The Unit Circle and Circular Functions

Study Sheet

The Unit Circle and Circular Functions

Special angles, reference angles, signs, periodicity, even/odd, and terminal points

The Unit Circle and the Six Functions

Concept
Definition from a point on the circle

The unit circle is the circle x2+y2=1x^2+y^2=1 centered at the origin with radius 11. If the terminal side of an angle θ\theta (measured counterclockwise from the positive xx-axis) meets the unit circle at the point (x,y)(x,y), then

x=cos⁡θ,y=sin⁡θ.x=\cos\theta,\qquad y=\sin\theta.

The other four functions are built from these:

tan⁡θ=yx=sin⁡θcos⁡θ,cot⁡θ=xy=cos⁡θsin⁡θ,sec⁡θ=1x=1cos⁡θ,csc⁡θ=1y=1sin⁡θ.\tan\theta=\dfrac{y}{x}=\dfrac{\sin\theta}{\cos\theta},\quad \cot\theta=\dfrac{x}{y}=\dfrac{\cos\theta}{\sin\theta},\quad \sec\theta=\dfrac{1}{x}=\dfrac{1}{\cos\theta},\quad \csc\theta=\dfrac{1}{y}=\dfrac{1}{\sin\theta}.
Example
Worked example: reading all six functions off a point

The terminal side of θ\theta meets the unit circle at (−12,32)\left(-\tfrac{1}{2},\tfrac{\sqrt3}{2}\right). Find all six functions.

Here x=−12x=-\tfrac12 and y=32y=\tfrac{\sqrt3}{2}, so

cos⁡θ=−12,sin⁡θ=32,tan⁡θ=3/2−1/2=−3.\cos\theta=-\tfrac12,\qquad \sin\theta=\tfrac{\sqrt3}{2},\qquad \tan\theta=\dfrac{\sqrt3/2}{-1/2}=-\sqrt3.

Reciprocals:

sec⁡θ=1−1/2=−2,csc⁡θ=13/2=23=233,cot⁡θ=1−3=−33.\sec\theta=\dfrac{1}{-1/2}=-2,\qquad \csc\theta=\dfrac{1}{\sqrt3/2}=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3},\qquad \cot\theta=\dfrac{1}{-\sqrt3}=-\dfrac{\sqrt3}{3}.
Tip

Tip. On the unit circle the radius is 11, so there is no dividing by rr: the coordinate is the value. Cosine is the xx-coordinate, sine is the yy-coordinate. “Co-sine goes with the horizontal.”

Exact Values at the Special Angles (All Four Quadrants)

Concept
The special-angle circle

The special angles are multiples of 30∘=π630^\circ=\tfrac{\pi}{6} and 45∘=π445^\circ=\tfrac{\pi}{4}. Their coordinates use only 0, 12, 22, 32, 10,\ \tfrac12,\ \tfrac{\sqrt2}{2},\ \tfrac{\sqrt3}{2},\ 1. Learn Quadrant I, then attach the correct signs in the other quadrants.

1.25

Example
Worked example: values in Quadrants II and III

Evaluate cos⁡150∘\cos 150^\circ and sin⁡5π4\sin\dfrac{5\pi}{4}.

150∘150^\circ is in Quadrant II, where cosine is negative, and it corresponds to the point (−32,12)\left(-\tfrac{\sqrt3}{2},\tfrac12\right), so cos⁡150∘=−32\cos 150^\circ=-\dfrac{\sqrt3}{2}.

5π4=225∘\dfrac{5\pi}{4}=225^\circ is in Quadrant III at (−22,−22)\left(-\tfrac{\sqrt2}{2},-\tfrac{\sqrt2}{2}\right), so sin⁡5π4=−22\sin\dfrac{5\pi}{4}=-\dfrac{\sqrt2}{2}.

Tip

Tip. At the quadrantal angles 0,π2,π,3π20,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2} one coordinate is 00: watch for undefined values (tan⁡π2\tan\tfrac{\pi}{2} and sec⁡π2\sec\tfrac{\pi}{2} are undefined because x=0x=0; cot⁡0\cot 0 and csc⁡0\csc 0 are undefined because y=0y=0).

Reference Angles and Signs by Quadrant

Concept
Reference angle

The reference angle θ′\theta' is the acute angle between the terminal side and the xx-axis. A special angle and its reference angle share the same exact value up to sign.

Q I: θ′=θ,Q II: θ′=180∘−θ,Q III: θ′=θ−180∘,Q IV: θ′=360∘−θ.\text{Q I: } \theta'=\theta,\qquad \text{Q II: } \theta'=180^\circ-\theta,\qquad \text{Q III: } \theta'=\theta-180^\circ,\qquad \text{Q IV: } \theta'=360^\circ-\theta.

In radians replace 180∘180^\circ with π\pi and 360∘360^\circ with 2π2\pi.

Example
Worked example: reference angle then sign

Evaluate sin⁡210∘\sin 210^\circ.

210∘210^\circ is in Quadrant III, so its reference angle is 210∘−180∘=30∘210^\circ-180^\circ=30^\circ. The reference value is sin⁡30∘=12\sin 30^\circ=\tfrac12. In Quadrant III sine is negative, so

sin⁡210∘=−12.\sin 210^\circ=-\tfrac12.
Tip

Remember: ASTC (“All Students Take Calculus”), reading Quadrants I, II, III, IV. It tells you which functions are positive; every other function is negative there. A function and its reciprocal always share the same sign.

Domain, Range, and Periodicity

Concept
The six circular functions at a glance
2

1.3

Here nn is any integer. Periodic means f(θ+P)=f(θ)f(\theta+P)=f(\theta) for the period PP.

Amplitude is the height; period is one full cycle.

Example
Worked example: using periodicity to reduce a large angle

Evaluate cos⁡13π6\cos\dfrac{13\pi}{6}.

Since cosine has period 2π=12π62\pi=\dfrac{12\pi}{6}, subtract one full turn:

13π6−2π=13π6−12π6=π6.\dfrac{13\pi}{6}-2\pi=\dfrac{13\pi}{6}-\dfrac{12\pi}{6}=\dfrac{\pi}{6}.

Therefore cos⁡13π6=cos⁡π6=32\cos\dfrac{13\pi}{6}=\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}.

Tip

Tip. Tangent and cotangent repeat every π\pi, not 2π2\pi. To simplify tan⁡θ\tan\theta for a huge angle, subtract multiples of π\pi; for the other four functions subtract multiples of 2π2\pi.

Even and Odd (Symmetry) Properties

Concept
Even and odd functions

Reflecting an angle to −θ-\theta reflects the point (x,y)(x,y) across the xx-axis to (x,−y)(x,-y). So cosine (the xx-coordinate) is unchanged, while sine (the yy-coordinate) flips sign:

cos⁡(−θ)=cos⁡θ,sec⁡(−θ)=sec⁡θ⏟even,sin⁡(−θ)=−sin⁡θ,  csc⁡(−θ)=−csc⁡θ,  tan⁡(−θ)=−tan⁡θ,  cot⁡(−θ)=−cot⁡θ⏟odd.\underbrace{\cos(-\theta)=\cos\theta,\quad \sec(-\theta)=\sec\theta}_{\text{even}},\qquad \underbrace{\sin(-\theta)=-\sin\theta,\ \ \csc(-\theta)=-\csc\theta,\ \ \tan(-\theta)=-\tan\theta,\ \ \cot(-\theta)=-\cot\theta}_{\text{odd}}.
Example
Worked example: applying even/odd

Evaluate sin⁡ ⁣(−π6)\sin\!\left(-\dfrac{\pi}{6}\right) and simplify cos⁡(−θ) csc⁡(−θ)\cos(-\theta)\,\csc(-\theta).

Sine is odd: sin⁡ ⁣(−π6)=−sin⁡π6=−12\sin\!\left(-\dfrac{\pi}{6}\right)=-\sin\dfrac{\pi}{6}=-\dfrac12.

Cosine is even and cosecant is odd, so

cos⁡(−θ) csc⁡(−θ)=cos⁡θ⋅(−csc⁡θ)=−cos⁡θsin⁡θ=−cot⁡θ.\cos(-\theta)\,\csc(-\theta)=\cos\theta\cdot(-\csc\theta)=-\dfrac{\cos\theta}{\sin\theta}=-\cot\theta.
Tip

Tip. Only cosine and secant are even. The other four are odd, so a negative angle just pulls a minus sign out front.

Evaluating from a Point on the Terminal Side

Concept
When the point is not on the unit circle

If the terminal side passes through any point (x,y)≠(0,0)(x,y)\neq(0,0), let

r=x2+y2>0.r=\sqrt{x^2+y^2}>0.

Then

sin⁡θ=yr,cos⁡θ=xr,tan⁡θ=yx,csc⁡θ=ry,sec⁡θ=rx,cot⁡θ=xy.\sin\theta=\dfrac{y}{r},\quad \cos\theta=\dfrac{x}{r},\quad \tan\theta=\dfrac{y}{x},\quad \csc\theta=\dfrac{r}{y},\quad \sec\theta=\dfrac{r}{x},\quad \cot\theta=\dfrac{x}{y}.

The unit circle is the special case r=1r=1.

Example
Worked example: all six from a terminal point

The terminal side of θ\theta passes through (−3,4)(-3,4). Find all six functions.

First r=(−3)2+42=9+16=25=5r=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5. The point is in Quadrant II. Then

sin⁡θ=45,cos⁡θ=−35,tan⁡θ=4−3=−43,\sin\theta=\dfrac{4}{5},\quad \cos\theta=-\dfrac{3}{5},\quad \tan\theta=\dfrac{4}{-3}=-\dfrac{4}{3},
csc⁡θ=54,sec⁡θ=−53,cot⁡θ=−34.\csc\theta=\dfrac{5}{4},\quad \sec\theta=-\dfrac{5}{3},\quad \cot\theta=-\dfrac{3}{4}.

The signs match Quadrant II: only sine and cosecant are positive.

Tip

Tip. Always take rr positive. All the sign information comes from the signs of xx and yy (that is, from the quadrant). If an answer has a radical in the denominator, rationalize it, e.g. 213=21313\dfrac{2}{\sqrt{13}}=\dfrac{2\sqrt{13}}{13}.

Going Deeper: Advanced Unit-Circle Ideas

Concept
Combining exact values: products, sums, and powers
2

Once the special-angle coordinates are memorized, any expression built from them is just arithmetic with the numbers 0, 12, 22, 32, 10,\ \tfrac12,\ \tfrac{\sqrt2}{2},\ \tfrac{\sqrt3}{2},\ 1 and their reciprocals. Two cautions:

  • The power notation means “square the value,” so sin⁡2θ=(sin⁡θ)2\sin^2\theta=(\sin\theta)^2. For example sin⁡2π3=(32)2=34\sin^2\tfrac{\pi}{3}=\left(\tfrac{\sqrt3}{2}\right)^2=\tfrac34.
  • Evaluate each function first (with its correct sign), then multiply, add, or take reciprocals. A single wrong sign changes the whole answer.

A handy check is the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, which must hold at every angle.

Amplitude is the height; period is one full cycle.

Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Example
Worked example: an unusual product-and-sum

Evaluate

E=tan⁡π3 cot⁡π6  −  csc⁡2π4  +  cos⁡π.E=\tan\dfrac{\pi}{3}\,\cot\dfrac{\pi}{6}\;-\;\csc^2\dfrac{\pi}{4}\;+\;\cos\pi.

Take the pieces one at a time:

tan⁡π3=3,cot⁡π6=3,csc⁡π4=2,cos⁡π=−1.\tan\dfrac{\pi}{3}=\sqrt3,\qquad \cot\dfrac{\pi}{6}=\sqrt3,\qquad \csc\dfrac{\pi}{4}=\sqrt2,\qquad \cos\pi=-1.

So tan⁡π3 cot⁡π6=3⋅3=3\tan\tfrac{\pi}{3}\,\cot\tfrac{\pi}{6}=\sqrt3\cdot\sqrt3=3 and csc⁡2π4=(2)2=2\csc^2\tfrac{\pi}{4}=(\sqrt2)^2=2. Therefore

E=3−2+(−1)=0.E=3-2+(-1)=0.
Concept
Equally spaced angles: the cosines (and sines) sum to zero

If you place n≥2n\ge 2 points evenly around the unit circle, their xx-coordinates cancel and so do their yy-coordinates:

∑k=0n−1cos⁡ ⁣(θ+2πkn)=0,∑k=0n−1sin⁡ ⁣(θ+2πkn)=0.\sum_{k=0}^{n-1}\cos\!\left(\theta+\dfrac{2\pi k}{n}\right)=0, \qquad \sum_{k=0}^{n-1}\sin\!\left(\theta+\dfrac{2\pi k}{n}\right)=0.

Geometrically the tips of the nn equally spaced unit vectors form a regular polygon centered at the origin, so the vectors add to the zero vector; the two coordinate sums are just the horizontal and vertical parts of that fact.

Example
Worked example: three equally spaced angles

Evaluate cos⁡0+cos⁡2π3+cos⁡4π3\cos 0+\cos\dfrac{2\pi}{3}+\cos\dfrac{4\pi}{3}.

These are n=3n=3 equally spaced angles (120∘120^\circ apart). Reading the coordinates,

cos⁡0=1,cos⁡2π3=−12,cos⁡4π3=−12,\cos 0=1,\qquad \cos\dfrac{2\pi}{3}=-\dfrac12,\qquad \cos\dfrac{4\pi}{3}=-\dfrac12,

so the sum is

1+(−12)+(−12)=0,1+\left(-\dfrac12\right)+\left(-\dfrac12\right)=0,

exactly as the boxed rule predicts. The matching sine sum is 0+32+(−32)=00+\tfrac{\sqrt3}{2}+\left(-\tfrac{\sqrt3}{2}\right)=0 as well.

Tip

Reducing a huge or negative angle. To evaluate f(θ)f(\theta) for a giant or negative θ\theta, add or subtract whole periods until you land in [0,2π)[0,2\pi):

θreduced=θ−2π⌊θ2π⌋(use period π for tan⁡,cot⁡).\theta_{\text{reduced}}=\theta-2\pi\left\lfloor \dfrac{\theta}{2\pi}\right\rfloor \quad(\text{use period }\pi\text{ for }\tan,\cot).

In practice: keep adding 2π=12π62\pi=\tfrac{12\pi}{6} (or subtracting it) until the angle is a familiar special angle. Coterminal angles have identical function values.

Example
Worked example: a large negative angle

Evaluate sin⁡ ⁣(−17π6)\sin\!\left(-\dfrac{17\pi}{6}\right).

Add full turns of 2π=12π62\pi=\dfrac{12\pi}{6} until the angle is in [0,2π)[0,2\pi):

−17π6+12π6=−5π6,−5π6+12π6=7π6.-\dfrac{17\pi}{6}+\dfrac{12\pi}{6}=-\dfrac{5\pi}{6}, \qquad -\dfrac{5\pi}{6}+\dfrac{12\pi}{6}=\dfrac{7\pi}{6}.

Now 7π6=210∘\dfrac{7\pi}{6}=210^\circ is in Quadrant III with reference angle π6\dfrac{\pi}{6}, and sine is negative there, so

sin⁡ ⁣(−17π6)=sin⁡7π6=−12.\sin\!\left(-\dfrac{17\pi}{6}\right)=\sin\dfrac{7\pi}{6}=-\dfrac12.

(Check with the odd property: sin⁡ ⁣(−17π6)=−sin⁡17π6=−sin⁡5π6=−12\sin\!\left(-\tfrac{17\pi}{6}\right)=-\sin\tfrac{17\pi}{6}=-\sin\tfrac{5\pi}{6}=-\tfrac12.)

Concept
One value plus a sign condition (the twist)

A classic problem gives one function value together with a sign clue instead of a named quadrant, and often the value is a reciprocal function. Strategy:

  • Convert to sin⁡θ\sin\theta or cos⁡θ\cos\theta if you were handed sec⁡,csc⁡,cot⁡,\sec,\csc,\cot, or tan⁡\tan.
  • Use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 to get the partner coordinate, keeping it as a radical.
  • Let the two sign clues (or the quadrant) fix the signs of x=cos⁡θx=\cos\theta and y=sin⁡θy=\sin\theta.
  • Build the remaining functions as quotients and reciprocals; rationalize denominators.
Example
Worked example: all six from sec⁡θ=−3\sec\theta=-3 with sin⁡θ>0\sin\theta>0

The reciprocal relation gives cos⁡θ=1sec⁡θ=−13\cos\theta=\dfrac{1}{\sec\theta}=-\dfrac13. Since cos⁡θ<0\cos\theta<0 and sin⁡θ>0\sin\theta>0, the angle is in Quadrant II. From the Pythagorean identity,

sin⁡2θ=1−cos⁡2θ=1−19=89,sin⁡θ=+89=223(positive in Q II).\begin{aligned} \sin^2\theta &= 1-\cos^2\theta = 1-\dfrac19 = \dfrac89,\\ \sin\theta &= +\sqrt{\dfrac89}=\dfrac{2\sqrt2}{3}\quad(\text{positive in Q\,II}). \end{aligned}

The rest follow:

tan⁡θ=sin⁡θcos⁡θ=22/3−1/3=−22,cot⁡θ=−122=−24,\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{2\sqrt2/3}{-1/3}=-2\sqrt2,\qquad \cot\theta=-\dfrac{1}{2\sqrt2}=-\dfrac{\sqrt2}{4},
csc⁡θ=1sin⁡θ=322=324,sec⁡θ=−3,cos⁡θ=−13.\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{3}{2\sqrt2}=\dfrac{3\sqrt2}{4},\qquad \sec\theta=-3,\qquad \cos\theta=-\dfrac13.

Signs check against Quadrant II: only sin⁡θ\sin\theta and csc⁡θ\csc\theta are positive.

Example
Worked example: simplifying a long even/odd expression

Simplify, for a general angle θ\theta,

sin⁡(−θ) sec⁡(−θ)tan⁡(−θ)+cos⁡(−θ).\dfrac{\sin(-\theta)\,\sec(-\theta)}{\tan(-\theta)}+\cos(-\theta).

Pull each negative sign through using even/odd: sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta, sec⁡(−θ)=sec⁡θ\sec(-\theta)=\sec\theta, tan⁡(−θ)=−tan⁡θ\tan(-\theta)=-\tan\theta, and cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta. The fraction becomes

(−sin⁡θ)(sec⁡θ)−tan⁡θ=−sin⁡θcos⁡θ−tan⁡θ=−tan⁡θ−tan⁡θ=1.\dfrac{(-\sin\theta)(\sec\theta)}{-\tan\theta} =\dfrac{-\dfrac{\sin\theta}{\cos\theta}}{-\tan\theta} =\dfrac{-\tan\theta}{-\tan\theta}=1.

Adding the last term,

sin⁡(−θ) sec⁡(−θ)tan⁡(−θ)+cos⁡(−θ)=1+cos⁡θ.\dfrac{\sin(-\theta)\,\sec(-\theta)}{\tan(-\theta)}+\cos(-\theta)=1+\cos\theta.

Two odd factors (sin⁡\sin and tan⁡\tan) and one flipped sign in the denominator cancel, leaving no minus signs.

Concept
Parameterized terminal points

Writing P(θ)=(cos⁡θ, sin⁡θ)P(\theta)=(\cos\theta,\ \sin\theta) turns the unit circle into a machine: feed in the angle, read out the terminal point. Rotations and reflections of the angle move the point in predictable ways:

P(−θ)=(cos⁡θ, −sin⁡θ),P(θ+π)=(−cos⁡θ, −sin⁡θ),P ⁣(θ+π2)=(−sin⁡θ, cos⁡θ).P(-\theta)=(\cos\theta,\,-\sin\theta),\quad P(\theta+\pi)=(-\cos\theta,\,-\sin\theta),\quad P\!\left(\theta+\tfrac{\pi}{2}\right)=(-\sin\theta,\,\cos\theta).

So if you know one terminal point, you instantly know the antipodal point (+π+\pi), the reflection across the xx-axis (−θ-\theta), and the quarter-turn (+π2+\tfrac{\pi}{2})---no new radicals required.

Example
Worked example: reusing one terminal point

The terminal point of θ\theta on the unit circle is (53, 23)\left(\dfrac{\sqrt5}{3},\,\dfrac23\right). (It is valid because (53)2+(23)2=59+49=1\left(\tfrac{\sqrt5}{3}\right)^2+\left(\tfrac23\right)^2=\tfrac59+\tfrac49=1.) Find the terminal points of θ+π\theta+\pi and −θ-\theta, and evaluate cos⁡(θ+π)\cos(\theta+\pi).

Using the parameterization rules,

P(θ+π)=(−53, −23),P(−θ)=(53, −23).P(\theta+\pi)=\left(-\dfrac{\sqrt5}{3},\,-\dfrac23\right),\qquad P(-\theta)=\left(\dfrac{\sqrt5}{3},\,-\dfrac23\right).

Because cosine is the xx-coordinate of P(θ+π)P(\theta+\pi),

cos⁡(θ+π)=−cos⁡θ=−53.\cos(\theta+\pi)=-\cos\theta=-\dfrac{\sqrt5}{3}.

Formulas, Proofs & Tips

Tip
The Pythagorean identity
sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1

What it means. Knowing one of sin⁡\sin or cos⁡\cos (plus the quadrant) determines the other.

Example. If sin⁡θ=35\sin\theta=\tfrac35, then cos⁡θ=1−925=45\cos\theta=\sqrt{1-\tfrac{9}{25}}=\tfrac45.

Why it works. On the unit circle the point at angle θ\theta is (cos⁡θ, sin⁡θ)(\cos\theta,\ \sin\theta) and lies at distance 11 from the origin. The distance formula gives cos⁡2θ+sin⁡2θ=12\cos^2\theta+\sin^2\theta=1^2 — it is Pythagoras on a radius.

Tip. Dividing through by cos⁡2θ\cos^2\theta gives 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta; by sin⁡2θ\sin^2\theta gives 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta. Use the quadrant to pick the sign when you take the square root.