Building the Unit Circle

An angle, which can be expressed in both degrees and radians, measures the distance of turn in a rotation. Because rotations follow a circular path, angles can be used when analyzing circles and vice versa. In fact, there's a special relationship between right triangles and a circle of radius

1 .

Rule

Special Right Triangles

Right triangles with certain acute angle measures are considered noteworthy. These include the following angle relationships.

$3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$
$4 5^{\circ}$ - $4 5^{\circ}$ - $9 0^{\circ}$
$6 0^{\circ}$ - $3 0^{\circ}$ - $9 0^{\circ}$

Although the first and third relationships contain the same angles, they are considered different because the reference angle in each is different — $3 0^{\circ}$ in the first and $6 0^{\circ}$ in the third. The sine, cosine, and tangents of these triangles can be useful when finding unknown side lengths.

Angle $θ$	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$
$sin (θ)$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$
$cos (θ)$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$
$tan (θ)$	$\frac{3}{3}$	$1$	$3$

These special measures are justified below.

Derivation

Special Right Triangles 30-60-90

Consider the $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle. Suppose an equilateral triangle has side lengths of $1 .$

Bisecting the apex angle yields the following $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle.

The value of $h$ — the length of the third side — can be found using the Pythagorean Theorem. Here, $a = \frac{1}{2}, b = h,$ and $c = 1 .$

a^{2} + b^{2} = c^{2}

SubstituteValues

Substitute values

(\frac{1}{2})^{2} + h^{2} = 1^{2}

CalcPow

Calculate power

\frac{1}{4} + h^{2} = 1

SubEqn

$LHS - \frac{1}{4} = RHS - \frac{1}{4}$

h^{2} = \frac{3}{4}

SqrtEqn

$LHS = RHS$

h = \frac{3}{4}

DivSqrt

$\frac{a}{b} = \frac{a}{b}$

h = \frac{3}{4}

CalcRoot

Calculate root

h = \frac{3}{2}

Since, $h = \frac{3}{2},$ the $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle can be redrawn as follows.

Using $θ = 3 0^{\circ},$ the following relationships can be concluded.

$sin (3 0^{\circ}) = \frac{1}{2} / 1 = \frac{1}{2} cos (3 0^{\circ}) = \frac{3}{2} / 1 = \frac{3}{2} tan (3 0^{\circ}) = \frac{1}{2} / \frac{3}{2} = \frac{3}{3}$

Using the same triangle, the values for a $6 0^{\circ}$ - $3 0^{\circ}$ - $9 0^{\circ}$ triangle can be determined.

$sin (6 0^{\circ}) = \frac{3}{2} / 1 = \frac{3}{2} cos (6 0^{\circ}) = \frac{1}{2} / 1 = \frac{1}{2} tan (6 0^{\circ}) = \frac{3}{2} / \frac{1}{2} = 3$

Derivation

Special Right Triangles 45-45-90

Suppose an isosceles triangle has a hypotenuse of $1$ and base angles that measure $4 5^{\circ} .$

Because the triangle is isosceles, its legs have equal measure. Because they are unknown, $x$ can be used to represent them. The Pythagorean Theorem can be used to determine the value of $x .$

a^{2} + b^{2} = c^{2}

SubstituteExpressions

Substitute expressions

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

SimpPowTerm

Simplify power and terms

2 x^{2} = 1

DivEqn

$LHS / 2 = RHS / 2$

x^{2} = \frac{1}{2}

SqrtEqn

$LHS = RHS$

x = \pm \frac{1}{2}

$x > 0$

x = \frac{1}{2}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

x = \frac{1}{2}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

x = \frac{2}{2}

The $4 5^{\circ}$ - $4 5^{\circ}$ - $4 5^{\circ}$ triangle can be redrawn as follows.

Using $θ = 4 5^{\circ},$ the following relationships can be concluded.

$sin (4 5^{\circ}) = \frac{2}{2} / 1 = \frac{2}{2} cos (4 5^{\circ}) = \frac{2}{2} / 1 = \frac{2}{2} tan (4 5^{\circ}) = \frac{2}{2} / \frac{2}{2} = 1$

As was shown above, because the hypotenuse of these special right triangles is $1,$

$sin (θ) = opp and cos (θ) = adj, for θ = 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ} .$

Therefore, these values can be used to determine unknown side lengths.

Concept

Building the Unit Circle

By placing right triangles with one of their legs along the $x$ -axis, the coordinates of the vertex not on the axis can be found using the legs of the triangle.

In the first quadrant, the coordinates are the lengths of the triangle's legs. If the length of the hypotenuse is $1$ the legs of the special right triangles are given directly by the sine and cosine value of the angle at the origin and the coordinates are also the sine and cosine values.

Since the hypotenuse of these triangles is $1,$ the marked points all lie on a circle with radius $1$ and a center at the origin.

This is called the unit circle. The coordinates of any point on this circle are the cosine and sine values of the angle the segment from the point to the origin creates with the positive $x$ -axis.

$x = cos (θ) and y = sin (θ)$

Concept

Radian

A radian, like a degree, is an angle unit. One radian is defined as the measure of the central angle that intercepts an arc equal in length to the radius of the circle. It corresponds to roughly

57 . 3^{\circ} .

Animation of forming an arc that measures 1 radian

If the arc length is

2

radii, the measure of the corresponding central angle is

2

radians, and so on. Therefore, radians describe the number of radii an angle creates on a circle.

It can be observed that a semicircle corresponds to an arc length of

π

radii and the circumference of a circle corresponds to an arc length of

2 π

radii.

In calculations, even if the angle is given in radians, rad is seldom written. Instead, no unit marker indicates radians. Consider two expressions.

cos 6 4^{\circ} and cos 5

The first angle is given in degrees, and the other is given in radians. At first glance, radians might seem inconvenient, but they make calculations simpler in certain circumstances,. Radians are also the SI unit for angles.

Determine the trigonometric values

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Exercise

Use special right triangles and the unit circle to find the following trigonometric values. $sin (4 5^{\circ}) tan (π) cos (15 0^{\circ})$

Show Solution

Solution

Let's find the trigonometric values one at a time.

Example

$sin (4 5^{\circ})$

The value

sin (4 5^{\circ})

is the

y

-coordinate of the point on the unit circle that, together with the positive

x

-axis, creates the angle

4 5^{\circ} .

Here, we have a right triangle with the angles $4 5^{\circ}$ and $4 5^{\circ} .$ The hypotenuse is $1,$ so we know that the legs of the triangle are both $\frac{2}{2} .$ That means the point is $(\frac{2}{2}, \frac{2}{2}) .$

Therefore,

sin (4 5^{\circ})

is

\frac{2}{2} .

Example

$tan (π)$

Notice there is no degree symbol. This means the angle is given in radians.

π

radians is half a revolution, meaning

18 0^{\circ} .

tan (θ)

does not correspond with a point on the unit circle, but it can be found using

sin (θ)

and

cos (θ) .

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

The point corresponding to

π

rad has the coordinates

(- 1, 0) .

Thus,

cos (π) = - 1

and

sin (π) = 0 .

tan (π) = \frac{sin ( π )}{cos ( π )}

${\textstyle \ifnumequal{180}{0}{\cos\left(0\right)=1}{}\ifnumequal{180}{30}{\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{45}{\cos\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{60}{\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}}{}\ifnumequal{180}{90}{\cos\left(\dfrac{\pi}{2}\right)=0}{}\ifnumequal{180}{120}{\cos\left(\dfrac{2\pi}{3}\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{180}{135}{\cos\left(\dfrac{3\pi}{4}\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{150}{\cos\left(\dfrac{5\pi}{6}\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{180}{\cos\left(\pi\right)=\text{-} 1}{}\ifnumequal{180}{210}{\cos\left(\dfrac{7\pi}6\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{180}{225}{\cos\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{240}{\cos\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {1}2}{}\ifnumequal{180}{270}{\cos\left(\dfrac{3\pi}{2}\right)=0}{}\ifnumequal{180}{300}{\cos\left(\dfrac{5\pi}3\right)=\dfrac{1}2}{}\ifnumequal{180}{315}{\cos\left(\dfrac{7\pi}4\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{330}{\cos\left(\dfrac{11\pi}6\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{180}{360}{\cos\left(2\pi\right)=1}{}}$ , ${\textstyle \ifnumequal{180}{0}{\sin\left(0\right)=0}{}\ifnumequal{180}{30}{\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{180}{45}{\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{60}{\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{90}{\sin\left(\dfrac{\pi}{2}\right)=1}{}\ifnumequal{180}{120}{\sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{135}{\sin\left(\dfrac{3\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{150}{\sin\left(\dfrac{5\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{180}{180}{\sin\left(\pi\right)=0}{}\ifnumequal{180}{210}{\sin\left(\dfrac{7\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{180}{225}{\sin\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{240}{\sin\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{180}{270}{\sin\left(\dfrac{3\pi}{2}\right)=\text{-} 1}{}\ifnumequal{180}{300}{\sin\left(\dfrac{5\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{180}{315}{\sin\left(\dfrac{7\pi}4\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{330}{\sin\left(\dfrac{11\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{180}{360}{\sin\left(2\pi\right)=0}{}}$

tan (π) = \frac{0}{- 1}

CalcQuot

Calculate quotient

tan (π) = 0

The tangent value of

π

is

0 .

Example

$cos (15 0^{\circ})$

Let's draw the angle

15 0^{\circ}

from the positive

x

-axis and mark the point it corresponds to on the unit circle.

An angle of $18 0^{\circ}$ would span from the positive $x$ -axis to the negative. Hence, a $3 0^{\circ}$ special right triangle can be drawn from the negative $x$ -axis to the terminal side of the $15 0^{\circ}$ angle. Recall that this triangle has legs that measure $\frac{1}{2}$ and $\frac{3}{2} .$

Since $cos (15 0^{\circ})$ is the $x$ -coordinate of the point, $cos (15 0^{\circ}) = - \frac{3}{2} .$

Rule

Relation between Radians and Degrees

Since both degrees and radians are used to measure angles, it's useful to be able to translate between them.

Rule

18 0^{\circ} = π rad

The arc length for one revolution around a circle is also known as the circumference of the circle, $2 π r .$ By dividing this with the length of the arc corresponding to one radian, $r,$ the number of radians for one revolution is obtained. $\frac{2 π r}{r} = 2 π$ Thus, one revolution is $2 π$ radians. In degrees, this is $36 0^{\circ},$ leading to the relation $36 0^{\circ} = 2 π$ rad. Dividing both sides by $2$ gives $18 0^{\circ} = π rad .$ Thus, $18 0^{\circ}$ correspond to $π$ radians.

Rule

Relations

From the relation

18 0^{\circ} = π

rad it's possible to find two rules, by dividing both sides by either

180

or

π .

Rule

1^{\circ} = \frac{π}{1 8 0} rad

To find an expression for $1^{\circ},$ divide both sides by $180 .$

18 0^{\circ} = π rad

DivEqn

$LHS / 180 = RHS / 180$

\frac{1 8 0}{1 8 0}^{\circ} = \frac{π}{1 8 0} rad

CalcQuot

Calculate quotient

1^{\circ} = \frac{π}{1 8 0} rad

$1^{\circ}$ corresponds $\frac{π}{1 8 0} \approx 0.017$ radians. That means that $2^{\circ} = 2 \cdot \frac{π}{1 8 0} rad 3^{\circ} = 3 \cdot \frac{π}{1 8 0} rad,$ and so forth.

Rule

1 rad = \frac{1 8 0 ^{\circ}}{π}

To get an expression for $1$ rad, divide both sides by $π .$

18 0^{\circ} = π rad

DivEqn

$LHS / π = RHS / π$

\frac{1 8 0 ^{\circ}}{π} = \frac{π}{π} rad

CalcQuot

Calculate quotient

\frac{1 8 0 ^{\circ}}{π} = 1 rad

RearrangeEqn

Rearrange equation

1 rad = \frac{1 8 0 ^{\circ}}{π}

$1$ radian corresponds to $\frac{1 8 0}{π} \approx 57 . 3^{\circ} .$

Convert between degrees and radians

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Exercise

Convert $4 5^{\circ}$ to radians and $\frac{π}{2} rad$ to degrees.

Show Solution

Solution

To convert an angle from degrees to radians we can use $1^{\circ} = \frac{π}{1 8 0} rad .$ $4 5^{\circ}$ should be multiplied with $\frac{π}{1 8 0}$ rad.

4 5^{\circ}

$1^{\circ} = \frac{π}{1 8 0} rad$

45 \cdot \frac{π}{1 8 0} rad

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

\frac{4 5 π}{1 8 0} rad

ReduceFrac

$\frac{a}{b} = \frac{a / 4 5}{b / 4 5}$

\frac{π}{4} rad

Thus,

4 5^{\circ}

is equal to

\frac{π}{4}

radians. We could use the caluclator to get a numerical value here, but there are infinitely many decimals, so we'll keep it in its exact form:

4 5^{\circ} = \frac{π}{4} rad .

To convert an angle from radians to degrees we can use that

1 rad = \frac{1 8 0 ^{\circ}}{π} .

Therefore, if we multiply

\frac{π}{2}

with

\frac{1 8 0 ^{\circ}}{π},

we'll get how many degrees it corresponds to.

\frac{π}{2} \cdot \frac{1 8 0 ^{\circ}}{π}

MultFrac

Multiply fractions

\frac{π \cdot 1 8 0 ^{\circ}}{2 π}

CancelCommonFac

Cancel out common factors

\frac{1 8 0 ^{\circ}}{2}

CalcQuot

Calculate quotient

9 0^{\circ}

Thus,

\frac{π}{2}

radians correspond to

9 0^{\circ} .