Explain how the unit circle extends sine, cosine and tangent to all real numbers

Learning goal: Explain how the unit circle in the coordinate plane enables the extension of the trigonometric functions to all real numbers, where each real number is interpreted as the radian measure of an angle traversed counterclockwise from (1, 0) around the unit circle, and find the sign of each function from the quadrant.

Where right-triangle trigonometry stops. In a right triangle, sin θ = opposite/hypotenuse and cos θ = adjacent/hypotenuse. That definition only makes sense for an acute angle: θ must be strictly between 0 and π/2, because a right triangle cannot have a second angle of 90° or more, let alone a negative angle or an angle of 7 radians. Yet real situations turn through more than a quarter turn all the time: a wheel rotates through many full turns, a pendulum swings to both sides of vertical, and a tide rises and falls forever. We need sine and cosine of every real number, and the unit circle provides the definition.

The definition. Place the unit circle, x² + y² = 1, in the coordinate plane with center at the origin. Given any real number t, start at the point (1, 0) and travel along the circle a distance |t|: counterclockwise if t is positive, clockwise if t is negative. By the previous lesson, traveling an arc of length t counterclockwise is the same as turning through an angle of t radians, so t is the radian measure of the angle traversed. Call the point where you stop P(t). Then define cos t to be the x-coordinate of P(t) and sin t to be the y-coordinate of P(t), and define tan t = sin t / cos t whenever cos t ≠ 0. Every real number t gives exactly one point P(t), so every real number has exactly one sine and one cosine: the domain of both functions is all real numbers.

Why this agrees with the triangle definition. For an acute angle t, drop a perpendicular from P(t) to the x-axis. That makes a right triangle with hypotenuse 1 (the radius), horizontal leg equal to the x-coordinate and vertical leg equal to the y-coordinate. So opposite/hypotenuse = y/1 = y and adjacent/hypotenuse = x/1 = x. The new definition gives the old values in the old cases and gives values in all the new cases as well. It also explains why the radius must be 1: with radius 1 the ratios are the coordinates, with no division needed.

A fully worked model. Find cos(3π/2) and sin(3π/2). Starting at (1, 0) and traveling counterclockwise a distance of 3π/2, which is three quarters of the circumference 2π, we pass (0, 1) at a quarter turn and (−1, 0) at a half turn and stop at (0, −1). So cos(3π/2) = 0 and sin(3π/2) = −1, and tan(3π/2) is undefined because it would require dividing by cos(3π/2) = 0. No triangle has an angle of 3π/2, but the circle has no difficulty with it.

Angles beyond one turn and negative angles. For t = 7π/3, note that 7π/3 = 2π + π/3: one full turn plus a third of a half turn. A full turn brings you back to (1, 0), so P(7π/3) = P(π/3), and cos(7π/3) = cos(π/3) = 1/2 and sin(7π/3) = √3/2. For t = −π/2, travel a quarter turn clockwise from (1, 0) to (0, −1): cos(−π/2) = 0 and sin(−π/2) = −1. For a number like t = 10, which is not a multiple of π, subtract full turns: 10 − 2π ≈ 3.717, which lies between π ≈ 3.142 and 3π/2 ≈ 4.712, so P(10) is in the third quadrant. A calculator in radian mode gives cos 10 ≈ −0.839 and sin 10 ≈ −0.544, both negative as the quadrant predicts, and (−0.839)² + (−0.544)² ≈ 1.000, which confirms that P(10) lies on the unit circle.

Signs by quadrant. Because cos t and sin t are coordinates, their signs are the signs of x and y in each quadrant. Quadrant I (0 < t < π/2): both positive. Quadrant II (π/2 < t < π): cos negative, sin positive. Quadrant III (π < t < 3π/2): both negative. Quadrant IV (3π/2 < t < 2π): cos positive, sin negative. Tangent, being a quotient, is positive where sine and cosine share a sign, quadrants I and III, and negative in quadrants II and IV. For example, t = 2 lies in quadrant II (π/2 ≈ 1.571 < 2 < 3.142), and the calculator agrees: cos 2 ≈ −0.416, sin 2 ≈ 0.909, tan 2 ≈ −2.185.

Where tangent fails. Tangent is undefined exactly where cos t = 0, at the points (0, 1) and (0, −1), which are reached at t = π/2, 3π/2, and in general π/2 + kπ for any integer k. Everywhere else tan t is a real number. So the domain of tangent is all real numbers except these odd multiples of π/2, while sine and cosine have no exceptions at all.

The misconception. Students sometimes say "sin t is the y-coordinate of the angle." An angle does not have a coordinate; the point P(t) on the unit circle does, and the definition depends on the circle having radius 1 and on traveling counterclockwise from (1, 0). A second error is to treat t as degrees when the problem gives a bare number: sin 2 means the sine of 2 radians, about 0.909, not the sine of 2°, which is about 0.035. When a problem writes a bare real number, radian measure is meant; that convention is exactly what lets sine and cosine be functions of real numbers.

Worked, a quadrantal angle: Find sin π, cos π and tan π. Traveling a distance π counterclockwise from (1, 0) is a half turn, which ends at (−1, 0). So cos π = −1, sin π = 0 and tan π = 0/(−1) = 0. No right triangle contains an angle of π, but the unit circle gives the values immediately.

Worked, an input in quadrant III: Find the quadrant and the signs of the functions at t = 4, then evaluate with a calculator. Since π ≈ 3.142 < 4 < 3π/2 ≈ 4.712, P(4) is in quadrant III, so cos 4 < 0, sin 4 < 0 and tan 4 > 0. Calculator in radian mode: cos 4 ≈ −0.654, sin 4 ≈ −0.757, tan 4 ≈ 1.158. The signs agree with the prediction, and (−0.654)² + (−0.757)² ≈ 0.428 + 0.573 = 1.001 ≈ 1, so the point is on the unit circle.

Worked, a negative input: Find cos(−π/2), sin(−π/2), and tan(−π/2). A negative number means travel clockwise. A quarter turn clockwise from (1, 0) ends at (0, −1). So cos(−π/2) = 0, sin(−π/2) = −1, and tan(−π/2) is undefined because the cosine is 0. Notice that P(−π/2) is the same point as P(3π/2), so the two inputs have the same sine and cosine.

Worked, from a coordinate to the functions: The point P(t) is (−0.28, 0.96). Which quadrant is t in, and what are cos t, sin t and tan t? Check the point is on the unit circle: (−0.28)² + 0.96² = 0.0784 + 0.9216 = 1. Then cos t = −0.28, sin t = 0.96 and tan t = 0.96/(−0.28) ≈ −3.43. A negative x and positive y put P(t) in quadrant II, so t lies between π/2 and π (plus any number of full turns).

Representation task: Draw the unit circle on a coordinate grid where 1 unit is 5 cm. Mark the points reached at t = 0, π/2, π, 3π/2, 2π and write the coordinates of each. Then mark, by estimating arc length, the points P(1), P(2), P(3), P(4), P(5) and P(6), using the fact that one radian is a little less than 60°. For each of the six, record the quadrant and the predicted signs of cos t and sin t, then check with a calculator in radian mode and record the values to three decimal places. Finally compute cos²t + sin²t for two of them and record that each is 1 to rounding.

Independent performance: Write an explanation, with a labeled diagram, of how the unit circle makes sine and cosine functions of all real numbers. It must state where you start, which direction is counterclockwise, what the input t measures, what point P(t) is, how cos t and sin t are read from it, and why this agrees with opposite/hypotenuse and adjacent/hypotenuse when t is acute. Include one evaluation for a t between π and 2π, one for a negative t, one for a t greater than 2π, and one explanation of why tan t is undefined at π/2. Success criteria: every value is correct and its sign matches its quadrant; the diagram shows the arc for at least one t and the coordinates of P(t); and the text names the counterclockwise convention and radian measure explicitly.