Circular & Periodic Functions

The unit circle is a circle of radius 11 centered at the origin, and it's the machine that turns any angle into a sine and cosine. Draw an angle θ\theta from the positive xx-axis, going counterclockwise. The spot where the angle's terminal side crosses the circle is the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) — cosine is the xx-coordinate, sine is the yy-coordinate. That single fact is the entire unit circle.

Everything else — exact values like sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2}, why sine is negative in Quadrant III, why trig works for angles bigger than 90∘90^\circ — falls out of reading coordinates off this circle. You don't memorize sixteen unrelated facts; you learn a handful of first-quadrant points and a sign rule for the other three quadrants.

Cosine is x, sine is y

Because the radius is 11, the right-triangle definitions collapse into coordinates. For an angle θ\theta, the hypotenuse of the reference triangle is the radius, which is 11 — so the adjacent side is just cos⁡θ\cos\theta and the opposite side is just sin⁡θ\sin\theta. The point on the circle is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).

This makes the axis angles free. At 0∘0^\circ the point is (1,0)(1, 0), so cos⁡0∘=1\cos 0^\circ = 1 and sin⁡0∘=0\sin 0^\circ = 0. At 90∘90^\circ the point is (0,1)(0, 1); at 180∘180^\circ it's (−1,0)(-1, 0); at 270∘270^\circ it's (0,−1)(0, -1). No triangle needed — just read the coordinates.

Tangent comes along for the ride: tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}, the yy-coordinate divided by the xx-coordinate. Wherever cosine is 00 (at 90∘90^\circ and 270∘270^\circ), tangent is undefined.

The diagram below shows the setup: the radius to the point on the circle is the angle's terminal side, and the point's coordinates are (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) — cosine measured horizontally, sine measured vertically.

11
sin⁡θ\sin\theta
cos⁡θ\cos\theta
θ\theta
(cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)

The three points worth memorizing

In the first quadrant there are three special angles, and their coordinates use only two values: 12\dfrac{1}{2}, 22\dfrac{\sqrt{2}}{2}, and 32\dfrac{\sqrt{3}}{2}. At 30∘30^\circ the point is (32,12)\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right). At 45∘45^\circ it's (22,22)\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right). At 60∘60^\circ it's (12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right).

A pattern makes these easy to keep straight: as the angle grows from 30∘30^\circ to 60∘60^\circ, the point climbs the circle — so xx shrinks and yy grows. The small value 12\dfrac{1}{2} and the big value 32\dfrac{\sqrt{3}}{2} simply trade places, and 45∘45^\circ is the balanced middle where both coordinates match.

In radians, these angles are π6\dfrac{\pi}{6}, π4\dfrac{\pi}{4}, and π3\dfrac{\pi}{3}. Same points, same values — radians are just a different unit for measuring the rotation, where a full circle is 2π2\pi instead of 360∘360^\circ.

Reference angles and signs in the other quadrants

Every angle outside the first quadrant borrows its values from a first-quadrant twin called the reference angle — the acute angle between the terminal side and the xx-axis. For 150∘150^\circ the reference angle is 180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ; for 225∘225^\circ it's 225∘−180∘=45∘225^\circ - 180^\circ = 45^\circ; for 300∘300^\circ it's 360∘−300∘=60∘360^\circ - 300^\circ = 60^\circ.

The reference angle supplies the size of the sine and cosine; the quadrant supplies the signs. In Quadrant II, xx is negative and yy is positive, so cosine is negative and sine is positive. In Quadrant III both coordinates are negative, so both sine and cosine are negative. In Quadrant IV, xx is positive and yy is negative, so cosine is positive and sine is negative.

So evaluating something like cos⁡225∘\cos 225^\circ is a two-part job: reference angle 45∘45^\circ gives the size 22\dfrac{\sqrt{2}}{2}, and Quadrant III makes it negative — cos⁡225∘=−22\cos 225^\circ = -\dfrac{\sqrt{2}}{2}.

Worked examples

Example 1: a first-quadrant angle

Find cos⁡60∘\cos 60^\circ and sin⁡60∘\sin 60^\circ.

Locate 60∘60^\circ on the unit circle — its point is(12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)
Cosine is the xx-coordinatecos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}
Sine is the yy-coordinatesin⁡60∘=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}

Answer: cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2} and sin⁡60∘=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}

Example 2: a Quadrant II angle

Find cos⁡135∘\cos 135^\circ.

Find the reference angle180∘−135∘=45∘180^\circ - 135^\circ = 45^\circ
The reference angle gives the sizecos⁡45∘=22\cos 45^\circ = \dfrac{\sqrt{2}}{2}
In Quadrant II, xx is negative, so cosine is negativecos⁡135∘=−22\cos 135^\circ = -\dfrac{\sqrt{2}}{2}

Answer: cos⁡135∘=−22\cos 135^\circ = -\dfrac{\sqrt{2}}{2}

Example 3: a radian angle on an axis

Find sin⁡3π2\sin \dfrac{3\pi}{2}.

Convert to a position: 3π2\dfrac{3\pi}{2} is three quarters of a full turn3π2=270∘\dfrac{3\pi}{2} = 270^\circ
The terminal side points straight down; the point on the circle is(0,−1)(0, -1)
Sine is the yy-coordinatesin⁡3π2=−1\sin \dfrac{3\pi}{2} = -1

Answer: sin⁡3π2=−1\sin \dfrac{3\pi}{2} = -1

Example 4: tangent from the coordinates

Find tan⁡30∘\tan 30^\circ.

The point at 30∘30^\circ is(32,12)\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)
Tangent is yy divided by xxtan⁡30∘=123/2\displaystyle \tan 30^\circ = \dfrac{\frac{1}{2}}{\sqrt{3}/2}
The halves canceltan⁡30∘=13\tan 30^\circ = \dfrac{1}{\sqrt{3}}
Rationalize the denominatortan⁡30∘=33\tan 30^\circ = \dfrac{\sqrt{3}}{3}

Answer: tan⁡30∘=33\tan 30^\circ = \dfrac{\sqrt{3}}{3}

Try one yourself

210∘210^\circ

Common questions

Why is the radius exactly 11?

So the coordinates equal the trig values with no extra division. On a circle of radius rr, the point is (rcos⁡θ,rsin⁡θ)(r\cos\theta, r\sin\theta); setting r=1r = 1 strips away the rr and leaves (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). It's a convenience choice that makes the circle a direct lookup table.

Do I have to memorize the whole circle?

No. Memorize the three first-quadrant points — 30∘30^\circ, 45∘45^\circ, 60∘60^\circ — plus the four axis points, and derive everything else with reference angles and quadrant signs. That's seven pieces of information doing the work of dozens.

What do I do with an angle bigger than 360∘360^\circ or a negative angle?

Add or subtract full turns of 360∘360^\circ (or 2π2\pi) until you land between 0∘0^\circ and 360∘360^\circ — the trig values don't change, because you end at the same point on the circle. For example, 390∘390^\circ lands where 30∘30^\circ does, and −45∘-45^\circ lands where 315∘315^\circ does.

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