Ⅳ Functions · Stage 25 — Trigonometry · 25.3 The Unit CircleAll lessons →
Stage 25 · Trigonometry

Trig on the Unit Circle

Plant the ray at the centre of a unit circle: sin θ is the height, cos θ the across.

Ages 14–18 · Reasoning, one step at a time
The unit circle. The terminal side of θ meets the circle at P, whose coordinates are (cos θ, sin θ). The horizontal cos leg is "the across"; the vertical sin leg is "the height." Every angle — acute, obtuse, reflex, negative — gets a point, so every angle gets a sine and a cosine.

In Right-triangle trig a sine was a ratio of sides — but a triangle only has acute angles, so that definition stops dead at 90°. Here we set it free. Draw a circle of radius 1 centred at the origin, and let the rotating ray from the last lesson sweep out the angle θ. Wherever the ray crosses the circle, it pins a single point P — and we simply define cos θ to be the x-coordinate of P and sin θ to be its y-coordinate. For an acute angle this agrees exactly with the old triangle ratios; for any other angle it just keeps working. From that one picture flow the signs of every quadrant and a family of reduction formulas you can read straight off the circle's symmetry.

25.3.1 Sine and cosine, reborn on the circle

Pick an acute angle, say θ = 50°, and drop a right triangle inside the unit circle: the radius to P is the hypotenuse, the horizontal leg runs along the x-axis, the vertical leg rises to P. Because the hypotenuse is 1, the old ratios collapse to something beautiful:

cos θ = adjacent1 = adjacent = x(P)   and   sin θ = opposite1 = opposite = y(P).

So on the unit circle the cosine is the across and the sine is the height. There is no division left to do — the coordinates of P simply are the two trig values. And nothing in "x-coordinate" or "y-coordinate" demands that the angle be acute. Spin the ray to 130°, to 250°, to −40°, to 800°: it still lands on the circle at exactly one point, and we read off cos θ = x(P), sin θ = y(P) as before. That single move — turning a ratio into a coordinate — is what extends sine and cosine to every angle.

For the acute angle θ = 50°, the unit-circle definition matches the right-triangle one: the across is cos 50° ≈ 0.64 and the height is sin 50° ≈ 0.77. The radius is the hypotenuse, and it equals 1 — so the legs are the cosine and sine directly.
Key idea

For the terminal side of θ meeting the unit circle at P: cos θ = x(P), sin θ = y(P), and tan θ = sin θcos θ = yx. Because P sits on a circle of radius 1, its coordinates always satisfy cos2θ + sin2θ = 1 — that is just x2 + y2 = 1, the equation of the circle, in disguise.

The tangent is the slope of the terminal side: rise over run, y/x. That is why tan θ = sin θ / cos θ — and why it goes haywire exactly when the run x = cos θ is zero. Now dial the angle yourself and watch the three values, the quadrant, and the signs change together.

Try it THE UNIT-CIRCLE DRIVER — dial θ all the way round

Spin θ from 0° past 360°. The point P traces the circle; cos θ is its across, sin θ its height, tan θ its slope — every number computed, never measured off the picture.

angle θ 40°
Watch out

tan θ is undefined wherever cos θ = 0 — that is at θ = 90° and θ = 270° (the ray points straight up or straight down, so the run is 0 and the slope is vertical). The driver shows "tan θ = undefined" there, not a huge number and not zero.

25.3.2 Signs by quadrant — "All Students Take Calculus"

Once sine and cosine are coordinates, their signs are no mystery at all — they are just the signs of x and y in each quadrant. In both x and y are positive, so everything is positive. In we have moved left (x < 0) but stayed up (y > 0), so only sine survives. In both coordinates are negative, so sine and cosine are both negative but their ratio, the tangent, is positive. In we are right (x > 0) but down (y < 0), so only cosine is positive.

Which functions are positive in each quadrant. : All. : Sine only. : Tangent only. : Cosine only. Reading counter-clockwise from Ⅰ spells A·S·T·C — "All Students Take Calculus."
Example

θ = 210° lands in quadrant . There only the tangent is positive, so sin 210° and cos 210° are both negative while tan 210° is positive. Indeed cos 210° = −√32, sin 210° = −12, and tan 210° = +1√3. The mnemonic predicted all three signs before we computed a thing.

25.3.3 Reduction formulas from symmetry

Here is the real payoff of the picture. Related angles put P at symmetric spots on the circle, and a reflection only flips a coordinate's sign — so each reduction formula is a sign-flip you can see. Take the angle π − θ (that is 180° − θ). Its point is the mirror image of P across the y-axis: the height is unchanged, the across is negated. Reading off coordinates:

sin(π − θ) = sin θ     cos(π − θ) = −cos θ

Now π + θ rotates P a half-turn through the origin — both coordinates flip:

sin(π + θ) = −sin θ     cos(π + θ) = −cos θ

And −θ reflects P across the x-axis — the across stays, the height flips. This says cosine is even and sine is odd:

cos(−θ) = cos θ     sin(−θ) = −sin θ

The dial below shows θ and its partner side by side. Toggle the relationship and watch how the two y's stay equal or turn opposite, and likewise the two x's — proof by symmetry, not by arithmetic.

Try it THE REFLECTION PAIR — see the reduction in the mirror

Choose a partner of θ and dial the base angle. Compare the green point (θ) and the amber point (its partner): equal heights ⇒ same sine, opposite heights ⇒ negated sine.

angle θ 35°

Two more reductions come from reflecting across the line y = x, which swaps the across and the height — so they trade sine for cosine. These are the co-function relations:

sin(π2 − θ) = cos θ     cos(π2 − θ) = sin θ

That is the original meaning of "co-sine": the sine of the complementary angle. The same complementary swap is why a 30° height equals a 60° across, a fact you first met back in the right triangle.

25.3.4 One memory rule, then evaluate by reference angle

All seven formulas above are special cases of one rule for sin or cos of (n·90° ± θ):

Key idea — the 90° rule

For n·90° ± θ: if n is odd, the function swaps its co-name (sin ↔ cos); if n is even, the name is kept. Then attach the sign of the original expression, found by treating θ as a small positive acute angle and asking which quadrant n·90° ± θ falls in. ("Odd swaps, even keeps; sign from the quadrant.")

For example, sin(90° + θ): here n = 1 is odd, so sine swaps to cosine; 90° + θ sits in quadrant Ⅱ where sine is positive, so the sign is +. Hence sin(90° + θ) = cos θ. To actually evaluate a specific angle, reduce it to its acute reference angle — the acute angle between the terminal side and the x-axis (computed by GX.refAngle) — then attach the sign from its quadrant.

Example — sin 210° step by step

210° lands in quadrant , where sine is negative. Its reference angle is 210° − 180° = 30°. So sin 210° = −sin 30° = 12. Likewise cos 330° lands in Ⅳ (cosine positive), reference 360° − 330° = 30°, giving cos 330° = +cos 30° = √32. Every "big" angle is one sign and one acute lookup away.

What to carry forward

On the unit circle a trig value is a coordinate: cos θ = x(P), sin θ = y(P), tan θ = y/x. That single idea extends sine and cosine to every angle, makes cos2θ + sin2θ = 1 automatic, and turns sign-and-symmetry questions into glances at the picture.

QuadrantⅠ (0–90°)Ⅱ (90–180°)Ⅲ (180–270°)Ⅳ (270–360°)
positive hereAllSineTangentCosine
sin θ++
cos θ++
tan θ++

Co-function / sign rule: for n·90° ± θodd n swaps the co-name (sin ↔ cos), even n keeps it; the sign comes from the quadrant of the original angle. To evaluate: reduce to the acute reference angle, then attach that sign.

Exercises

  1. The terminal side of θ meets the unit circle at P(−0.6, 0.8). Find cos θ, sin θ, and tan θ, and name the quadrant.

    Answer

    cos θ = x = −0.6, sin θ = y = 0.8, tan θ = y/x = 0.8/(−0.6) = −4/3. Since x < 0 and y > 0, P is in quadrant (check: (−0.6)2 + 0.82 = 0.36 + 0.64 = 1 ✓).

  2. Without a calculator, state the sign of each: sin 200°, cos 160°, tan 300°, cos 95°.

    Answer

    200° is Ⅲ → sin negative. 160° is Ⅱ → cos negative. 300° is Ⅳ → tan negative. 95° is Ⅱ → cos negative. (ASTC: only sine survives in Ⅱ, only tangent in Ⅲ, only cosine in Ⅳ.)

  3. Evaluate exactly using a reference angle: sin 135° and cos 240°.

    Answer

    135° is Ⅱ (sine positive), reference 180° − 135° = 45°: sin 135° = +sin 45° = √22. 240° is Ⅲ (cosine negative), reference 240° − 180° = 60°: cos 240° = −cos 60° = 12.

  4. Use a reduction formula to simplify cos(180° + θ) + cos(180° − θ).

    Answer

    cos(180° + θ) = −cos θ and cos(180° − θ) = −cos θ. Their sum is −2 cos θ. (Both partners reflect/rotate P to the left half, so both cosines flip sign.)

  5. Given sin θ = 3/5 with θ in quadrant , find cos θ and tan θ.

    Answer

    cos2θ = 1 − sin2θ = 1 − 9/25 = 16/25, so cos θ = ±4/5. In Ⅱ cosine is negative, so cos θ = −4/5. Then tan θ = sin θ/cos θ = (3/5)/(−4/5) = −3/4.

  6. Show, by symmetry on the circle, why sin(π/2 − θ) = cos θ. Then check it at θ = 30°.

    Answer

    Reflecting P across the line y = x swaps its coordinates: the point for (π/2 − θ) is the mirror of the point for θ, so its height equals the across of θ — that is sin(π/2 − θ) = cos θ. Check: sin(90° − 30°) = sin 60° = √32, and cos 30° = √32 ✓.

🎯 Quick check

Six questions to lock it in. Tap the answer you think is right.

§ For teachers and parents

This lesson covers the unit-circle extension of the trigonometric functions and the reduction (symmetry) identities — CCSS HSF-TF.A.2 (explain how the unit circle enables sine and cosine to be extended to all real numbers) and HSF-TF.A.3 / HSF-TF.C.8 (special angles, the Pythagorean identity cos2θ + sin2θ = 1, and using symmetry to relate angles). The two interactive widgets compute every value with the shared GX trig helpers, so the figure and the readout can never disagree. Next, in 25.4, we unroll this rotating point into the sine wave; the algebra of these symmetries returns as identities in 25.5.

eastmath.com · Stage 25 · 25.3 The Unit Circle · Reasoning, one step at a time