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
θ(cos θ, sin θ)cos θsin θ1
The radius is 11, so the legs of the triangle are cos⁡θ\cos\theta and sin⁡θ\sin\theta.

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}.

For an acute angle this is SOH-CAH-TOA with a hypotenuse of 11: the radius to PP is the hypotenuse, the horizontal leg is x1=cos⁡θ\frac{x}{1}=\cos\theta, and the vertical leg is y1=sin⁡θ\frac{y}{1}=\sin\theta. Past 90∘90^\circ there is no right triangle containing θ\theta, but the point PP still exists, so its coordinates become the definition. Signs now come for free: in Quadrant II the point is left of the yy-axis, so cos⁡θ<0\cos\theta<0.

Example
Worked example: reading all six functions off a point
(−1/2, √3/2)−1/2
The point sits left of the yy-axis, so cosine, tangent, secant and cotangent are negative.

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}.
Side note
Where it goes wrong

Getting a sign wrong in quadrants II–IV. Use the reference angle for the size and the quadrant (All Students Take Calculus) for the sign.

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.”

Try it
Quick check

The terminal side of θ\theta meets the unit circle at (35,−45)\left(\tfrac35,-\tfrac45\right). Find all six trigonometric functions of θ\theta.

Show the answer

Answer: cos⁡θ=35\cos\theta=\tfrac35, sin⁡θ=−45\sin\theta=-\tfrac45, tan⁡θ=−4/53/5=−43\tan\theta=\dfrac{-4/5}{3/5}=-\tfrac43, sec⁡θ=53\sec\theta=\tfrac53, csc⁡θ=−54\csc\theta=-\tfrac54, cot⁡θ=−34\cot\theta=-\tfrac34. The point is in Quadrant IV (x>0x>0, y<0y<0), where only cosine and secant are positive, and that is what the signs show. The point really is on the circle: 925+1625=1\tfrac{9}{25}+\tfrac{16}{25}=1.

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.

Learn the three Quadrant I points; every other point is one of them with signs changed.

Example
Worked example: values in Quadrants II and III
150°225°30°45°
Each point makes its reference angle with the negative xx-axis.

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).

Try it
Quick check

Evaluate exactly: sin⁡300∘\sin 300^\circ, cos⁡3π4\cos\dfrac{3\pi}{4}, and tan⁡7π6\tan\dfrac{7\pi}{6}.

Show the answer

Answer: −32-\dfrac{\sqrt3}{2}, −22-\dfrac{\sqrt2}{2}, 33\dfrac{\sqrt3}{3}. 300∘300^\circ is the 60∘60^\circ point reflected into Quadrant IV, (12,−32)\left(\tfrac12,-\tfrac{\sqrt3}{2}\right), so its sine is −32-\tfrac{\sqrt3}{2}. 3π4=135∘\tfrac{3\pi}{4}=135^\circ is (−22,22)\left(-\tfrac{\sqrt2}{2},\tfrac{\sqrt2}{2}\right). 7π6=210∘\tfrac{7\pi}{6}=210^\circ is (−32,−12)\left(-\tfrac{\sqrt3}{2},-\tfrac12\right), and tan⁡=−1/2−3/2=13=33\tan=\dfrac{-1/2}{-\sqrt3/2}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}: both coordinates are negative, so the quotient is positive.

Reference Angles and Signs by Quadrant

Concept
Reference angle
40°θ = 40°40°140°40°220°40°320°
Four angles, one per quadrant, all with reference angle 40∘40^\circ (measured to the xx-axis).

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.

Reflecting a point across the xx-axis, the yy-axis, or both changes only the signs of its coordinates, never their sizes. The four points at θ′\theta', 180∘−θ′180^\circ-\theta', 180∘+θ′180^\circ+\theta' and 360∘−θ′360^\circ-\theta' are exactly such reflections of each other, so their sines and cosines agree up to sign. The reference angle finds the Quadrant I twin; the quadrant supplies the sign.

Example
Worked example: reference angle then sign
30°210°(−√3/2, −1/2)
The point for 210∘210^\circ lies below the axis, so its sine is −12-\tfrac12.

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.

Try it
Quick check

(a) Find the reference angle of 305∘305^\circ. (b) Evaluate cos⁡4π3\cos\dfrac{4\pi}{3} exactly.

Show the answer

(a) 55∘55^\circ. 305∘305^\circ is in Quadrant IV, so θ′=360∘−305∘=55∘\theta'=360^\circ-305^\circ=55^\circ. (b) −12-\tfrac12. 4π3\dfrac{4\pi}{3} is in Quadrant III with reference angle 4π3−π=π3\dfrac{4\pi}{3}-\pi=\dfrac{\pi}{3}; cos⁡π3=12\cos\dfrac{\pi}{3}=\tfrac12, and cosine is negative in Quadrant III.

Domain, Range, and Periodicity

Concept
The six circular functions at a glance
fdomainrangeperiodsin⁡all reals[−1,1]2πcos⁡all reals[−1,1]2πtan⁡x≠π2+nπall realsπcot⁡x≠nπall realsπsec⁡x≠π2+nπ∣y∣≥12πcsc⁡x≠nπ∣y∣≥12π\begin{array}{c|c|c|c} f & \text{domain} & \text{range} & \text{period}\\ \hline \sin & \text{all reals} & [-1,1] & 2\pi\\ \cos & \text{all reals} & [-1,1] & 2\pi\\ \tan & x\neq \tfrac{\pi}{2}+n\pi & \text{all reals} & \pi\\ \cot & x\neq n\pi & \text{all reals} & \pi\\ \sec & x\neq \tfrac{\pi}{2}+n\pi & |y|\ge 1 & 2\pi\\ \csc & x\neq n\pi & |y|\ge 1 & 2\pi \end{array}

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

Every row reads straight off the circle. Sine and cosine are coordinates of a point on a circle of radius 11, so they never leave [−1,1][-1,1]. Tangent and secant divide by x=cos⁡θx=\cos\theta, so they are undefined where the point is on the yy-axis (θ=π2+nπ\theta=\tfrac{\pi}{2}+n\pi); cotangent and cosecant divide by yy and fail on the xx-axis. Tangent repeats after half a turn because the point (−x,−y)(-x,-y) opposite (x,y)(x,y) gives the same ratio yx\tfrac{y}{x}.

Example
Worked example: using periodicity to reduce a large angle
13π/6same point as π/6
One full turn plus π6\tfrac{\pi}{6} lands on the π6\tfrac{\pi}{6} point.

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.

Try it
Quick check

Evaluate tan⁡19π4\tan\dfrac{19\pi}{4} exactly.

Show the answer

Answer: −1-1. Tangent has period π\pi, so subtract four half-turns: 19π4−4π=19π−16π4=3π4\dfrac{19\pi}{4}-4\pi=\dfrac{19\pi-16\pi}{4}=\dfrac{3\pi}{4}. The point for 3π4\dfrac{3\pi}{4} is (−22,22)\left(-\tfrac{\sqrt2}{2},\tfrac{\sqrt2}{2}\right), so tan⁡3π4=−1\tan\dfrac{3\pi}{4}=-1.

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
(√3/2, 1/2)(√3/2, −1/2)π/6−π/6
Negating the angle reflects the point: same xx, opposite yy.

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.

Try it
Quick check

Evaluate cos⁡ ⁣(−5π3)+sin⁡ ⁣(−π2)\cos\!\left(-\dfrac{5\pi}{3}\right)+\sin\!\left(-\dfrac{\pi}{2}\right).

Show the answer

Answer: −12-\tfrac12. Cosine is even, so cos⁡ ⁣(−5π3)=cos⁡5π3=12\cos\!\left(-\tfrac{5\pi}{3}\right)=\cos\tfrac{5\pi}{3}=\tfrac12 (Quadrant IV, reference angle π3\tfrac{\pi}{3}). Sine is odd, so sin⁡ ⁣(−π2)=−sin⁡π2=−1\sin\!\left(-\tfrac{\pi}{2}\right)=-\sin\tfrac{\pi}{2}=-1. The sum is 12−1=−12\tfrac12-1=-\tfrac12.

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.

The point (x,y)(x,y) at distance rr and the unit-circle point on the same ray form similar right triangles with the xx-axis; the larger is the smaller scaled by rr. So the unit-circle coordinates are (xr,yr)\left(\tfrac{x}{r},\tfrac{y}{r}\right), which is exactly (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). Any point on the ray gives the same answers, because the scale factor cancels.

Example
Worked example: all six from a terminal point
(−3, 4)r = 5−34θ
Divide the coordinates by r=5r=5 to land on the unit circle (dashed).

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.

Side note
Where it goes wrong

Forgetting to divide by rr. For the point (−3,4)(-3,4) it is tempting to write sin⁡θ=4\sin\theta=4, reading the yy-coordinate as if the point were on the unit circle. But a sine can never exceed 11; the point is 55 units out, so sin⁡θ=45\sin\theta=\tfrac{4}{5}. Only on the unit circle, where r=1r=1, are the coordinates themselves the values.

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}.

Try it
Quick check

The terminal side of θ\theta passes through (5,−12)(5,-12). Find sin⁡θ\sin\theta, cos⁡θ\cos\theta, tan⁡θ\tan\theta, and csc⁡θ\csc\theta.

Show the answer

Answer: r=25+144=13r=\sqrt{25+144}=13, so sin⁡θ=−1213\sin\theta=-\dfrac{12}{13}, cos⁡θ=513\cos\theta=\dfrac{5}{13}, tan⁡θ=−125\tan\theta=-\dfrac{12}{5}, csc⁡θ=−1312\csc\theta=-\dfrac{13}{12}. The point is in Quadrant IV, so sine, tangent and cosecant are negative, and cosine is positive.

Going Deeper: Advanced Unit-Circle Ideas

Concept
Combining exact values: products, sums, and powers

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.

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
(1, 0)(−1/2, √3/2)(−1/2, −√3/2)
Three unit vectors 120∘120^\circ apart: the xx-coordinates 1−12−121-\tfrac12-\tfrac12 cancel.

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
(−1/3, 2√2/3)−1/3
cos⁡θ=−13\cos\theta=-\tfrac13 and sin⁡θ>0\sin\theta>0 put the point in Quadrant II.

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
P(θ)P(θ + π)P(−θ)
Adding π\pi sends the point through the origin; negating θ\theta reflects it across the xx-axis.

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}.

Why It Works

For an acute angle the unit-circle definition agrees with SOH-CAH-TOA: the radius is the hypotenuse of length 11, so the xx-coordinate is the adjacent side (cos⁡θ\cos\theta) and the yy-coordinate the opposite side (sin⁡θ\sin\theta). Defining the functions as coordinates for ALL angles keeps every identity true beyond the right triangle, and periodicity is free: going around 2π2\pi returns to the same point. Reference angles work because reflecting a point across an axis only changes signs of coordinates.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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.