) on [0, 2π).
ThinkReplace sin(2θ) with 2sin(θ)cos(θ). Move everything to one side. Factor.
Solve 2sin(θ)cos(θ) = sin(θ) 2sin(θ)cos(θ) − sin(θ) = 0 sin(θ)(2cos(θ) − 1) = 0
Cases sin(θ) = 0: θ = 0, π 2cos(θ) − 1 = 0: cos(θ) = 1/2 → θ = π/3, 5π/3

Answer: θ = 0, π/3, π, 5π/3

Never Divide Both Sides by a Trig Function

In the double-angle example above, if you had divided both sides by sin(θ), you would have obtained 2cos(θ) − 1 = 0 and lost the solutions θ = 0 and θ = π. When a trigonometric function is a factor on both sides, move it to one side and factor — never divide it away.

Practice 6.13Solving Equations on Intervals

Solve each equation on [0, 2π). Show all steps.

sin(θ) = −√2/2
2cos(θ) − √3 = 0
tan(θ) = −√3
2sin²(θ) + 3sin(θ) + 1 = 0
cos²(θ) + cos(θ) = 0
sec²(θ) − 2 = 0
sin(2θ) = cos(θ)
cos(2θ) = 1 − sin(θ) [Hint: use cos(2θ) = 1 − 2sin²(θ)]
tan²(θ) − 3 = 0
4sin²(θ) = 3

Section 6.14

Common Identity and Equation Mistakes

Mistake 1 — Treating the sine function as additive

sin(A + B) = sin(A) + sin(B)

Correct: sin(A + B) = sin(A)cos(B) + cos(A)sin(B). Test: sin(π/2) = sin(π/4 + π/4) ≠ sin(π/4) + sin(π/4) = √2, but sin(π/2) = 1. These are not equal.

Mistake 2 — Confusing sin(θ) + cos(θ) = 1 with the Pythagorean identity

sin(θ) + cos(θ) = 1

Correct: sin²(θ) + cos²(θ) = 1. The identity involves squares, not the functions themselves. At θ = π/4: sin + cos = √2/2 + √2/2 = √2 ≠ 1.

Mistake 3 — Dividing by a trigonometric function when solving equations

sin(θ)(2cos(θ)−1) = 0 → divide by sin(θ) → 2cos(θ)−1 = 0

Correct: Set each factor equal to zero separately. Dividing loses the solutions where sin(θ) = 0. Always factor, never divide.

Mistake 4 — Ignoring the ± sign in half-angle identities

sin(θ/2) = √((1−cos(θ))/2) always (ignoring the ±)

Correct: The sign depends on the quadrant of θ/2. If θ = 300°, then θ/2 = 150° is in QII where sin > 0, but if θ = 300° is considered differently, always evaluate the quadrant of θ/2 first.

Mistake 5 — Cross-multiplying or adding to both sides during verification

To verify: A = B. Add 1 to both sides: A+1 = B+1. Therefore A = B. ✓ (circular)

Correct: Work on one side only. Manipulate the more complex side using identities until it matches the other side. You may not perform operations on both sides simultaneously.

Mistake 6 — Forgetting multiple solutions when solving trigonometric equations

sin(θ) = 1/2 → θ = π/6 only

Correct: Sine equals 1/2 in both QI (θ = π/6) and QII (θ = 5π/6). Always use the unit circle to find all solutions in the given interval, accounting for both relevant quadrants.

Practice 6.14Error Analysis

Each problem shows a student's work. Identify the error and write the correct solution.

A student writes cos(A − B) = cos(A) − cos(B). Identify the error and state the correct formula.
A student solves sin²(θ) + cos²(θ) = 1 and concludes "therefore sin(θ) + cos(θ) = 1 for all θ." What is wrong?
A student is asked to verify that (1 − cos²(θ)) / sin(θ) = sin(θ). The student writes: "(1 − cos²(θ))/sin(θ) = sin(θ), so multiplying both sides by sin(θ) gives 1 − cos²(θ) = sin²(θ). True by the Pythagorean identity, so verified." Explain why this reasoning is flawed. Write a correct verification.
A student solving 2cos²(θ) = cos(θ) on [0, 2π) divides both sides by cos(θ) and gets 2cos(θ) = 1, finding only θ = π/3 and 5π/3. What solutions were lost? Solve correctly.
A student evaluates sin(150°/2) = sin(75°) as sin(75°) = −√((1−cos(150°))/2), choosing the negative sign "because 150° is in QII." Identify the error. What sign should be used?

Chapter 6 — Back Matter

Chapter Summary

Sec 6.1A trigonometric identity is true for all values in its domain. A conditional equation is true only for specific values. Identities are tools for rewriting expressions, not problems to solve.
Sec 6.2The reciprocal identities define csc, sec, and cot as reciprocals of sin, cos, and tan respectively. They follow from the unit circle definitions and allow mixed expressions to be unified.
Sec 6.3The quotient identities state tan(θ) = sin(θ)/cos(θ) and cot(θ) = cos(θ)/sin(θ). These follow directly from the y/x and x/y ratios of unit circle coordinates.
Sec 6.4The three Pythagorean identities — sin² + cos² = 1, 1 + tan² = sec², 1 + cot² = csc² — all derive from the equation x² + y² = 1 for a point on the unit circle. Their alternate forms are used constantly in simplification.
Sec 6.5Cosine is an even function: cos(−θ) = cos(θ). Sine and tangent are odd: sin(−θ) = −sin(θ), tan(−θ) = −tan(θ). These reflect the unit-circle symmetry across the x-axis.
Sec 6.6Cofunction identities express each trigonometric function in terms of its cofunction at a complementary angle. They arise from right-triangle geometry and generalize to all angles.
Sec 6.7To simplify: convert to sin and cos, apply Pythagorean identities, factor, combine fractions, or multiply by a conjugate. Look for new patterns after each step.
Sec 6.8To verify an identity, work on one side only. Choose the more complex side. Use known identities and algebra to transform it into the other side. Never cross the equals sign.
Sec 6.9Sum and difference identities for sin, cos, and tan express the function of a sum or difference of angles in terms of functions of each angle alone. They enable exact values at non-special angles and are the foundation for double-angle and half-angle identities.
Sec 6.10Double-angle identities: sin(2θ) = 2sin(θ)cos(θ); cos(2θ) has three forms derived from the Pythagorean identity. They come directly from sum identities with A = B = θ. The power-reducing forms are used in calculus integration.
Sec 6.11Half-angle identities express sin(θ/2), cos(θ/2), and tan(θ/2) in terms of cos(θ). The ± sign is determined by the quadrant of θ/2. They are derived from the double-angle identities for cosine.
Sec 6.12Solving a trigonometric equation: isolate the function, find the reference angle using an inverse function, determine all correct-quadrant solutions on [0, 2π), and write the general solution by adding full periods.
Sec 6.13On a restricted interval, list only those solutions that fall within the interval. Quadratic-type equations can be factored after substituting a single variable. Identity substitution reduces mixed-function equations. Double-angle equations use the double-angle identity and then factor — never divide by a trig function.
Sec 6.14Six key mistakes: treating sin as additive; confusing sin + cos = 1 with the Pythagorean identity; dividing by a trig function and losing solutions; ignoring ± in half-angle identities; performing operations on both sides during verification; finding only one solution when two exist.

Key Terms Review

identity

An equation that is true for all values of the variable in the domain where both sides are defined.

conditional equation

An equation that is true only for specific values of the variable; solving it means finding those values.

reciprocal identities

csc = 1/sin, sec = 1/cos, cot = 1/tan. The reciprocal functions defined in terms of the primary three.

quotient identities

tan(θ) = sin(θ)/cos(θ) and cot(θ) = cos(θ)/sin(θ). Derived from the unit circle coordinate ratios.

Pythagorean identities

Three identities derived from x² + y² = 1 on the unit circle. The most fundamental: sin²(θ) + cos²(θ) = 1.

even function

A function satisfying f(−x) = f(x). Cosine and secant are even trigonometric functions.

odd function

A function satisfying f(−x) = −f(x). Sine, tangent, cosecant, and cotangent are odd.

cofunction identities

Identities relating a function at angle θ to its cofunction at the complementary angle π/2 − θ.

simplifying

Rewriting a trigonometric expression in an equivalent but shorter, more useful form.

verifying an identity

Proving that two expressions are always equal by transforming one side using identities and algebra.

sum and difference identities

Formulas expressing sin(A ± B), cos(A ± B), and tan(A ± B) in terms of individual function values.

double-angle identities

Formulas for sin(2θ), cos(2θ), and tan(2θ) derived from sum identities with A = B = θ.

half-angle identities

Formulas for sin(θ/2), cos(θ/2), and tan(θ/2) derived by solving double-angle identities.

power-reducing formulas

Rearrangements of the double-angle cosine identity: sin²(θ) = (1−cos(2θ))/2, cos²(θ) = (1+cos(2θ))/2.

principal value

The value returned by an inverse trigonometric function; it lies in a specific restricted range.

reference angle

The positive acute angle formed by the terminal side of θ and the nearest x-axis. Used to find all solutions.

general solution

All solutions of a trigonometric equation, expressed using the period and an integer k.

period

The length of one complete cycle: 2π for sin and cos, π for tan and cot.

Concept Review Questions

Explain in your own words why an identity is different from a conditional equation. Give an example of each from this chapter.
Explain where each of the three Pythagorean identities comes from. Which is most fundamental?
Why is it invalid to "verify" an identity by performing the same operation on both sides?
Explain in words what the even-odd identities say about the symmetry of sine and cosine graphs.
Why does sin(A + B) ≠ sin(A) + sin(B)? Give a numerical counterexample.
The double-angle formula for cos(2θ) has three forms. When would you choose each one?
Explain why half-angle identities require a ± sign and how you determine which sign to use.
Explain why you should never divide both sides of a trigonometric equation by a trigonometric function when solving.
When solving sin(θ) = c on [0, 2π), under what conditions are there two solutions? One solution? No solution?
How do power-reducing formulas (derived from the double-angle cosine identity) appear in calculus?

Identity Practice

Part AFundamental Identity Values

Use the given information to find all six trigonometric function values.

sin(θ) = 5/13, QI
cos(θ) = −3/5, QII
tan(θ) = −7/24, QIV
csc(θ) = −2, QIII

Part BSimplification Practice

Simplify each expression fully. Show every step.

csc(θ) · cos(θ) · tan(θ)
sin²(θ) / (1 − cos(θ))
tan(θ) + cot(θ)
(sec(θ) − tan(θ))(sec(θ) + tan(θ))
sin²(θ)·cot²(θ) + cos²(θ)·tan²(θ)
(1 + tan²(θ))·cos²(θ)
(sin(θ) + cos(θ))² − 1
csc²(θ)·(1 − cos²(θ))
sin(θ)/(1 + cos(θ)) + (1 + cos(θ))/sin(θ)
1/(sec(θ) − 1) + 1/(sec(θ) + 1)

Part CVerification Practice

Verify each identity. Work on one side only. Show all steps.

sin(θ)·cot(θ)·sec(θ) = 1
cot²(θ)/(1 + cot²(θ)) = cos²(θ)
(1 − tan²(θ))/(1 + tan²(θ)) = cos(2θ) [Hint: use cos(2θ) = cos² − sin² and simplify the left side.]
sin(θ)/(1 − cos(θ)) = (1 + cos(θ))/sin(θ) [Hint: multiply numerator and denominator of left side by (1 + cos(θ)).]
(sec(θ) − cos(θ)) / tan(θ) = sin(θ)
cos⁴(θ) − sin⁴(θ) = cos(2θ)
tan(θ/2) = (1 − cos(θ))/sin(θ) [Verify using the alternate form of the half-angle tangent identity.]
2csc(2θ) = sec(θ)csc(θ) [Hint: expand 2csc(2θ) = 2/sin(2θ) and use the double-angle formula.]

Part DAdvanced Identities — Exact Values

Find exact values using the given identities. Show all steps.

sin(255°) [Difference identity]
cos(7π/12) [Sum or difference]
tan(15°) [Difference identity]
sin(2θ) if cos(θ) = −7/25, QIII
cos(2θ) if sin(θ) = √5/5, QI
sin(θ/2) if cos(θ) = −1/3, QII
cos(22.5°) [Half-angle]
tan(67.5°) [Half-angle]

Part EEquation-Solving Practice

Solve each equation on [0, 2π) unless otherwise stated. Show all steps.

2sin(θ) + √3 = 0
2cos²(θ) − cos(θ) = 0
sin(2θ) + sin(θ) = 0
2cos²(θ) − 3cos(θ) + 1 = 0
tan²(θ) − tan(θ) = 0
sin²(θ) = cos²(θ)
cos(2θ) + cos(θ) = 0
2sin²(θ) + sin(θ) = 0
sec²(θ) − 2sec(θ) = 0
sin(θ)cos(θ) = √3/4 [Hint: use sin(2θ).]

General solution problems (all real θ):

cos(θ) = 0
sin(θ) = −√3/2
tan(θ) = 1
2sin(θ) + 1 = 0

Error Analysis

Each problem shows a student's error. Identify it and write the correct solution.

Problem 1

cos(π/4 + π/3) = cos(π/4) + cos(π/3) = √2/2 + 1/2

Identify the error. What is the correct exact value of cos(7π/12)?

Problem 2

Verification of tan(θ) · csc(θ) = sec(θ): Both sides equal sin/cos · 1/sin = 1/cos. Therefore both sides equal sec(θ). Verified. [Student worked on both sides simultaneously.]

Explain why this verification is not valid. Write a correct verification working on the left side only.

Problem 3

Solve 2sin(θ)cos(θ) = cos(θ) → divide by cos(θ) → 2sin(θ) = 1 → sin(θ) = 1/2 → θ = π/6, 5π/6

Which solutions are missing? Solve correctly on [0, 2π).

Problem 4

If sin(θ) = 0.6 and cos(θ) = 0.8, then sin²(θ) + cos²(θ) = 0.36 + 0.64 = 1 ✓. Therefore sin(θ) + cos(θ) = 1.

What is the logical error? Compute the actual value of sin(θ) + cos(θ).

Problem 5

cos(2·π/3) = 2cos(π/3) = 2 · (1/2) = 1

Identify the error. Compute cos(2π/3) correctly using the double-angle identity.

Chapter 6 Checkpoint

Complete without assistance — assesses all key skills and concepts from Chapter 6

Section 1 — Vocabulary and Concepts

Define trigonometric identity and explain how it differs from a conditional trigonometric equation. Give one example of each.
State the three Pythagorean identities. Which one is most fundamental, and how are the other two derived from it?
Explain what it means for a function to be even or odd. Which trigonometric functions are even? Which are odd?

Section 2 — Fundamental Identities

Given sin(θ) = −5/13 and θ in QIV, find the values of all six trigonometric functions.
Simplify: tan(θ)·csc(θ)
Simplify: (1 − cos²(θ))·sec²(θ)
Simplify: sin(−θ)·cot(−θ)
Simplify: (sec²(θ) − 1)/tan(θ)

Section 3 — Verification

Verify: sin(θ)·sec(θ) = tan(θ)
Verify: (1 − sin²(θ))·csc²(θ) = cot²(θ)
Verify: (tan(θ) + cot(θ))·sin(θ)·cos(θ) = 1

Section 4 — Advanced Identities

Find the exact value of sin(165°) using a sum or difference identity.
If sin(θ) = 4/5, θ in QI, find sin(2θ), cos(2θ), and tan(2θ).
Find the exact value of cos(π/8) using a half-angle identity.
Find sin(A − B) if cos(A) = 3/5 in QI and sin(B) = 5/13 in QI.

Section 5 — Solving Equations

Solve 2cos(θ) + 1 = 0 for all real θ. Write the general solution.
Solve 2sin²(θ) − sin(θ) = 0 on [0, 2π).
Solve sin(2θ) = √3/2 on [0, 2π). [Hint: Find all values of 2θ in [0, 4π) first, then halve.]

Section 6 — Error Analysis

A student simplifies sin²(θ) + cos²(θ) + tan²(θ) as "1 + tan²(θ) = sec²(θ)." Is this right? Verify or correct.
A student verifies the identity A = B by adding 1 to both sides and concluding "both sides equal B + 1." Explain the flaw and write what correct verification requires.

Bridge Note — Looking Ahead

Chapter 7: Triangle Trigonometry and Calculus Simplification

Chapter 7 will return to triangles — right triangles first, then oblique triangles through the Law of Sines and Law of Cosines. The identity skills you have built here will reappear immediately: the identity sin²(θ) + cos²(θ) = 1 underlies every formula you will use when moving from a unit-circle representation of an angle to a triangle-ratio representation.

In Calculus I, the identities in this chapter become indispensable. Simplifying products of trig functions before differentiating requires the quotient and Pythagorean identities. The power-reducing forms of the double-angle cosine identity — sin²(θ) = (1 − cos(2θ))/2 and cos²(θ) = (1 + cos(2θ))/2 — are the standard technique for integrating even powers of sine and cosine in Calculus II. In physics, sum and difference identities appear in the analysis of wave superposition, interference, and beats. In signal processing and Fourier analysis, double-angle and half-angle identities appear in every derivation. The work you have done in this chapter will continue to compound in value throughout your mathematical career.

MGU Library Connection

Appendix G: Trigonometric Identities Reference — full identity sheet for this textbook
Appendix F: Unit Circle Reference — coordinates, exact values, quadrant signs
Chapter 4: The Unit Circle and Trigonometric Functions — derivation of all fundamental identities
Chapter 5: Graphs of Trigonometric Functions — even/odd properties and their graph interpretations
Chapter 7: Triangle Trigonometry — applications of identity simplification in Law of Sines and Cosines
Calculus I: Derivatives of trigonometric functions; uses Pythagorean and quotient identities throughout
Calculus II: Integration of trigonometric functions; power-reducing formulas from Chapter 6.10 are required
Physics: Wave superposition, interference, and beats — direct application of sum and difference identities
MGU Dictionary entries: identity, Pythagorean identity, double-angle identity, half-angle identity, general solution

Chapter 6:Trigonometric Identitiesand Equations

Purpose and Overview

Learning Objectives

Sections at a Glance

What a Trigonometric Identity Is

Reciprocal Identities

Quotient Identities

Pythagorean Identities

Even and Odd Identities

Cofunction Identities

Simplifying Trigonometric Expressions

Verifying Trigonometric Identities

Sum and Difference Identities

Double-Angle Identities

Half-Angle Identities

Solving Basic Trigonometric Equations

Solving Trigonometric Equations on Intervals

Common Identity and Equation Mistakes

Chapter Summary

Key Terms Review

Concept Review Questions

Identity Practice

Error Analysis

Chapter 6 Checkpoint

Chapter 6:
Trigonometric Identities
and Equations