MHF4U The Unit Circle and Exact Values: Notes and Practice

Key things to remember

1On the unit circle, the point at angle

θ

(cos θ, sin θ)

, and

tan θ = \frac{sin θ}{cos θ}

2Because

x^{2} + y^{2} = 1

on the circle,

sin^{2} θ + cos^{2} θ = 1

for every angle.

3Related angles:

π - θ

in QII,

θ - π

in QIII,

2 π - θ

in QIV.

4CAST: all positive in QI, only sine in QII, only tangent in QIII, only cosine in QIV.

5Coterminal angles differ by a multiple of

2 π

and share identical trig values.

sin \frac{π}{6} = \frac{1}{2}

cos \frac{π}{6} = \frac{3}{2}

sin \frac{π}{4} = cos \frac{π}{4} = \frac{2}{2}

sin \frac{π}{3} = \frac{3}{2}

tan \frac{π}{4} = 1

SOH CAH TOA only ever described angles inside a right triangle, so angles between $0$ and $\frac{π}{2}$ . The unit circle rescues sine and cosine from that cage: it defines them for every angle, positive or negative, however many times around you go — and it does it without changing a single one of the values you already know.

The definition that replaces SOH CAH TOA

Draw a circle of radius $1$ centred at the origin. Put an angle $θ$ in standard position: its initial arm along the positive $x$ -axis, rotating counterclockwise. Wherever the terminal arm crosses the circle, that point is DEFINED to be $(cos θ, sin θ)$ . Cosine is the $x$ -coordinate and sine is the $y$ -coordinate — that is the whole definition, and $tan θ = \frac{y}{x} = \frac{sin θ}{cos θ}$ follows from it. Two things fall out immediately. First, because the point always sits on a circle of radius $1$ , both coordinates are trapped between $- 1$ and $1$ , which is why $sin θ$ and $cos θ$ can never exceed $1$ in size. Second, because $x^{2} + y^{2} = 1$ for any point on the unit circle, we get $cos^{2} θ + sin^{2} θ = 1$ for free — the Pythagorean identity is literally the equation of the circle.

Fig 1The unit circle labelled in radians, with the exact coordinates at every special angle.

Exact values from the special triangles

The unit circle tells you WHERE the point is, but the special triangles tell you the exact coordinates. Slice a square in half and you get the $\frac{π}{4}$ triangle with sides $1, 1, 2$ . Slice an equilateral triangle in half and you get the $\frac{π}{6}$ and $\frac{π}{3}$ triangle with sides $1, 3, 2$ . Drop a perpendicular from the unit-circle point to the $x$ -axis and you are looking at one of those two triangles, scaled so the hypotenuse is $1$ .

Fig 2The two triangles that generate every exact trig value on the circle.

$θ$	$sin θ$	$cos θ$	$tan θ$
$0$	$0$	$1$	$0$
$\frac{π}{6}$	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{1}{3} = \frac{3}{3}$
$\frac{π}{4}$	$\frac{2}{2}$	$\frac{2}{2}$	$1$
$\frac{π}{3}$	$\frac{3}{2}$	$\frac{1}{2}$	$3$
$\frac{π}{2}$	$1$	$0$	undefined

The five rows worth memorizing. Everything else on the circle is one of these with a sign attached.

Related angles and the CAST rule

For an angle outside the first quadrant, the RELATED ANGLE (also called the reference angle) is the acute angle between the terminal arm and the $x$ -axis. Its trig values have exactly the same size as the original angle's — the terminal arm is a mirror image, so the coordinates are the same numbers. Only the SIGNS change, and CAST records which signs survive in which quadrant.

Fig 3All positive in quadrant one, Sine in two, Tangent in three, Cosine in four.

Quadrant	$θ$ lies in	Related angle	Positive there
I	$(0, \frac{π}{2})$	$θ$	all three
II	$(\frac{π}{2}, π)$	$π - θ$	$sin$ only
III	$(π, \frac{3 π}{2})$	$θ - π$	$tan$ only
IV	$(\frac{3 π}{2}, 2 π)$	$2 π - θ$	$cos$ only

Related-angle formulas in radians. Notice 180° has simply become pi.

VideoShows how to find a reference angle, take the trig value from a special triangle, and attach the sign from the quadrant, which is the same routine this section uses.Watch on YouTube

Coterminal and negative angles

Two angles are COTERMINAL when their terminal arms land in the same place, which happens exactly when they differ by a whole number of revolutions. So adding or subtracting $2 π$ (as many times as you like) never changes any trig value: $sin (\frac{13 π}{6}) = sin (\frac{13 π}{6} - 2 π) = sin (\frac{π}{6}) = \frac{1}{2} .$ A NEGATIVE angle simply rotates clockwise instead, so $- \frac{π}{3}$ lands in the same place as $- \frac{π}{3} + 2 π = \frac{5 π}{3}$ , down in quadrant four.

Loading interactive…

Drag the angle around and watch the coordinates, the related angle, and the signs update together.

A three-step recipe for any exact value

One: decide which quadrant the angle is in. Two: find the related angle and look its value up in the special-triangle table. Three: attach the sign that CAST demands for that quadrant. Every exact-value question in this course is those three steps and nothing more.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Find the exact value of

sin \frac{2 π}{3}

Example 2

Find the exact value of

cos (- \frac{5 π}{4})

Example 3

State the exact coordinates of the point on the unit circle at

θ = \frac{7 π}{6}

Example 4

Solve

sin θ = - \frac{3}{2}

for

0 \leq θ < 2 π

Where marks get lost

Swapping the coordinates: writing the unit-circle point as

(sin θ, cos θ)

. Cosine comes first because cosine is the horizontal (

x

) coordinate.

Finding the related angle correctly and then forgetting the sign, so

cos \frac{5 π}{6}

comes out as

+ \frac{3}{2}

instead of

- \frac{3}{2}

Using the wrong related-angle formula for the quadrant — applying

π - θ

in quadrant III instead of

θ - π

produces a negative related angle, which is an immediate signal that you picked the wrong row.

Assuming a negative angle automatically gives a negative answer. The sign depends only on which quadrant the arm lands in:

cos (- \frac{π}{3}) = + \frac{1}{2}

, because

- \frac{π}{3}

is in quadrant IV where cosine is positive.

Giving only one solution when solving an equation over

[0, 2 π)

. Except for the values 1 and −1, every sine or cosine value between −1 and 1 occurs at exactly two angles in

[0, 2 π)

Key things to remember

Worked examples

Where marks get lost

Flashcards 9

Practice