Trigonometric identities and equations

Prepare to prove an equality and solve an equation

Be comfortable reading P = (cos θ, sin θ), measuring turns in radians and finding the signs of circle coordinates. The preceding unit-circle lesson explains those ideas. Algebraic preparation is also useful: identities versus equations, nonzero denominators, factoring and the effect of squaring an equation. The linked lessons below let you revisit those steps.

For a first reading, follow the circle identity, cancelled denominator holes and complete basic sine, cosine and tangent families. Then study factoring and original checks, and try the second and third worked examples. The rotation proof of the addition formulas and the first worked example form a second path: return to them once the coordinate definitions and identity-versus-equation distinction are clear. Both paths retain the original domain and use radians.

An identity holds throughout a domain; an equation selects inputs

At θ = π/6, the unit-circle point is (√3/2, 1/2). Squaring its coordinates gives 3/4 + 1/4 = 1. The same squared-sum equality holds as the point moves anywhere around the circle: it is an identity. Asking for sin θ = 1/2 is different. It selects only the points at height 1/2, reached at π/6 and 5π/6 in one turn. This lesson explains how to prove the first kind of statement and find every solution of the second.

The previous lesson defined P = (cos θ, sin θ) on the unit circle for every real directed radian angle θ. Here, all angle inputs are real and measured in radians unless another unit is stated. A trigonometric identity asserts that two expressions are equal for every input in a stated common domain: the inputs where both expressions exist. A trigonometric equation asks which permitted inputs make a particular equality true.

For example, sin² θ + cos² θ = 1 is an identity on all real θ. In this notation sin² θ means (sin θ)², not sin(θ²). The equation sin θ = 1 instead selects the angles whose terminal point is the top of the circle. Both statements use trigonometric functions, but they ask different questions.

Record the original domain before doing algebra. A denominator such as cos θ requires cos θ ≠ 0 even if a later simplification no longer displays it. An identity on a smaller common domain does not claim that the original expression exists at an excluded input. For an equation, its solution set must remain inside its original domain.

To prove an identity, derive one expression from the other using definitions, proved identities and algebra valid for every stated input. A finite list of matching values or a graph can illustrate the result; it cannot establish equality at every real input. One exact counterexample can disprove a universal claim, but several successful checks cannot replace its proof.

The circle equation gives identities with different domains

Every terminal point satisfies x² + y² = 1, so substituting x = cos θ and y = sin θ proves sin² θ + cos² θ = 1 for every real θ, including the axes. This is a statement about the whole unit circle, not a pattern inferred from special-angle values.

If cos θ ≠ 0, divide the entire squared-sum identity by cos² θ. The result is 1 + tan² θ = sec² θ. If sin θ ≠ 0, division by sin² θ instead gives 1 + cot² θ = csc² θ. These divided identities inherit the corresponding nonzero conditions; the original squared-sum identity needs neither condition.

The quotient and reciprocal identities follow directly from the definitions. Cotangent is cos θ/sin θ on sin θ ≠ 0. It agrees with 1/(tan θ) only when both sin θ and cos θ are nonzero: tangent must first exist and then be nonzero. For instance, cot(π/2) = 0 is defined, but 1/(tan(π/2)) is undefined. The narrower reciprocal expression cannot define cotangent everywhere.

The complete zero sets from the unit circle are sin θ = 0 exactly when θ = kπ, and cos θ = 0 exactly when θ = π/2 + kπ, where k is any integer. They turn the denominator conditions into complete angle exclusions. Keep the conditions even when using a memorised formula.

Simplification does not repair a missing input

Consider (1 − cos² θ)/sin θ. Its original domain is sin θ ≠ 0. On that domain, the circle identity changes the numerator to sin² θ, and cancellation gives sin θ. Thus (1 − cos² θ)/sin θ = sin θ is an identity on sin θ ≠ 0. At θ = 0, the simplified expression has value 0 but the original quotient has denominator 0 and remains undefined.

The same distinction matters when solving (1 − cos² θ)/sin θ = 0. Within the original domain the equation becomes sin θ = 0, yet every angle satisfying that condition is excluded by the original denominator. Therefore the original equation has no real solution. Treating the simplified expression as a new equation on all real inputs would wrongly add every kπ.

A square root also carries information that a square has hidden. For every real t, √(t²) = |t|, so √(cos² θ) = |cos θ|. If sin θ is known, the circle identity gives cos θ = ±√(1 − sin² θ), with the sign determined by the terminal location when the root is nonzero. Do not assume that the principal nonnegative square root is the original coordinate.

For example, cos θ = −3/5 and a quadrant II terminal point imply sin θ = +4/5: its square is 16/25 and that quadrant has positive vertical coordinate. The same cosine in quadrant III would give sin θ = −4/5. The coordinate magnitude alone cannot choose between those signs.

Rotation preserves distances and proves the addition formulas

Take arbitrary real angles α and β, with terminal points Pα = (cos α, sin α) and Pβ = (cos β, sin β). Rotate both points about the origin through −β. Their images are Pα−β and U = (1, 0). Rotation preserves the squared distance between any two points, including coincident or opposite points. No assumption that α or β is acute is needed.

Expanding the squared distance before rotation gives (cos α − cos β)² + (sin α − sin β)² = 2 − 2(cos α cos β + sin α sin β), because each endpoint has squared coordinates summing to 1. After rotation, it is (cos(α − β) − 1)² + sin²(α − β) = 2 − 2cos(α − β). Equating these expressions proves cos(α − β) = cos α cos β + sin α sin β for every real α and β.

Reflection across the horizontal axis gives cos(−β) = cos β and sin(−β) = −sin β. Replace β by −β in the proved difference formula to obtain cos(α + β) = cos α cos β − sin α sin β. These signs follow from the geometric definitions rather than from separate guesses about each quadrant.

Reflection across the line y = x exchanges the circle coordinates and sends the terminal angle θ to π/2 − θ modulo full turns. Therefore cos(π/2 − θ) = sin θ and sin(π/2 − θ) = cos θ for every real θ. Apply the cosine difference formula to (π/2 − α) − β: its cosine is sin(α + β), and its expanded form is sin α cos β + cos α sin β. Replacing β by −β then gives the sine difference formula. All four formulas have all real α and β as their domain.

Substitution proves double-angle formulas; division adds guards

Put α = β = θ in the addition formulas. This proves sin(2θ) = 2sin θ cos θ and cos(2θ) = cos² θ − sin² θ for every real θ. Substituting sin² θ + cos² θ = 1 gives two more equal forms: cos(2θ) = 1 − 2sin² θ = 2cos² θ − 1. Choose the form that uses the coordinate already known; none requires a nonzero denominator.

For tangent addition, first require cos α ≠ 0 and cos β ≠ 0 so both individual tangents exist. Divide the sine addition formula by the cosine addition formula, then divide their numerator and denominator by cos α cos β. The result is tan(α + β) = (tan α + tan β)/(1 − tan α tan β), provided 1 − tan α tan β ≠ 0. Under the first two conditions, that final guard is equivalent to cos(α + β) ≠ 0.

The subtraction derivation gives tan(α − β) = (tan α − tan β)/(1 + tan α tan β), requiring cos α ≠ 0, cos β ≠ 0 and 1 + tan α tan β ≠ 0. The formulas are identities on their common domains, not shortcuts that make an undefined individual tangent valid.

Setting both angles to θ yields tan(2θ) = 2tan θ/(1 − tan² θ), with cos θ ≠ 0 and 1 − tan² θ ≠ 0. At θ = π/2, tan(2θ) = tan π = 0 exists, but the displayed right side does not because tan θ is undefined. At θ = π/4, its final denominator is zero and tan(2θ) is also undefined. The sine and cosine double-angle formulas remain valid at both inputs.

Circle intersections account for every basic solution

For sin θ = u, first check −1 ≤ u ≤ 1. If this fails there is no real solution. Otherwise let a = arcsin u, the unique principal angle in [−π/2, π/2]. A horizontal line y = u meets the unit circle at (√(1 − u²), u) and (−√(1 − u²), u). These are precisely all points of that height. Their angles are a and π − a modulo full turns, so the two families below include every solution and only solutions.

For cos θ = u with −1 ≤ u ≤ 1, let b = arccos u in [0, π]. The vertical line x = u meets the circle at (u, √(1 − u²)) and (u, −√(1 − u²)). Their angles are b and −b modulo full turns. If |u| > 1, there is no intersection and no real solution. At the extreme outputs ±1, the two listed intersection points coincide; the repeated family must be deduplicated.

For tan θ = v, any real v is possible. Its definition requires cos θ ≠ 0. The condition sin θ/cos θ = v is then equivalent to y = vx on the circle with x ≠ 0. Substitution into x² + y² = 1 gives x = ±1/√(1 + v²) and y = vx. Both x values are nonzero, so there are exactly two antipodal permitted points. The principal angle c = arctan v lies in (−π/2, π/2); their angles give the single family θ = c + kπ.

In every family, k ranges independently over all integers. Principal inverse names return one branch angle, not all solutions by themselves. Arcsin and arccos take inputs only in [−1, 1], while arctan accepts every real input. Their exact expressions can describe solutions even when there is no special-angle name; decimal approximations are optional and must not replace a missing branch.

Merge equal roots while respecting interval endpoints

Two formulas can describe the same family. For sin θ = 1, a = π/2 and π − a = π/2, so write the family once. For sin θ = −1, the two starting angles −π/2 and 3π/2 differ by a full turn and give the same family. For cos θ = −1, π + 2πk and −π + 2πk likewise describe the same set. This is deduplication of equal real inputs across formulas, not removal of different rotations.

At output 0, two full-turn families can be combined into one half-turn family: sin θ = 0 gives θ = kπ, and cos θ = 0 gives θ = π/2 + kπ. A union of valid branches may be written more compactly when the equality of those sets is justified. Do not describe every repeated formula as an extra solution.

After finding all real families, apply the requested interval. Square brackets include an endpoint and round brackets exclude it. Thus cos θ = 1 has roots 0 and 2π on [0, 2π], but only 0 on [0, 2π). The two included inputs on the closed interval are distinct real angles, even though their terminal circle point is identical. Similarly, cos θ = −1 includes both −π and π on [−π, π].

For a family θ = a + kT with T > 0 on [L, R), solve L ≤ a + kT < R for integer k, then list the resulting angles and remove any duplicate values from other branches. This inequality method also works for negative or multi-turn intervals. Never decide endpoint membership merely from whether the diagram shows one terminal marker.

Solve for the whole argument before returning to θ

For an argument u = mθ + b with real m ≠ 0, solve the trigonometric equation completely in u first, then apply θ = (u − b)/m to every family. The linear change is reversible, so it neither loses nor adds roots. An interval restriction on θ must also be translated correctly into a restriction on u; if m is negative, inequality directions reverse.

For example, tan(2θ) = −1 gives 2θ = −π/4 + kπ, hence θ = −π/8 + kπ/2 for every integer k. Each listed argument has nonzero cosine, so every family member belongs to the original tangent domain. If 0 ≤ θ < π, the integer condition is 0 ≤ −π/8 + kπ/2 < π, or 1/4 ≤ k < 9/4. Thus k = 1 or 2, giving θ = 3π/8 or 7π/8.

The original interval here maps to 0 ≤ 2θ < 2π. Searching only the first principal tangent branch for 2θ would miss a solution. The principal angle starts the complete family; the period and interval filtering finish the equation. A zero coefficient m would produce a constant argument and is outside this reversible-change rule.

Factoring gives a union of branches; division may remove one

For real, defined factors A and B, AB = 0 is equivalent to A = 0 or B = 0. Solve both branches on the original domain and take their union. If a root occurs in both branches, list that real input once. If a factor is undefined, the zero-product rule cannot turn the original expression into a valid zero.

The equation sin θ(2cos θ − 1) = 0 is defined for all real θ. Its branches are sin θ = 0, giving θ = kπ, and cos θ = 1/2, giving θ = π/3 + 2πk or θ = −π/3 + 2πk. On [0, 2π), their union is 0, π/3, π and 5π/3. Dividing the original equation by sin θ would discard its entire zero-sine branch; it is equivalent only after separately restricting to sin θ ≠ 0.

When an equation is a polynomial in one coordinate, use a temporary variable such as u = sin θ. Solve its algebraic equation, then reject any proposed u outside [−1, 1] before finding angles. Each remaining value has its own complete circle-intersection families. Algebraic factoring and the trigonometric range check perform different necessary jobs.

Squaring supplies candidates, not automatic equivalence

For real expressions A and B, A = B implies A² = B². Conversely, A² = B² means (A − B)(A + B) = 0, so A = B or A = −B. Squaring therefore permits an additional sign branch unless other conditions rule it out. Keep the original domain, find every candidate from the squared equation and substitute each candidate into the original equality.

For sin θ + cos θ = 1, squaring gives 1 + 2sin θ cos θ = 1, hence sin θ cos θ = 0. On [0, 2π), the candidates are 0, π/2, π and 3π/2. Their original sums are 1, 1, −1 and −1, respectively. Only 0 and π/2 solve the original equation. Over all real inputs, the solutions are θ = 2πk or θ = π/2 + 2πk.

This list is complete because every original solution must satisfy the squared equation, both zero-factor branches of that equation were solved completely, and the original check removed exactly the failing candidates. A second geometric check is available: the line x + y = 1 intersects x² + y² = 1 only at (1, 0) and (0, 1). Substituting y = 1 − x gives 2x(x − 1) = 0, establishing those are all its intersections.

A plot or a few guessed values cannot replace the completeness argument. Likewise, an accepted squared candidate cannot excuse ignoring an original zero denominator. Reversible steps preserve the solution set on the retained domain; irreversible steps such as squaring require a final original check.

Inspect exact identities, complete families and original checks

Predict before changing a case. In Identities, compare cot θ with 1/(tan θ) at π/2. Which expression has a value? Check its own denominator and then the common domain. Next compare the absolute and signed square-root claims at 2π/3: a positive root need not be a negative coordinate.

In Angle formulas, choose (α, β) = (π/2, π/4). Predict whether the sine and cosine sum formulas still hold at an axis angle, then compare their exact sides. These formulas are valid for all real angles. The guarded tangent formulas are proved in the written section; this scene displays only sine and cosine equalities.

In Complete equation families, choose sine equal to 1/2 and predict both terminal points before reading the roots. Compare a boundary output and an out-of-range output. Read each displayed interval: two distinct included angles, such as 0 and 2π on a closed interval, can share one point. The table keeps those inputs separate; the diagram groups coincident positions.

In Original-equation checks, compare the cancelled quotient equal to 1 with the quotient equal to 0. Predict which simplified candidates are original sine-denominator holes. Then square sin θ = 1/2 and predict which of the four candidates fail the original sign. Use the original-value table to explain each rejection. These 42 fixed cases illustrate the proofs and help test predictions; they do not prove identities by sampling.

Worked examples

Try it yourself

1. Prove (1 − sin² θ)/cos θ = cos θ on its exact common domain. Does the original left side have a value at θ = π/2?

   Show the answer and reasoning

   The identity holds when cos θ ≠ 0, equivalently θ ≠ π/2 + kπ for any integer k. At π/2 the original quotient is undefined, even though the simplified right side is 0.

   The circle identity gives 1 − sin² θ = cos² θ. Division then gives cos² θ/cos θ = cos θ only where the original denominator is nonzero. Record that condition before cancellation and retain it afterward. Equality on the common domain does not extend the original quotient to its missing inputs.

2. Suppose cos θ = −3/5 and the terminal point lies in quadrant II. Find sin θ, sin(2θ), cos(2θ) and tan(2θ) exactly, checking the required signs and denominator.

   Show the answer and reasoning

   sin θ = 4/5, sin(2θ) = −24/25, cos(2θ) = −7/25 and tan(2θ) = 24/7.

   The circle identity gives sin² θ = 16/25; quadrant II requires the positive root 4/5. Thus sin(2θ) = 2(4/5)(−3/5) = −24/25 and cos(2θ) = 9/25 − 16/25 = −7/25. The latter is nonzero, so tangent is their quotient (−24/25)/(−7/25) = 24/7. The quadrant chooses the original sine sign, while the derived double-angle coordinates determine the new signs.

3. Solve these two equations separately: sin θ = −√3/2 over all real θ and on [0, 2π); tan(2θ) = −1 over all real θ and on [0, π).

   Show the answer and reasoning

   For sine: θ = 4π/3 + 2πk or θ = 5π/3 + 2πk; interval roots 4π/3 and 5π/3. For tangent: θ = −π/8 + kπ/2; interval roots 3π/8 and 7π/8. Each k is any integer.

   The horizontal line y = −√3/2 meets the circle in quadrants III and IV with reference angle π/3, giving both sine families. For tangent, solve the entire argument: 2θ = −π/4 + kπ, then divide by 2. Its argument cosine is never zero on that family. Filtering 0 ≤ θ < π gives integer k = 1 or 2. These are two separate solution sets, not a demand that both equations hold simultaneously.

4. Solve sin θ = 1 and cos θ = 1 separately over all real θ and on the closed interval [0, 2π]. Explain why the cosine endpoints remain two roots despite sharing a circle point.

   Show the answer and reasoning

   For sine: θ = π/2 + 2πk, with interval root π/2. For cosine: θ = 2πk, with interval roots 0 and 2π. Each k is any integer; 0 and 2π are distinct included inputs.

   At sine output 1, the two general sine branches coincide at the top circle point, so write one family. Cosine output 1 has one terminal point, but every full-turn input reaches it. On the stated closed interval both k = 0 and k = 1 are permitted. Deduplicate equal angle values arising from formulas; do not merge different real inputs merely because their terminal coordinates agree.

5. Solve sin θ(2sin θ − 1) = 0 over all real θ, then on [0, 2π), retaining both zero-factor branches.

   Show the answer and reasoning

   θ = kπ, θ = π/6 + 2πk or θ = 5π/6 + 2πk, with each k any integer. On [0, 2π): 0, π/6, 5π/6 and π.

   The factors give sin θ = 0 or sin θ = 1/2. Both coordinate values lie in [−1, 1]. Their complete circle families give the stated union. Every listed root makes an original factor zero, and the zero-product equivalence accounts for all roots. Dividing by sin θ would remove 0 and π in the interval, as well as their complete all-real family.

6. Solve (1 − cos² θ)/sin θ = 0 over the reals. Explain why solving only its simplified expression would give a wrong answer.

   Show the answer and reasoning

   There is no real solution. The apparent roots θ = kπ of the simplified equation are all excluded by the original denominator.

   The original domain requires sin θ ≠ 0. Within it, 1 − cos² θ = sin² θ and cancellation makes the left side sin θ. For that to be zero would require sin θ = 0, contradicting the retained domain. At each apparent kπ root, the original expression has zero denominator and no real value; cancellation does not make it defined.

7. Solve sin θ = cos θ + 1 over all real θ and on [0, 2π). Find every squared candidate and check it in the original equation.

   Show the answer and reasoning

   All real solutions are θ = π/2 + 2πk or θ = π + 2πk, with k any integer. On [0, 2π): π/2 and π. Squaring also proposes 3π/2, which fails the original equality.

   Squaring gives sin² θ = cos² θ + 2cos θ + 1. Substitute sin² θ = 1 − cos² θ to obtain 2cos θ(cos θ + 1) = 0. The complete candidates on [0, 2π) are π/2, 3π/2 and π. At π/2, 1 = 0 + 1; at π, 0 = −1 + 1; at 3π/2, −1 ≠ 0 + 1. Retain only the π/2 and π full-turn families. Every original root was a squared candidate, and the original checks restore exactly the required equality.

What to carry forward

• An identity asserts equality for every input in its stated common domain. An equation selects the permitted inputs that satisfy it. Finite checks illustrate a universal identity but do not prove it.
• The circle equation proves sin² θ + cos² θ = 1 for all real θ. Quotient, reciprocal and divided Pythagorean identities retain their nonzero denominators; cot = 1/tan holds only on a narrower common domain.
• Cancellation does not restore an original hole. A square determines magnitude, while a quadrant or location condition may be needed to choose the original coordinate sign.
• Rotation and circle-coordinate reflections prove the sum and difference identities for all real angles. Double-angle sine and cosine formulas inherit that domain; tangent formulas add individual and combined denominator guards.
• Horizontal, vertical and through-origin line intersections give complete basic sine, cosine and tangent families. Principal inverse angles start the families; every integer period shift and every required second branch must be included.
• Deduplicate equal root descriptions at boundary outputs, then filter by the exact interval endpoints. Distinct included inputs such as 0 and 2π remain distinct roots even when their terminal point is the same.
• Factoring requires the union of every zero-factor branch on the original domain. Squaring supplies candidates that must be checked in the original equation; completeness needs both all candidate branches and the final checks.