The unit circle and trigonometric functions

Bring coordinates, angles and function inputs

You need to read an ordered pair (x, y), distinguish negative coordinates from positive lengths, and recognise a function input and its output. The coordinate-plane and function-domain lessons provide that preparation. The Pythagorean theorem explains why a point one unit from the origin satisfies x² + y² = 1.

The preceding angle-measure lesson explains radians and acute right-triangle ratios. Recall that one full turn is 2π radians, a half turn is π and a quarter turn is π/2. For special values such as √3/2, keep the square root exact; its decimal is only a drawing approximation.

Choose a first reading, then return for the proofs

First follow the endpoint definition, its right-triangle bridge, quadrant signs and the four denominator checks. Compare the sine and cosine graphs with the circle, then try the worked axis example. These steps show what each function means before you need every general proof.

The complete-range constructions and least-period proofs explain why the rules hold for every permitted input. Keep them for a slower second pass if the first ideas are new. The inverse section introduces one chosen angle interval; solving for every possible angle comes later.

One angle chooses one endpoint

Start at the right-hand point (1, 0) on a circle of radius 1. A positive quarter turn, θ = π/2, reaches the top point (0, 1). Its horizontal coordinate is 0 and its vertical coordinate is 1. We call those two coordinates cos(π/2) and sin(π/2), respectively.

The unit circle has centre O = (0, 0), radius 1 and equation x² + y² = 1 in a perpendicular coordinate frame with the same unit on both axes. Start a ray along the positive x-axis and rotate it through a real angle θ measured in radians. Counterclockwise is positive and clockwise is negative. Its terminal ray meets the circle at one point P = (x, y). Define cos θ = x and sin θ = y, so P = (cos θ, sin θ).

This definition permits every real θ, including zero, negative angles, angles on the axes and rotations containing several turns. A terminal ray has exactly one intersection in its outward direction with this positive-radius circle, so each permitted angle has exactly one sine and one cosine. The coordinates are dimensionless numerical values in the chosen unit-radius frame. The input is a radian angle unless a degree symbol or another unit is explicitly supplied.

A full turn changes the accumulated rotation by 2π while preserving the terminal point. Thus θ and θ + 2π can be different real inputs with the same pair of outputs. In particular, 0 and 2π both give P = (1, 0). The functions record the terminal coordinates, not the direction of travel or the number of full turns. Keep that information in the original angle when a question asks about the rotation.

The new definitions agree with the acute triangle ratios

For 0 < θ < π/2, the unit-circle point lies in quadrant I, so x > 0 and y > 0. Drop its perpendicular to the horizontal axis at H = (x, 0). Triangle OHP is nondegenerate and right at H, with adjacent leg x, opposite leg y and hypotenuse OP = 1 relative to θ at O. Its cosine ratio is x/1 = x and its sine ratio is y/1 = y. The coordinate definitions therefore agree with the earlier acute definitions.

For a similar right triangle of positive hypotenuse h, the adjacent and opposite lengths are h cos θ and h sin θ. Dividing by h recovers the same dimensionless ratios. This bridge assumes the right triangle, common units and acute angle already hold; it does not authorise treating a non-right triangle's side quotients as these functions.

The previous lesson proved the 30°, 45° and 60° ratios with a halved equilateral triangle and an isosceles right triangle. Scaling their positive hypotenuse to 1 gives the exact first-quadrant coordinates below. Keep √2 and √3 symbolic: a decimal drawing position is an approximation to the exact value, not a replacement for the proof.

A reference angle supplies magnitudes; the quadrant supplies signs

First reduce only the terminal direction to φ with 0 ≤ φ < 2π, keeping the original θ separately. In quadrant I, both x and y are positive; in quadrant II, x is negative and y positive; in quadrant III, both are negative; in quadrant IV, x is positive and y negative. Since cosine is x and sine is y, these signs determine the signs of the two functions.

When the terminal point lies strictly inside a quadrant, its reference angle α is the acute angle between the terminal ray and the horizontal axis. For quadrant I, α = φ; for quadrant II, α = π − φ; for quadrant III, α = φ − π; for quadrant IV, α = 2π − φ. In each case 0 < α < π/2. A point on an axis is handled by its exact axis coordinates rather than by an acute reference triangle.

Reflection across the horizontal or vertical axis preserves the circle and changes the corresponding coordinate sign. Therefore the coordinate magnitudes at θ are cos α and sin α, with the signs of the original quadrant. For example, 4π/3 lies in quadrant III with reference angle π/3, giving P = (−1/2, −√3/2). The reference angle supplies positive leg magnitudes; it is not a coterminal replacement for θ.

Reflecting the terminal ray of θ across the horizontal axis produces the terminal ray of −θ. Consequently cos(−θ) = cos θ and sin(−θ) = −sin θ for every real θ: cosine is even and sine is odd. These follow from the coordinate transformation for the entire circle. Checking a few opposite-angle values illustrates the argument but does not prove it for all real angles.

Prove both the bounds and the attainment of every output

Sine and cosine each have domain all real θ because every real directed angle determines a terminal point. The circle equation gives (cos θ)² + (sin θ)² = 1 for every such θ. Both squares are nonnegative, so each is at most 1. Hence −1 ≤ cos θ ≤ 1 and −1 ≤ sin θ ≤ 1. This proves that no output lies outside the proposed range.

To prove there are no missing outputs inside that interval, take any real u with −1 ≤ u ≤ 1. The point (u, √(1 − u²)) lies on the unit circle, so some real angle reaches it and has cosine u. Similarly, (√(1 − u²), u) lies on the circle and is reached by an angle with sine u. The radicands are nonnegative, and both constructions include u = −1, 0 and 1. Thus the complete range of each function is the closed interval [−1, 1].

Knowing one coordinate determines the square of the other but may leave its sign undecided. If sin θ = −3/5, then (cos θ)² = 1 − 9/25 = 16/25, giving cos θ = ±4/5. A quadrant or another location condition is needed to choose the sign. A graph's apparent position or a rounded value cannot supply missing exact information.

The equation (cos θ)² + (sin θ)² = 1 now applies on the axes and in every quadrant, extending the acute identity by the circle definition. It expresses a property of every point on that whole circle, rather than a pattern inferred from the handful of special points in a table.

Four more functions require their own denominator checks

For P = (x, y) = (cos θ, sin θ), define tan θ = y/x and sec θ = 1/x when x ≠ 0. Define cot θ = x/y and csc θ = 1/y when y ≠ 0. The full names are tangent, secant, cotangent and cosecant. These are real-valued functions only on their stated domains; an excluded angle remains a valid rotation and still has sine and cosine.

Cosine is zero exactly at the top and bottom unit-circle points, so cos θ = 0 exactly when θ = π/2 + kπ for an integer k. Sine is zero exactly at the right and left points, so sin θ = 0 exactly when θ = kπ. These complete zero sets give the domain exclusions in the table. In every row, k ranges over all integers, including negative integers and zero.

Cotangent must be defined as cos θ/sin θ on sin θ ≠ 0. The expression 1/(tan θ) agrees with it only where tangent is defined and nonzero. At θ = π/2, cot θ = 0/1 = 0 while tangent is undefined, so that reciprocal expression cannot define cotangent on its whole domain. Likewise, rearranging or cancelling an expression cannot restore an input excluded by its original denominator.

Within any open quadrant, secant has the cosine sign, cosecant the sine sign, and tangent and cotangent are positive when x and y have the same sign and negative when their signs differ. On the axes, use the definitions individually: a zero numerator can give a valid zero quotient, but a zero denominator never gives a real output. Infinity is not the value of a trigonometric function at an excluded input.

Construct a permitted circle point for each claimed output

On their domains, tangent and cotangent produce real quotients. To show tangent attains every real number v, choose P = (1/√(1 + v²), v/√(1 + v²)). Since 1 + v² > 0, both coordinates are real, their squares add to 1 and the horizontal coordinate is positive. A real angle reaches P, has nonzero cosine and produces tan θ = y/x = v. This includes v = 0, positive values and negative values.

For cotangent, instead choose P = (v/√(1 + v²), 1/√(1 + v²)). Again the squared coordinates add to 1, but now the vertical coordinate is positive and nonzero. A reaching angle belongs to the cotangent domain and gives x/y = v. Thus cotangent also has all real numbers as its range, including zero. This proof does not use a reciprocal of an undefined tangent.

For secant, 0 < |cos θ| ≤ 1 on its domain, so |sec θ| ≥ 1. Conversely, given any real v with |v| ≥ 1, choose x = 1/v and y = √(1 − 1/v²). The point lies on the circle and x ≠ 0, so a reaching angle gives sec θ = 1/x = v. Both v = 1 and v = −1 are attained; no output lies strictly between −1 and 1.

The cosecant proof exchanges the two coordinates: 0 < |sin θ| ≤ 1 implies |csc θ| ≥ 1, and P = (√(1 − 1/v²), 1/v) attains every v with |v| ≥ 1. Its vertical coordinate is nonzero. Therefore the complete range of each reciprocal is (−∞, −1] ∪ [1, ∞). The symbol ∪ means the union of the two intervals; the infinite endpoints indicate no finite bound, not outputs called infinity.

A period repeats the whole function, not just one value

A positive number T is a period when translating the domain by T leaves it unchanged and f(θ + T) = f(θ) for every permitted θ. The least positive such T is the fundamental period. One pair of equal outputs does not establish this universal translation property: sin(π/6) = sin(5π/6), but those points have different cosines and are not the same terminal point.

Adding 2π returns every terminal point to itself, so sine and cosine repeat for every real θ. Secant and cosecant repeat too, and their denominator-zero exclusions are preserved by the same shift. Adding π sends every point (x, y) to its antipode (−x, −y). Both coordinate signs reverse. Consequently tan(θ + π) = (−y)/(−x) = tan θ whenever x ≠ 0, and cot(θ + π) = (−x)/(−y) = cot θ whenever y ≠ 0; these domain conditions are preserved as well.

The same half-turn reverses secant and cosecant rather than preserving them: sec(θ + π) = −sec θ and csc(θ + π) = −csc θ on their domains. Sine and cosine also change sign. Thus a half-turn is a period of tangent and cotangent even though it gives a different terminal point, and it is not a period of the other four functions.

The least-period claims require an additional argument. If T > 0 is a cosine period, its value at 0 forces cos T = 1, which occurs only at the right-hand circle point, so T is a positive integer multiple of 2π. For a sine period, use the value 1 at π/2: sin(π/2 + T) = 1 forces the same top point, hence again T is a positive integer multiple of 2π. The same tests with sec 0 = 1 and csc(π/2) = 1 prove their least positive periods are also 2π.

For a tangent period, its value at 0 forces tan T = 0 at a permitted input, hence sin T = 0 and T is a positive integer multiple of π. For a cotangent period, its value 0 at π/2 forces cos(π/2 + T) = 0 at a permitted input, again making T a positive integer multiple of π. Since the half-turn argument already proves π works, tangent and cotangent have least positive period π. These proofs establish complete repeating rules, not just agreement among selected angles.

A function graph has angle on its horizontal axis

The circle diagram uses horizontal coordinate cos θ and vertical coordinate sin θ. A graph of f instead plots (θ, f(θ)): its horizontal position is the radian input and its vertical position is one output. Thus at θ = π/2 the circle point is (0, 1), the sine-graph point is (π/2, 1) and the cosine-graph point is (π/2, 0). These are different coordinate descriptions.

Over 0 ≤ θ ≤ 2π, the sine graph passes through (0, 0), (π/2, 1), (π, 0), (3π/2, −1) and (2π, 0). The cosine graph passes through (0, 1), (π/2, 0), (π, −1), (3π/2, 0) and (2π, 1). Rotation moves the circle coordinates continuously, so the graphs are continuous curves rather than straight segments specified only by those five points. The circle definitions determine their values between the landmarks.

Both basic graphs have horizontal midline 0 and amplitude 1, meaning the greatest vertical distance from that midline is 1. Their full domain is all real angles, range [−1, 1] and fundamental period 2π. Sine zeros occur at every kπ and cosine zeros at every π/2 + kπ; these follow from the complete axis conditions. A plotting window shows only a bounded portion of those infinite sets.

Read a horizontal scale carefully: a tick labelled π/2 is an angle input, while a tick labelled 90° uses a different unit. A numerical tool must use the matching mode. Finite plotted samples and rounded marker positions help us see the function, but do not prove its whole domain, complete range or period. Those statements came from the unit-circle and algebraic arguments.

Excluded inputs separate branches of quotient graphs

Tangent and secant have vertical asymptote lines at θ = π/2 + kπ; cotangent and cosecant have them at θ = kπ. Here a vertical asymptote means outputs become unbounded in magnitude as permitted inputs approach the excluded angle from either side. It is not a graph point or an output at that angle. Draw separate branches rather than a segment connecting across the missing input.

The definitions explain the unbounded behaviour. As a circle point approaches a top or bottom point, its horizontal coordinate tends to zero while its vertical coordinate tends to 1 or −1. Thus |y/x| and |1/x| grow without bound. More precisely, for any proposed bound M > 0, sufficiently close permitted points have |y| > 1/2 and 0 < |x| < 1/(2M), which makes |tan θ| > M and |sec θ| > M. Near a right or left point, exchange x and y to obtain the cotangent and cosecant argument.

Tangent and cotangent have no finite maximum or minimum over their full domains and no finite amplitude. Secant and cosecant have outputs only at or beyond −1 and 1; their graphs never enter the open band −1 < f(θ) < 1. A vertical crop may hide much of an unbounded branch, and a finite set of samples may miss an entire interval near an asymptote. Use the exact domain and range rather than the displayed crop.

A decimal evaluation near π/2 may produce a very large number for tangent because the input is a nearby permitted angle, or because a numerical approximation to π/2 has nonzero computed cosine. It does not make tan(π/2) defined. Check exact denominators before interpreting a calculator result; an arbitrary tolerance should not turn a genuinely nonzero exact denominator into zero either.

An inverse returns one chosen angle rather than every possible rotation

The global functions repeat outputs, so reversing input and output does not give a unique angle. To define principal inverse functions, first restrict sine, cosine and tangent to intervals where each output occurs once. Arcsin uses the sine interval [−π/2, π/2], arccos uses the cosine interval [0, π], and arctan uses the tangent interval (−π/2, π/2). The open tangent endpoints retain its zero-denominator exclusions.

The unit circle explains the uniqueness. On the right semicircle, each height from −1 to 1 appears exactly once as θ goes from −π/2 to π/2. On the upper semicircle, each horizontal coordinate from 1 down to −1 appears exactly once from 0 to π. For tangent, each line y = vx has exactly one circle point with x > 0, giving one angle strictly between −π/2 and π/2. These restrictions also attain all outputs needed for their inverses.

An expression such as arcsin u means the unique angle in its declared principal interval with sine u. It differs from the reciprocal csc θ = 1/(sin θ). The alternative notation sin⁻¹ u is commonly used for arcsin u, but the superscript −1 in that notation does not mean a reciprocal. We use the arc names here to keep those operations distinct.

For example, sin(7π/6) = −1/2, but arcsin(−1/2) = −π/6, not 7π/6. The returned angle belongs to the principal interval; the original rotation need not. Recovering all angles from a trigonometric value is a later equation-solving task. This introduction gives the branch restrictions and their meaning without claiming that an inverse recovers every unrestricted input.

Predict an endpoint or a missing value, then check it

In Coordinates, choose π/3 and predict which coordinate is larger before reading the exact table. Compare it with π/6. The projection H = (cos θ, 0) shows the horizontal component; away from the axes, the right triangle has leg lengths |cos θ| and |sin θ|. On an axis the triangle can collapse while sine and cosine remain defined.

In Quadrants, choose 4π/3. Its reference angle is π/3, but both coordinates are negative. Compare a quadrant-II and a quadrant-IV case to see which sign changes. The reference angle determines positive coordinate magnitudes, not the original terminal direction.

In Signed turns and periods, compare an angle with the added half and full turns. Opposite endpoints can give equal tangent values; equal endpoints give the same sine and cosine. For a clockwise quarter turn, −π/2, the positive terminal representative is 3π/2. The original direction and completed-turn count stay separate from the decomposition integer k.

In Six functions, choose π/2 and predict which denominators vanish. Tangent and secant are undefined there, while cotangent is 0 and cosecant is 1. Each scene has eight exact choices. The pictures and finite choices illustrate the rules; the earlier written arguments establish their complete domains, ranges and periods.

Worked examples

Try it yourself

1. For θ = 17π/6, find its terminal representative in [0, 2π), quadrant, reference angle, sine, cosine and tangent. State what the reduction preserves.

   Show the answer and reasoning

   The representative is 5π/6 in quadrant II, with reference angle π/6. sin θ = 1/2, cos θ = −√3/2 and tan θ = −√3/3. Reduction preserves the terminal point and these values, while omitting one positive full turn.

   Subtract 2π = 12π/6 to obtain 5π/6. In quadrant II the horizontal coordinate is negative and vertical coordinate positive. Use the proved π/6 magnitudes √3/2 and 1/2. The tangent quotient is (1/2)/(−√3/2) = −1/√3 = −√3/3, with nonzero cosine.

2. Find all six function values for θ = −π/3 using its terminal quadrant and reference angle. Keep the values exact.

   Show the answer and reasoning

   sin θ = −√3/2, cos θ = 1/2, tan θ = −√3, cot θ = −√3/3, sec θ = 2 and csc θ = −2√3/3.

   The representative 5π/3 is in quadrant IV with reference angle π/3. Its horizontal coordinate is 1/2 and vertical coordinate −√3/2. Both are nonzero, so all functions are defined. Divide the coordinates for tangent and cotangent, and take their individual reciprocals for secant and cosecant. The clockwise sign of the original rotation does not mean every function value is negative.

3. Suppose sin θ = −3/5 and the terminal point lies in quadrant IV. Find cosine and the four quotient or reciprocal values. What would be missing without the quadrant condition?

   Show the answer and reasoning

   cos θ = 4/5, tan θ = −3/4, cot θ = −4/3, sec θ = 5/4 and csc θ = −5/3. Without the quadrant condition, cosine could be 4/5 or −4/5.

   The circle identity gives (cos θ)² = 1 − 9/25 = 16/25. Quadrant IV has positive x, so choose +4/5 rather than −4/5. Both coordinates are nonzero, making all four further functions defined. Their signs and exact values follow from the quotients −3/4 and −4/3 and reciprocals 5/4 and −5/3.

4. Evaluate all six functions at θ = 3π/2. Which function graphs have a point at that angle, and which have a vertical asymptote there?

   Show the answer and reasoning

   sin θ = −1, cos θ = 0, cot θ = 0 and csc θ = −1 are defined graph values. Tangent and secant are undefined and have vertical asymptotes at θ = 3π/2.

   The circle point is (0, −1). The horizontal denominator excludes tan and sec, while the nonzero vertical denominator permits cot = 0/(−1) = 0 and csc = 1/(−1) = −1. Sine and cosine have all real inputs. A vertical asymptote records nearby unbounded values, not a point at the excluded input.

5. Can secant output −3? Construct an exact unit-circle point that attains it. Can secant output 1/2? Explain both answers using its complete range rather than a plot.

   Show the answer and reasoning

   Yes: P = (−1/3, 2√2/3) lies on the unit circle and gives sec θ = −3. No: 1/2 is outside the secant range, which contains exactly values v ≤ −1 or v ≥ 1.

   The chosen squares are 1/9 and 8/9, summing to 1, and x = −1/3 is nonzero. Therefore a real angle reaches that point and sec θ = 1/x = −3. For every permitted secant input, 0 < |x| ≤ 1, so |1/x| ≥ 1; an output 1/2 is impossible. The general construction x = 1/v with a matching circle coordinate proves every value of absolute magnitude at least 1 is attained.

6. Compare θ = π/4 and θ + π = 5π/4. Find their unit points and tangent values. Does matching tangent imply the same terminal point? Is π also a period of sine?

   Show the answer and reasoning

   The points are (√2/2, √2/2) and (−√2/2, −√2/2), and both tangents are 1. They are antipodal, not identical. π is a tangent period but is not a sine period; the two sine values here are opposite and nonzero.

   A half-turn negates both coordinates, so their quotient is unchanged wherever cosine is nonzero. This gives the general tangent-period proof with its preserved domain. The sine outputs in this example are √2/2 and −√2/2, disproving π as a sine period. Sine's least positive period is 2π, justified by the general unit-circle argument rather than this pair alone.

7. Evaluate arcsin(sin(7π/6)), arccos(cos(5π/3)) and arctan(tan(3π/4)) exactly. Is arcsin 2 a real value? Distinguish these inverses from reciprocal functions.

   Show the answer and reasoning

   The principal results are −π/6, π/3 and −π/4, respectively. Arcsin 2 is undefined over the reals. Arc functions return angles in their principal intervals; reciprocals such as csc return reciprocal coordinate values instead.

   The inner values are −1/2, 1/2 and −1. Choose their matching angles in [−π/2, π/2], [0, π] and (−π/2, π/2), respectively. None of the original three angles is in its corresponding principal interval. The input 2 is outside the arcsin domain [−1, 1], so it cannot produce a real principal angle. A reciprocal of sine is csc, not arcsin.

What to carry forward

• For every real radian angle θ, the terminal unit-circle point is (cos θ, sin θ). The definitions agree with acute right-triangle ratios and also include axes, negative angles and multi-turn rotations.
• Use an acute reference angle for coordinate magnitudes inside a quadrant, then assign the quadrant signs. Axis values come directly from the circle; the original signed rotation stays separate from its terminal representative.
• Sine and cosine have domain all real numbers and complete range [−1, 1]. The unit-circle equation proves their squared-sum identity, and coordinate constructions prove every claimed output is attained.
• Tangent and secant exclude θ = π/2 + kπ; cotangent and cosecant exclude θ = kπ for integers k. Define cot as cos/sin: 1/tan agrees only on a narrower common domain.
• Tangent and cotangent attain every real output. Secant and cosecant attain exactly outputs at or beyond −1 and 1. Prove both the bounds and attainment with nonzero denominator guards.
• The least positive periods are 2π for sin, cos, sec and csc and π for tan and cot. A period preserves the domain and every output; one matching pair or an antipodal pair need not mean the same terminal point.
• A function graph plots angle against output, and an asymptote is not an output at an excluded angle. Principal inverses use restricted angle intervals; finite samples and numerical drawings do not replace the full definitions and proofs.