The ray which determines the size of an angle in standard position. Can be formed by rotating the initial side counter-clockwise about the origin by the desired angle measure.

A point P at (x, y) plotted on a first quadrant coordinate plane without numbers. A segment with length r is drawn from the origin to P. A dashed vertical segment with length y is drawn from P to a point on the positive x axis. A dashed horizontal segment with length x is drawn from the origin to a point along the positive x axis, and directly below point P. The segment with length r makes an angle labeled theta with respect to the positive x axis.

Let \theta be an angle in standard form and P(x,y) be a point on the terminal side of \theta. The distance from P to the origin, r, can be found using the formula: x^2+y^2=r^2This is called the Pythagorean theorem.

Based on the right triangle,

\qquad r is the hypotenuse

\qquad x is the adjacent side of the angle \theta

\qquad y is the opposite side of the angle \theta

A trigonometric ratio is a relationship between an angle and a pair of sides in a right triangle. For the triangle formed by angle \theta above, we have:

Sine (sin)

The ratio between the side opposite to a given angle \theta and the hypotenuse of a right triangle.

\sin \theta=\dfrac{y}{r}

Cosine (cos)

The ratio between the side adjacent to a given angle \theta and the hypotenuse of a right triangle.

\cos \theta=\dfrac{x}{r}

Tangent (tan)

The ratio between the sides opposite and adjacent to a given angle \theta of a right triangle.

\tan \theta=\dfrac{y}{x}

We can also rearrange these ratios to get x=r\cos\theta and y=r\sin\theta. In this way, we can describe any point on the coordinate plane by an angle \theta and a distance r by\left(x,y\right)=\left(r\cos\theta,r\sin\theta\right)where \theta is the angle of the point in standard form.

For any fixed value of r, varying the size of \theta from 0\degree to 360\degree describes a circle centered at the origin with radius r. The set of points that are obtained by r = 1 are known as the unit circle.

Unit circle

A circle with a radius of 1 unit, typically centered at the origin.

-1

Each angle has a reference angle, which is the acute angle formed by the terminal side of the initial angle and the x-axis. The measure of a reference angle for angle \theta can be calculated in the different quadrants as follows:

\text{Quadrant}	\text{Reference angle } (\theta \gt 0\degree)	\text{Reference angle } (\theta \lt 0\degree)
\text{I}	\theta	360\degree-\|\theta\|
\text{II}	180\degree-\theta	\|\theta\|-180\degree
\text{III}	\theta-180\degree	180\degree-\|\theta\|
\text{IV}	360\degree-\theta	\|\theta\|

Every angle has many equivalent representations in the coordinate plane. These are known as coterminal angles.

Coterminal angles

Angles that have a common terminal side when in standard position

Coterminal angles are formed by one or more rotations of 360 \degree either clockwise or counter-clockwise from the initial angle.

\displaystyle \beta=360\degree n+\theta

\bm{\beta}

A coterminal angle of angle \theta

\bm{\theta}

The initial angle in standard position

\bm{n}

An integer

Worked examples

Example 1

Consider the angle \theta = 210 \degree.

Sketch the angle on the coordinate plane.

Approach

To sketch the given angle, we first express the angle as a simplified fraction within one rotation of the unit circle, which measures 360 \degree.

The vertex is at the origin and the initial side lies along the positive x-axis.

If the angle is positive, then we move counter-clockwise from the initial side to the terminal side.

If the angle is negative, then we move clockwise from the initial side to the terminal side.

Solution

Clearly, 210 \degree is a positive angle. So we move counter-clockwise from the initial side to the terminal side. If we divide 210 \degree by 360 \degree, we get \dfrac{210 \degree}{360 \degree}=\dfrac{7}{12}

Observe that \dfrac{7}{12}=\dfrac{7}{3}\left(\dfrac{1}{4}\right). This means that 210 \degree can be obtained by rotating counter-clockwise through two quadrants and one-third of another quadrant on the coordinate plane.

That is,

Find the measure of the reference angle.

Approach

To find the reference angle, we first determine which quadrant the terminal side of the angle \theta lies within. Then we use the correct formula in the reference angle table, or simply find the measure of the acute angle that lies between the x-axis and the terminal side of the angle.

Solution

From part (a), we know that the terminal side of angle 210 \degree lies in Quadrant III.

Since 210 \degree>0\degree, the reference angle is given by \theta-180\degree=210\degree-180\degree=30\degree

Therefore, the reference angle of 210\degree is 30\degree.

Reflection

We can verify the answer from the graph in part (a).

The reference angle is the acute angle formed by the terminal side and the negative x-axis. This angle is one-third of a rotation through the third quadrant. The measure of one quadrant is 90\degree. So the reference angle is \dfrac{1}{3}\left(90\degree\right)=30\degree

We obtain similar result that the reference angle of 210\degree is 30\degree.

Find another coterminal angle.

Approach

To find a coterminal angle, we simply add any number of full clockwise or counter-clockwise rotations to the given angle.

Solution

Note that there are infinitely many coterminal angles for 210\degree, formed by the expression \beta=360\degree n + 210\degree for different values of n.

To find one, we can let n=1. Geometrically, this represents one full rotation counter-clockwise from the terminal side of 210\degree.

Substituting n=1 we get: \beta=360\degree (1)+210\degree=570\degree

Therefore, 210\degree and 570\degree are coterminal angles.

Reflection

We can find another coterminal angles by substituting different values for n. For example:

If n=-1, \beta=360\degree (-1)+210\degree=-150\degree

If n=2, \beta=360\degree (2)+210\degree=930\degree

This means that 210\degree,570\degree,-150\degree and 930\degree are all coterminal angles.

Example 2

Consider the point on the graph that has coordinates \left(8, 15\right).

-2

Find r, the distance from the point to the origin.

Approach

First, we construct a right triangle based on the given point. Then we use the Pythagorean Theorem to find r.

Solution

With the given point P\left(8, 15\right), we can construct a right triangle.

-2

By the Pythagorean Theorem, the distance from a point P(x,y) to the origin, r, can be found using the formula: x^2+y^2=r^2

Taking the square root of both sides, we obtain r=\pm\sqrt{x^2+y^2}

Since the distance r is always positive we will only keep the positive root, r=\sqrt{x^2+y^2}

Substituting x=8 and y=15, we get r=\sqrt{(8)^2+(15)^2}=\sqrt{289}=17

Therefore, r=17 units.

Find \sin \theta.

Approach

To find \sin \theta, we first determine the side opposite to the angle \theta. Then we use the correct ratio for \sin \theta.

Solution

Note that the trigonometric ratio for sine (sin) is given by \sin \theta=\dfrac{\text{opposite}}{\text{hypotenuse}} From part (a), we know that the side opposite to angle \theta is y and the hypotenuse is r. So we have \sin \theta=\dfrac{y}{r}

Since y=15 and r=17, we obtain \sin \theta=\dfrac{15}{17}

Find \cos \theta.

Approach

To find \cos \theta, we first determine the side adjacent to the angle \theta. Then we use the correct ratio for \cos \theta.

Solution

Note that the trigonometric ratio for cosine (cos) is given by \cos \theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} From part (a), we know that the side adjacent to angle \theta is x and the hypotenuse is r. So we have \cos \theta=\dfrac{x}{r}

Since x=8 and r=17, we obtain \cos \theta=\dfrac{8}{17}

Find \tan \theta.

Approach

To find \tan \theta, we first determine the opposite side and the adjacent side to the angle \theta. Then we use the correct ratio for \tan \theta.

Solution

Note that the trigonometric ratio for tangent (tan) is given by \tan \theta=\dfrac{\text{opposite}}{\text{adjacent}} From part (a), we know that the opposite side to angle \theta is y and the adjacent side to angle \theta is x. So we have \tan \theta=\dfrac{y}{x}

Since y=15 and x=8, we have \tan \theta=\dfrac{15}{8}

Example 3

The graph shows an angle \theta in standard position with its terminal side intersecting the unit circle at P\left(-\dfrac{12}{37},\dfrac{35}{37}\right).

-1

Find \sin \theta.

Approach

First, we determine the hypotenuse r when the terminal side of angle \theta intersects the unit circle at the given point P. Then we use the simplified ratio for \sin \theta.

Solution

Note that \sin \theta=\dfrac{y}{r}.

Since the the terminal side of angle \theta intersects the unit circle at the point P\left(-\dfrac{12}{37},\dfrac{35}{37}\right), the distance from P to the origin, r, should be the radius of the unit circle which is 1. So we have \sin \theta=y

Since y=\dfrac{35}{37} , \sin \theta=\dfrac{35}{37}

Find \cos \theta.

Approach

First, we determine the hypotenuse r when the terminal side of angle \theta intersects the unit circle at the given point P. The we use the simplified ratio for \cos \theta.

Solution

Note that \cos \theta=\dfrac{x}{r}.

Since x=-\dfrac{12}{37} , \cos \theta=-\dfrac{12}{37}

Find \tan \theta.

Approach

We use the simplified ratios for \cos \theta and \sin \theta. Then we use the ratio for \tan \theta.

Solution

From part (a), we get \sin \theta=y while from part (b), we get \cos \theta=x.

Note that \tan \theta=\dfrac{y}{x}. Since y=\sin \theta and x=\cos \theta, \tan \theta=\dfrac{\sin \theta}{\cos \theta}.

Since \sin \theta=\dfrac{35}{37} and \cos \theta=-\dfrac{12}{37}, \tan \theta=\dfrac{\dfrac{35}{37}}{-\dfrac{12}{37}}

Multiplying the fraction in the numerator by the reciprocal of the fraction in the denominator, we have \tan \theta=-\dfrac{35}{12}

Example 4

Suppose that \sin \theta = -\dfrac{6}{7}, where 270 \degree < \theta < 360 \degree.

Find \cos \theta.

Approach

First, we find the value of the adjacent side using the Pythagorean Theorem. Then we substitute the values to the ratio \cos \theta. We need to consider the sign of \cos \theta based on the location of the terminal side of angle \theta.

Solution

We consider the signs of the coordinates when \sin \theta = -\dfrac{6}{7} such that 270 \degree < \theta < 360 \degree. The terminal side of angle \theta lies in Quadrant IV. This means that the x-coordinate must be positive and the y-coordinate must be negative.

Note that \sin \theta =\dfrac{y}{r}. So we have y=-6 and r=7. Now, we find the value of x using the Pythagorean Theorem. x^2+y^2=r^2

Adding -y^2 to both sides, we get x^2=r^2-y^2

Taking the square root of both sides, we have x=\pm \sqrt{r^2-y^2}

Since x is positive, we choose x=\sqrt{r^2-y^2}

Now,

\displaystyle x	\displaystyle =	\displaystyle \sqrt{r^2-y^2}	Pythagorean theorem, solved for x
	\displaystyle =	\displaystyle \sqrt{7^2-(-6)^2}	Substitute r=7 and y=-6
	\displaystyle =	\displaystyle \sqrt{13}	Simplify

Note that \cos \theta=\dfrac{x}{r}. Substituting x=\sqrt{13} and r=7, we obtain\cos \theta=\dfrac{\sqrt{13}}{7}

Find \tan \theta.

Solution

Note that \tan \theta = \dfrac{y}{x}. From part (a), we know that y=-6 and x=\sqrt{13}.

Substituting y=-6 and x=\sqrt{13}, we get \tan \theta = -\dfrac{6}{\sqrt{13}}

We rationalize the denominator by multiplying both the numerator and the denominator by \sqrt{13} so, we have \tan \theta = -\dfrac{6\sqrt{13}}{13}

10.08 Trigonometry in the coordinate plane

Concept summary

Worked examples

Example 1

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Reflection

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Reflection

Example 2

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Example 3

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Example 4

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Outcomes

MA.912.T.1.1

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