7.2 Sine and Cosine Functions

Vanessa Coffelt

7.2 Sine and Cosine Functions

7.2.1 Right Triangle Definitions

This section focuses on right triangles, triangles in which one angle measures $90^{\circ}$ . Consider the right triangle below, where, as usual, the small square $\qed$ denotes the right angle, the labels ` $a$ ,’ ` $b$ ,’ and ` $c$ ‘ denote the lengths of the sides of the triangle, and $\alpha$ and $\beta$ represent the (measure of) the non-right angles. As you may recall, the side opposite the right angle is called the hypotenuse of the right triangle. Also note that because the sum of the measures of all angles in a triangle must add to $180^{\circ}$ , we have $\alpha + \beta + 90^{\circ}= 180^{\circ}$ , or $\alpha + \beta = 90^{\circ}$ . Said differently, the non-right angles in a right triangle are complements.

A right triangle with horizontal side, a, vertical side, b, and hypotenuse, c . The angle between sides a and c is labeled beta. The angle between sides c and b is labeled alpha. The right angle is between sides a and b. — Right Triangle

We now state and prove the most famous result about right triangles: The Pythagorean Theorem.

Theorem 7.1 The Pythagorean Theorem

The square of the length of the hypotenuse of a right triangle is equal to the sums of the squares of the other two sides.

More specifically, if $c$ is the length of the hypotenuse of a right triangle and $a$ and $b$ are the lengths of the other two sides, then $a^2 + b^2 = c^2$ .

There are several proofs of the Pythagorean Theorem,^[1] but the one we choose to reproduce here showcases a nice interplay between algebra and geometry. Consider taking four copies of the right triangle below on the left and arranging them as seen below on the right.

A triangle and a square. The triangle is a right triangle with horizontal side a, vertical side b, and hypotenuse c. The angle between sides a and c is beta and the angle alpha is between sides b and c. The square has sides of length a plus b. There is a second square inside the first square rotated in such a manner that it divides each side of the large square into length b and a. The smaller square has side lengths of c. The regions inside the larger square, but outside the smaller square are the same triangle as to the left. — Geometric Proofs of the Pythagorean Theorem

It should be clear that we have produced a large square with a side length of $(a+b)$ . What is also true, but may not be obvious, is that the inner quadrilateral is also a square. We can readily see the inner quadrilateral has equal sides of length $c$ . Moreover, because $\alpha + \beta = 90^{\circ}$ , we get the interior angles of the inner quadrilateral are each $90^{\circ}$ . Hence, the inner quadrilateral is indeed a square.

We finish the proof by computing the area of the large square in two ways. First, we square the length of its side: $(a+b)^2$ . Next, we add up the areas of the four triangles, each having area $\frac{1}{2} ab$ along with the area of the inner square, $c^2$ . Equating these to expressions gives: $(a+b)^2 = 4 \left( \frac{1}{2} ab\right)+c^2$ . As a result of $(a+b)^2 = a^2+2ab+b^2$ and $4 \left( \frac{1}{2} ab\right) = 2ab$ , we have $a^2+2ab+b^2 = 2ab + c^2$ or $a^2+b^2 = c^2$ , as required.

It should be noted that the converse of the Pythagorean Theorem is also true. That is if $a$ , $b$ , and $c$ are the lengths of sides of a triangle and $a^2+b^2 = c^2$ , then the triangle is a right triangle.^[2]

A list of integers $(a,b,c)$ which satisfy the relationship $a^2+b^2 = c^2$ is called a Pythagorean Triple. Some of the more common triples are: $(3,4,5)$ , $(5,12,13)$ , $(7,24,25)$ , and $(8,15,17)$ . We leave it to the reader to verify these integers satisfy the equation $a^2+b^2 = c^2$ and suggest committing these triples to memory.

Next, we set about defining characteristic ratios associated with acute angles. Given any acute angle $\theta$ , we can imagine $\theta$ being an interior angle of a right triangle as seen below.

A right triangle with horizontal side, a, vertical side, b, and hypotenuse, c . The angle between sides a and c is labeled theta. The angle between sides c and b is not labeled. The right angle is between sides a and b. — Right Triangle with angle theta

Focusing on the arrangement of the sides of the triangle with respect to the angle $\theta$ , we make the following definitions: the side with length $a$ is called the side of the triangle which is adjacent to $\theta$ and the side with length $b$ is called the side of the triangle opposite $\theta$ . As usual, the side labeled ` $c$ ‘ (the side opposite the right angle) is the hypotenuse. Using this diagram, we define two important trigonometric ratios of $\theta$ .

Definition 7.2

Suppose $\theta$ is an acute angle residing in a right triangle as depicted above.

The sine of $\theta$ , denoted $\sin(\theta)$ is defined by the ratio: $\sin(\theta) = \dfrac{b}{c}$ , or $\dfrac{\text{`length of opposite'}}{\text{`length of hypotenuse'}}$ .
The cosine of $\theta$ , denoted $\cos(\theta)$ is defined by the ratio: $\cos(\theta) = \dfrac{a}{c}$ , or $\dfrac{\text{`length of adjacent'}}{\text{`length of hypotenuse'}}$ .

For example, consider the angle $\theta$ indicated in the 3-4-5 triangle given. Using Definition 7.2, we get $\sin(\theta) = \frac{4}{5}$ and $\cos(\theta) = \frac{3}{5}$ . One may well wonder if these trigonometric ratios we’ve found for $\theta$ change if the triangle containing $\theta$ changes. For example, if we scale all the sides of the 3-4-5 triangle on the left by a factor of $2$ , we produce the similar triangle below in the middle.^[3] Using this triangle to compute our ratios for $\theta$ , we find $\sin(\theta) = \frac{8}{10} = \frac{4}{5}$ and $\cos(\theta) = \frac{6}{10} = \frac{3}{5}$ . Note that the scaling factor, here $2$ , is common to all sides of the triangle, and, hence, divides out of the numerator and denominator when simplifying each of the ratios.

Three right triangles side by side. The first has sides of 3 and 4 with a hypotenuse of 5. The next right triangle has sides of 6 and 8 with a hypotenuse of 10, the last right triangle has sides of 3r and 4r with hypotenuse 5r. — Examples of Pythagorean Triples

In general, thanks to the Angle Angle Similarity Postulate, any two right triangles which contain our angle $\theta$ are similar which means there is a positive constant $r$ so that the sides of the triangle are $3r$ , $4r$ , and $5r$ as seen above on the right. Hence, regardless of the right triangle in which we choose to imagine $\theta$ , $\sin(\theta) = \frac{4r}{5r} = \frac{4}{5}$ and $\cos(\theta) = \frac{3r}{5r} = \frac{3}{5}$ . Generalizing this same argument to any acute angle $\theta$ assures us that the ratios as described in Definition 7.2 are independent of the triangle we use.

Our next objective is to determine the values of $\sin(\theta)$ and $\cos(\theta)$ for some of the more commonly used angles. We begin with $45^{\circ}$ . In a right triangle, if one of the non-right angles measures $45^{\circ}$ , then the other measures $45^{\circ}$ as well. It follows that the two legs of the triangle must be congruent. We may choose any right triangle containing a $45^{\circ}$ angle for our computations, thus we choose the length of one (hence both) of the legs to be $1$ . The Pythagorean Theorem gives the hypotenuse is: $c^2 = 1^2+1^2 = 2$ , so $c = \sqrt{2}$ . (We take only the positive square root here as $c$ represents the length of the hypotenuse here, so, necessarily $c>0$ .) From this, we obtain the values below, and suggest committing them to memory.

A right triangle with sides of 1 and hypotenuse of square root of 2. The two non-right angles are 45 degrees. — A 40-45-90 Triangle

$\sin\left(45^{\circ}\right) = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$
$\cos\left(45^{\circ}\right) = \dfrac{1}{\sqrt{2}} =\dfrac{\sqrt{2}}{2}$

Note that we have `rationalized’ here to avoid the irrational number $\sqrt{2}$ appearing in the denominator. This is a common convention in trigonometry, and we will adhere to it unless extremely inconvenient.

Next, we investigate $60^{\circ}$ and $30^{\circ}$ angles. Consider the equilateral triangle each of whose sides measures $2$ units. Each of its interior angles is necessarily $60^{\circ}$ , so if we drop an altitude, we produce two $30^{\circ} - 60^{\circ} - 90^{\circ}$ triangles each having a base measuring $1$ unit and a hypotenuse of $2$ units. Using the Pythagorean Theorem, we can find the height, $h$ of these triangles: $1^2+h^2 = 2^2$ so $h^2 = 3$ or $h = \sqrt{3}$ . Using these, we can find the values of the trigonometric ratios for both $60^{\circ}$ and $30^{\circ}$ . Again, we recommend committing these values to memory.

An equilateral triangle divided in half to make two equal right triangles. The sides of the right triangles 1 and square root of 3 with hypotenuse 2. The angle opposite the side of length 1 is 30 degrees. The other angle is 60 degrees. — 30-60-90 Triangles

Sine Values:

$\sin\left(30^{\circ}\right) = \dfrac{1}{2}$
$\sin\left(60^{\circ}\right) = \dfrac{\sqrt{3}}{2}$

Cosine Values:

$\cos\left(30^{\circ}\right) = \dfrac{\sqrt{3}}{2}$
$\cos\left(60^{\circ}\right) = \dfrac{1}{2}$

Recall $30^{\circ}$ and $60^{\circ}$ are complements, so the side adjacent to the $60^{\circ}$ angle is the side opposite the $30^{\circ}$ and the side opposite the $60^{\circ}$ angle is the side adjacent to the $30^{\circ}$ . This sort of `swapping’ is true of all complementary angles and will be generalized in Section 8.2, Theorem 8.6.

Note that the values of the trigonometric ratios we have derived for $30^{\circ}$ , $45^{\circ}$ , and $60^{\circ}$ angles are the exact values of these ratios. For these angles, we can conveniently express the exact values of their sines and cosines resorting, at worst, to using square roots. The reader may well wonder if, for instance, we can express the exact value of, say, $\sin\left(42^{\circ}\right)$ in terms of radicals. The answer in this case is `yes’ (see here), but, in general, we will not take the time to pursue such representations.^[4] Hence, if a problem requests an `exact’ answer involving $\sin\left(42^{\circ}\right)$ , we will leave it written as ` $\sin\left(42^{\circ}\right)$ ‘ and use a calculator to produce a suitable approximation as the situation warrants.

The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground), and whose terminal side is the line-of-sight to an object above the base-line. Schematically:

A diagram with a horizontal line segment labeled "base line". A dashed line starts at one end of the line segment and increases to a point above the other end of the line. The end of the dashed line has a star and the word object. The angle formed by the baseline and the dashed line is theta. — The angle of inclination from the base line to the object is theta

7.2.2 Unit Circle Definitions

In Section 7.1.3, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle $\theta$ in standard position and let $P$ denote the point where the terminal side of $\theta$ intersects the Unit Circle, as diagrammed below.

Two unit circles side by side. The first has the terminal side in the first quadrant and the angle is marked as theta. The second circle has the same terminal side and angle as the first picture. The point where the terminal side intersects the circle is labeled P( cosine theta, sine theta). — Sine and Cosine on a Unit Circle

By associating the point $P$ with the angle $\theta$ , we are assigning a position on the Unit Circle to the angle $\theta$ . For each angle $\theta$ , the terminal side of $\theta$ , when graphed in standard position, intersects The Unit Circle only once, so the mapping of $\theta$ to $P$ is a function.^[5] Because there is only one way to describe a point using rectangular coordinates,^[6] the mappings of $\theta$ to each of the $x$ and $y$ coordinates of $P$ are also functions. We give these functions names in the following definition.

Definition 7.3

Suppose an angle $\theta$ is graphed in standard position. Let $P(x,y)$ be the point of intersection of the terminal side of $P$ and the Unit Circle.

The $x$ -coordinate of $P$ is called the cosine of $\theta$ , written $\cos(\theta)$ .
The $y$ -coordinate of $P$ is called the sine of $\theta$ , written $\sin(\theta)$ .^[7]

You may have already seen definitions for the sine and cosine of an (acute) angle in terms of ratios of sides of a right triangle.^[8] While not incorrect, defining sine and cosine using right triangles limits the angles we can study to acute angles only. Definition 7.3, on the other hand, applies to all angles. These functions are defined in terms of points on the Unit Circle, thus they are called circular functions. Rest assured, Definition 7.3 specializes to Definition 7.2 when $\theta$ is an acute angle. We will see instances of this fact in the next example.

Example 7.2.1

Example 7.2.1.1

State the sine and cosine of the following angles.

$\theta = 270^{\circ}$

Solution:

State the sine and cosine of $\theta = 270^{\circ}$ .

To find $\cos\left(270^{\circ}\right)$ and $\sin\left(270^{\circ}\right)$ , we plot the angle $\theta =270^{\circ}$ in standard position and find the point on the terminal side of $\theta$ which lies on the Unit Circle. As $270^{\circ}$ represents $\frac{3}{4}$ of a counter-clockwise revolution, the terminal side of $\theta$ lies along the negative $y$ -axis.

Hence, the point we seek is $(0,-1)$ so that $\cos\left(270^{\circ}\right) = 0$ and $\sin\left(270^{\circ}\right) = -1$ .

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Example 7.2.1.2

State the sine and cosine of the following angles.

$\theta = - \pi$

Solution:

State the sine and cosine of $\theta = - \pi$ .

The angle $\theta=-\pi$ represents one half of a clockwise revolution so its terminal side lies on the negative $x$ -axis.

The point on the Unit Circle that lies on the negative $x$ -axis is $(-1,0)$ which means $\cos(-\pi) = -1$ and $\sin(-\pi) = 0$ .

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Example 7.2.1.3

State the sine and cosine of the following angles.

$\theta = 45^{\circ}$

Solution:

State the sine and cosine of $\theta = 45^{\circ}$ .

In this section, we derived values for $\cos\left(45^{\circ}\right)$ and $\sin\left(45^{\circ}\right)$ using Definition 7.2. In order to connect what we know from Section 7.2.1 with what we are asked to find per Definition 7.3, we sketch $\theta = 45^{\circ}$ in standard position and let $P(x,y)$ denote the point on the terminal side of $\theta$ which lies on the Unit Circle. If we drop a perpendicular line segment from $P$ to the $x$ -axis as shown, we obtain a $45^{\circ} - 45^{\circ} - 90^{\circ}$ right triangle whose legs have lengths $x$ and $y$ units with hypotenuse $1$ .

From our work in Section 7.2.1, we obtain the (familiar) values $x = \cos\left(45^{\circ}\right) = \frac{\sqrt{2}}{2}$ and $y = \sin\left(45^{\circ}\right) = \frac{\sqrt{2}}{2}$ .

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Example 7.2.1.4

State the sine and cosine of the following angles.

$\theta = \frac{\pi}{6}$

Solution:

State the sine and cosine of $\theta = \frac{\pi}{6}$ .

As before, the terminal side of $\theta = \frac{\pi}{6}$ does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting $P(x,y)$ denote the point on the terminal side of $\theta$ which lies on the Unit Circle, we drop a perpendicular line segment from $P$ to the $x$ -axis to form a $30^{\circ} - 60^{\circ} - 90^{\circ}$ right triangle.

Re-using some of our work from this section, we get $x = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $y=\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ .

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Example 7.2.1.5

State the sine and cosine of the following angles.

$\theta = \frac{5\pi}{6}$

Solution:

State the sine and cosine of $\theta = \frac{5\pi}{6}$ .

We plot $\theta = \frac{5\pi}{6}$ in standard position below on the left and, as usual, let $P(x,y)$ denote the point on the terminal side of $\theta$ which lies on the Unit Circle. In plotting $\theta$ , we find it lies $\frac{\pi}{6}$ radians short of one half revolution.

As we’ve just determined that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left( \frac{\pi}{6} \right) = \frac{1}{2}$ , we know the coordinates of the point $Q$ below on the right are $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ .

Using symmetry, the coordinates of $P$ are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ , so $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$ and $\sin\left( \frac{5\pi}{6} \right) = \frac{1}{2}$ .

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A few remarks are in order. First, after having re-used some of our work from this section in a few specific instances, we can reconcile Definition 7.3 with Definition 7.2 in the case $\theta$ is an acute angle. We situate $\theta$ in a right triangle with hypotenuse length $1$ , adjacent side length ` $x$ ,’ and the opposite side length ` $y$ ‘ as seen below on the left. Placing the vertex of $\theta$ at the origin and the adjacent side of $\theta$ along the $x$ -axis as seen below on the right effectively puts $\theta$ in standard position with $\theta$ ‘s adjacent side as the initial side of $\theta$ and the hypotenuse as the terminal side of $\theta$ . The hypotenuse of the triangle has length $1$ , thus we know the point $P(x,y)$ is on the Unit Circle.^[9]

A right triangle and a circle centered at the origin of the cartesian plane side by side. The right triangle has a horizontal side of x and vertical side of y with a hypotenuse of 1. The angle between the side of x and 1 is labeled theta. The circle has a radius of one. The triangle to the left is also inscribed in the circle with theta in standard position. The point P where the hypotenuse touches the circle is labeled with the coordinates (x,y). — Correlation between a Right Triangle and Unit Circle

Definition 7.2 gives $\cos(\theta) = \frac{x}{1} = x$ and $\sin(\theta) = \frac{y}{1} = y$ which exactly matches Definition 7.3. Hence, in the case of acute angles, the two definitions agree. In other words, the values of the trigonometric ratios} of acute angles are the same as the corresponding circular function values.

A second important take-away from Example 7.2.1 is use of symmetry in number 5. Indeed, we found the sine and cosine of $\frac{5\pi}{6}$ using the (acute) angle $\frac{\pi}{6}$ `for reference.’ As the Unit Circle is rife with symmetry, we would like to generalize this concept and exploit symmetry whenever possible. To that end, we introduce the notion of reference angle.

In general, for a non-quadrantal angle $\theta$ , the reference angle for $\theta$ (which we’ll usually denote $\alpha$ ) is the acute angle made between the terminal side of $\theta$ and the $x$ -axis. If $\theta$ is a Quadrant I or IV angle, $\alpha$ is the angle between the terminal side of $\theta$ and the positive $x$ -axis:

Two unit circles on coordinate planes side by side. The first circle has a positive angle alpha in the first quadrant. The second circle shows a negative angle alpha. On the first circle P=Q and on the second P is in the fourth quadrant with symmetry to Q across the x-axis. — Reference Angles in Quadrants I and IV

If $\theta$ is a Quadrant II or III angle, $\alpha$ is the angle between the terminal side of $\theta$ and thenegative $x$ -axis:

Two unit circles on coordinate planes side by side. The first circle has a terminal side in the second quadrant and a dashed line line in the first quadrant showing the angle to both lines from the x-axis as alpha. The second circle has a terminal side in the third quadrant and a dashed line line in the first quadrant showing the angle to both lines from the x-axis as alpha. On the first circle P is in the second quadrant and is symmetric across the y-axis. On the second circle P is in the third quadrant with symmetry to Q across the origin. — Reference Angles in Quadrants II and III

If we let $P$ denote the point $(\cos(\theta), \sin(\theta))$ , then $P$ lies on the Unit Circle. Due to the fact that the Unit Circle possesses symmetry with respect to the $x$ -axis, $y$ -axis and origin, regardless of where the terminal side of $\theta$ lies, there is a point $Q$ symmetric with $P$ which determines $\theta$ ‘s reference angle, $\alpha$ . The only difference between the points $P$ and $Q$ are the signs of their coordinates, $\pm$ . Hence, we have the following:

Theorem 7.2 Reference Angle Theorem

Suppose $\alpha$ is the reference angle for $\theta$ . Then:

$\cos(\theta) = \pm \cos(\alpha) \text{ and } \sin(\theta) = \pm \sin(\alpha),$

where the choice of the ( $\pm$ ) depends on the quadrant in which the terminal side of $\theta$ lies.

In light of Theorem 7.2, it pays to know the sine and cosine values for certain common Quadrant I angles as well as to keep in mind the signs of the coordinates of points in the given quadrants.

$\begin{array}{|c|c||c|c|} \hline \theta (\text{degrees}) & \theta (\text{radians}) & \cos(\theta) & \sin(\theta) \\ \hline 0^{\circ} & 0 & 1 & 0 \\ \hline 30^{\circ} & \frac{\pi}{6} & \frac{\sqrt{3}}{2} & \frac{1}{2} \\ [2pt] \hline 45^{\circ} & \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ [2pt] \hline 60^{\circ} & \frac{\pi}{3} & \frac{1}{2} & \frac{\sqrt{3}}{2} \\ [2pt] \hline 90^{\circ} & \frac{\pi}{2} & 0 & 1 \\ [2pt] \hline \end{array}$

A unit circle on a coordinate plane. There are 8 points marked on the circle. (1,0) where the circle meets the positive x-axis. (0,1) where the circle meets the positive y-axis. (-1,0) where the circle meets the negative x-axis and (0,-1) where the circle meets the negative y-axis. The fifth point is in the first quadrant and indicates that both the coordinates are positive. In the second quadrant a point is labeled where the first coordinate is negative and the second coordinate is positive. The seventh point is in the third quadrant noting that both coordinates are negative. The last point is in the fourth quadrant and the first coordinate is positive and the second coordinate is negative. — The Signs in Each Quadrant

Example 7.2.2

Example 7.2.2.1

Determine the sine and cosine of the following angles.

$\theta = 225^{\circ}$

Solution:

Determine the sine and cosine of $\theta = 225^{\circ}$ .

We begin by plotting $\theta = 225^{\circ}$ in standard position and find its terminal side overshoots the negative $x$ -axis to land in Quadrant III.

Hence, we obtain $\theta$ ‘s reference angle $\alpha$ by subtracting: $\alpha = \theta - 180^{\circ} = 225^{\circ} - 180^{\circ} = 45^{\circ}$ .

As $\theta$ is a Quadrant III angle, both $\cos(\theta) < 0$ and $\sin(\theta) < 0$ . The Reference Angle Theorem yields:

$\cos\left(225^{\circ}\right) = -\cos\left(45^{\circ}\right) = -\frac{\sqrt{2}}{2}$ and $\sin\left(225^{\circ}\right) = - \sin\left(45^{\circ}\right) = -\frac{\sqrt{2}}{2}$

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Example 7.2.2.2

Determine the sine and cosine of the following angles.

$\theta = \frac{11 \pi}{6}$

Solution:

Determine the sine and cosine of $\theta = \frac{11 \pi}{6}$ .

The terminal side of $\theta = \frac{11\pi}{6}$ , when plotted in standard position, lies in Quadrant IV, just shy of the positive $x$ -axis.

To find $\theta$ ‘s reference angle $\alpha$ , we subtract: $\alpha = 2\pi - \theta = 2\pi - \frac{11 \pi}{6} = \frac{\pi}{6}$ .

As $\theta$ is a Quadrant IV angle, $\cos(\theta) > 0$ and $\sin(\theta) < 0$ , so the Reference Angle Theorem gives:

$\cos\left(\frac{11 \pi}{6} \right) = \cos\left(\frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{11\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$

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Example 7.2.2.3

Determine the sine and cosine of the following angles.

$\theta = -\frac{5 \pi}{4}$

Solution:

Determine the sine and cosine of $\theta = -\frac{5 \pi}{4}$ .

To plot $\theta = -\frac{5\pi}{4}$ , we rotate clockwise an angle of $\frac{5 \pi}{4}$ from the positive $x$ -axis. The terminal side of $\theta$ , therefore, lies in Quadrant II making an angle of $\alpha = \frac{5 \pi}{4} - \pi = \frac{\pi}{4}$ radians with respect to the negative $x$ -axis.

As $\theta$ is a Quadrant II angle, $\cos(\theta) < 0$ and $\sin(\theta) > 0$ so the Reference Angle Theorem gives:

$\cos\left(-\frac{5 \pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ and $\sin\left(-\frac{5 \pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

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Example 7.2.2.4

Determine the sine and cosine of the following angles.

$\theta = \frac{7 \pi}{3}$

Solution:

Determine the sine and cosine of $\theta = \frac{7 \pi}{3}$ .

Given the angle $\theta = \frac{7 \pi}{3}$ measures more than $2 \pi = \frac{6 \pi}{3}$ , we find the terminal side of $\theta$ by rotating one full revolution followed by an additional $\alpha = \frac{7 \pi}{3} - 2\pi = \frac{\pi}{3}$ radians.

Hence, $\theta$ and $\alpha$ have the same terminal side,^[10] and so

$\cos\left(\frac{7\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ and $\sin\left(\frac{7\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

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A couple of remarks are in order. First off, the reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of $\pi$ with a denominator of $6$ have $\frac{\pi}{6}$ as a reference angle, those with a denominator of $4$ have $\frac{\pi}{4}$ as their reference angle, and those with a denominator of $3$ have $\frac{\pi}{3}$ as their reference angle.^[11]

Also note in number 4 of the last example, the angles $\frac{\pi}{3}$ and $\frac{7\pi}{3}$ are coterminal. As a result, have the same values for sine and cosine. It turns out that we can characterize coterminal angles in this manner, as stated below.

Theorem 7.3

Two angles $\alpha$ and $\beta$ are coterminal if and only if:

$\cos(\alpha) = \cos(\beta) \text{ and } \sin(\alpha) = \sin(\beta)$

Recall the phraseology `if and only if’ means there are two things to argue in Theorem 7.3: first, if $\alpha$ and $\beta$ are co-terminal, then $\cos(\alpha) = \cos(\beta)$ and $\sin(\alpha) = \sin(\beta)$ . This is immediate as coterminal share terminal sides, and, in particular, the (unique) point on the Unit Circle shared by said terminal side. Second, we need to argue that if $\cos(\alpha) = \cos(\beta)$ and $\sin(\alpha) = \sin(\beta)$ , then $\alpha$ and $\beta$ are coterminal.

To prove this second claim, note that when an angle is drawn in standard position, the terminal side of the angle is the ray that starts at the origin and is completely determined by any other point on the terminal side. If $\cos(\alpha) = \cos(\beta)$ and $\sin(\alpha) = \sin(\beta)$ , then their terminal sides share a point on the Unit Circle, namely $(\cos(\alpha), \sin(\alpha)) = (\cos(\beta), \sin(\beta))$ . Hence, $\alpha$ and $\beta$ are coterminal.

The Reference Angle Theorem in conjunction with the table of sine and cosine values given previously can be used to generate the figure below. We recommend committing it to memory.

A Unit Circle with the special angles between 0 and 2 pi, inclusively, labeled. At each angle there is the corresponding point on the unit circle labeled with the x and y coordinates. — Important Points on the Unit Circle

Our next example uses The Reference Angle Theorem in a slightly more sophisticated context.

Example 7.2.3

Example 7.2.3.1

Suppose $\alpha$ is an acute angle with $\cos(\alpha) = \frac{5}{13}$ .

Determine $\sin(\alpha)$ and use this to plot $\alpha$ in standard position.

Solution:

Suppose $\alpha$ is an acute angle with $\cos(\alpha) = \frac{5}{13}$ . Determine $\sin(\alpha)$ and use this to plot $\alpha$ in standard position.

Given $\alpha$ is an acute angle, we know $0 < \alpha < \frac{\pi}{2}$ , so the terminal side of $\alpha$ lies in Quadrant I.

Moreover, because $\cos(\alpha) = \frac{5}{13}$ , we know the $x$ -coordinate of the intersection point of the terminal side of $\alpha$ and the Unit Circle is $\frac{5}{13}$ .

To find $\sin(\alpha)$ , we need the $y$ -coordinate. Taking a cue from Example 7.2.1, we drop a perpendicular from the terminal side of $\alpha$ to the $x$ -axis as seen below on the right to form a right triangle with one leg measuring $\frac{5}{13}$ units and hypotenuse with a length of $1$ unit.

The Pythagorean Theorem gives $\left(\frac{5}{13}\right)^2 + y^2 = 1^2$ or $y = \frac{12}{13}$ .

Hence, $\sin(\alpha) = \frac{12}{13}$ .

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Example 7.2.3.2a

Suppose $\alpha$ is an acute angle with $\cos(\alpha) = \frac{5}{13}$ .

State the sine and cosine of the following angles:

$\theta = \pi + \alpha$

Solution:

State the sine and cosine of $\theta = \pi + \alpha$ .

To find the cosine and sine of $\theta = \pi + \alpha$ , we first plot $\theta$ in standard position.

We can imagine the sum of the angles $\pi + \alpha$ as a sequence of two rotations: a rotation of $\pi$ radians followed by a rotation of $\alpha$ radians.^[12] We see that $\alpha$ is the reference angle for $\theta$ .

By The Reference Angle Theorem, $\cos(\theta) = \pm \cos(\alpha) = \pm \frac{5}{13}$ and $\sin(\theta) = \pm \sin(\alpha) = \pm \frac{12}{13}$ . As the terminal side of $\theta$ falls in Quadrant III, both $\cos(\theta)$ and $\sin(\theta)$ are negative, so

$\cos(\theta) = - \frac{5}{13}$ and $\sin(\theta) = - \frac{12}{13}.$

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Example 7.2.3.2b

Suppose $\alpha$ is an acute angle with $\cos(\alpha) = \frac{5}{13}$ .

State the sine and cosine of the following angles:

$\theta = 2\pi - \alpha$

Solution:

State the sine and cosine of $\theta = 2\pi - \alpha$ .

Rewriting $\theta = 2\pi - \alpha$ as $\theta = 2\pi + (-\alpha)$ , we can plot $\theta$ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or `backing up,’ of $\alpha$ radians. Once again, we see that $\alpha$ is $\theta$ ‘s reference angle.

Because $\theta$ is a Quadrant IV angle, we choose the appropriate signs and get:

$\cos(\theta) = \frac{5}{13}$ and $\sin(\theta) = -\frac{12}{13}$ .

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Example 7.2.3.2c

Suppose $\alpha$ is an acute angle with $\cos(\alpha) = \frac{5}{13}$ .

State the sine and cosine of the following angles:

$\theta = 3\pi - \alpha$

Solution:

State the sine and cosine of $\theta = 3\pi - \alpha$ .

Taking a cue from the previous problem, we rewrite $\theta = 3\pi - \alpha$ as $\theta = 3\pi + (-\alpha)$ . The angle $3\pi$ represents one and a half revolutions counter-clockwise, so that when we `back up’ $\alpha$ radians, we end up in Quadrant II.

As $\alpha$ is the reference angle for $\theta$ , The Reference Angle Theorem gives

$\cos(\theta) = -\frac{5}{13}$ and $\sin(\theta) = \frac{12}{13}$ .

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Example 7.2.3.2d

Suppose $\alpha$ is an acute angle with $\cos(\alpha) = \frac{5}{13}$ .

State the sine and cosine of the following angles:

$\theta = \frac{\pi}{2} + \alpha$

Solution:

State the sine and cosine of $\theta = \frac{\pi}{2} + \alpha$ .

To plot $\theta = \frac{\pi}{2} + \alpha$ , we first rotate $\frac{\pi}{2}$ radians and follow up with $\alpha$ radians. The reference angle here is not $\alpha$ , so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.)

Let $Q(x,y)$ be the point on the terminal side of $\theta$ which lies on the Unit Circle so that $x = \cos(\theta)$ and $y = \sin(\theta)$ . Once we graph $\alpha$ in standard position, we use the fact from Geometry that equal angles subtend equal chords to show that the dotted lines in the figure below are equal.

Hence, $x = \cos(\theta) = -\frac{12}{13}$ . Similarly, we find $y = \sin(\theta) = \frac{5}{13}$ .

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A couple of remarks about Example 7.2.3 are in order. First, we note the right triangle we used to find $\sin(\alpha)$ is a scaled 5-12-13 triangle. Recognizing this Pythagorean Triple^[13] may have simplified our workflow. Along the same lines, because, the Unit Circle, by definition, is described by the equation $x^2+y^2 = 1$ , we could substitute $x = \frac{5}{13}$ in order to find $y$ . We leave it to the reader to show we get the exact same answer regardless of the approach used.

Our next example turns the tables and makes good use of the Unit Circle values given previously as well as Theorem 7.3 in a different way: instead of giving information about the angle and asking for sine or cosine values, we are given sine or cosine values and asked to produce the corresponding angles. In other words, we solve some rudimentary equations involving sine and cosine.^[14]

Example 7.2.4

Example 7.2.4.1

Determine all angles that satisfy the following equations. Express your answers in radians.^[15]

$\cos(\theta) = \dfrac{1}{2}$

Solution:

Determine all angles that satisfy $\cos(\theta) = \dfrac{1}{2}$ .

If $\cos(\theta) = \frac{1}{2}$ , then we know the terminal side of $\theta$ , when plotted in standard position, intersects the Unit Circle at $x = \frac{1}{2}$ . This means $\theta$ is a Quadrant I or IV angle.

Because $\cos(\theta) = \frac{1}{2}$ , we know from the values on the given Unit Circle that the reference angle is $\frac{\pi}{3}$ .

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One solution in Quadrant I is $\theta = \frac{\pi}{3}$ .

Per Theorem 7.3, all other Quadrant I solutions must be coterminal with $\frac{\pi}{3}$ . Recall from Section 7.1.2, two angles in radian measure are coterminal if and only if they differ by an integer multiple of $2\pi$ . Hence to describe all angles coterminal with a given angle, we add $2\pi k$ for integers $k = 0$ , $\pm 1$ , $\pm 2$ , \dots. Hence, we record our final answer as $\theta = \frac{\pi}{3} + 2\pi k$ for integers $k$ .

Proceeding similarly for the Quadrant IV case, we find the solution to $\cos(\theta) = \frac{1}{2}$ here is $\frac{5 \pi}{3}$ , so our answer in this Quadrant is $\theta = \frac{5\pi}{3} + 2\pi k$ for integers $k$ .

Example 7.2.4.2

Determine all angles that satisfy the following equations. Express your answers in radians.

$\sin(\alpha) = -\dfrac{1}{2}$

Solution:

Determine all angles that satisfy $\sin(\alpha) = -\dfrac{1}{2}$ .

If $\sin(\alpha) = -\frac{1}{2}$ , then when $\alpha$ is plotted in standard position, its terminal side intersects the Unit Circle at $y=-\frac{1}{2}$ . From this, we determine $\alpha$ is a Quadrant III or Quadrant IV angle with reference angle $\frac{\pi}{6}$ .

In Quadrant III, one solution is $\frac{7\pi}{6}$ , so we capture all Quadrant III solutions by adding integer multiples of $2\pi$ : $\alpha = \frac{7\pi}{6} + 2\pi k$ .

In Quadrant IV, one solution is $\frac{11\pi}{6}$ so all the solutions here are of the form $\alpha = \frac{11\pi}{6} + 2\pi k$ for integers $k$ .

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Example 7.2.4.3

Determine all angles that satisfy the following equations. Express your answers in radians.

$\cos(\beta) = 0$

Solution:

Determine all angles that satisfy $\cos(\beta) = 0$ .

If $\cos(\beta) = 0$ , then the terminal side of $\beta$ must lie on the line $x=0$ , also known as the $y$ -axis.

While, technically speaking, $\frac{\pi}{2}$ isn’t a reference angle (it’s not acute), we can nonetheless use it to find our answers.

If we follow the procedure set forth in the previous examples, we find $\beta = \frac{\pi}{2} + 2\pi k$ and $\beta = \frac{3\pi}{2} + 2\pi k$ for integers, $k$ .

While this solution is correct, it can be shortened to $\beta = \frac{\pi}{2} + \pi k$ for integers $k$ . The reader is encouraged to see the geometry using the diagram above on the left.

Example 7.2.4.4

Determine all angles that satisfy the following equations. Express your answers in radians.

$\sin(\gamma) = \dfrac{3}{2}$

Solution:

Determine all angles that satisfy $\sin(\gamma) = \dfrac{3}{2}$ .

We are asked to solve $\sin(\gamma) = \frac{3}{2}$ . As sine values are $y$ -coordinates on the Unit Circle, $\sin(\gamma)$ can’t be any larger than $1$ . Hence, $\sin(t) = \frac{3}{2}$ has no solutions.

One of the key items to take from Example 7.2.4 is that, in general, solutions to trigonometric equations consist of infinitely many answers. This is especially important when checking answers to the exercises.

For example, another Quadrant IV solution to $\sin(\theta) = -\frac{1}{2}$ is $\theta = -\frac{\pi}{6}$ . Hence, the family of Quadrant IV answers to number 2 in the last example could just have easily been written $\theta = -\frac{\pi}{6} + 2\pi k$ for integers $k$ . While on the surface, this family may look different than the stated solution of $\theta = \frac{11\pi}{6} + 2\pi k$ for integers $k$ , we leave it to the reader to show they represent the same list of angles.

It is also worth noting that when asked to solve equations in algebra, we are usually looking for real number solutions. Thanks to the work we did previously on the identifications, we are able to regard the inputs to the sine and cosine functions as real numbers by identifying any real number $t$ with an oriented angle $\theta$ measuring $\theta = t$ radians. That is, for each real number $t$ , we associate an oriented arc $t$ units in length with initial point $(1,0)$ and endpoint $P(\cos(t), \sin(t))$ .

Two circles side by side unit circles. The first unit circle has the angle theta with terminal side in the first quadrant. The length of the arc created by theta as t. That same length is drawn parallel to the y-axis as starting at the point (1,0). The second unit circle has the same angle as the first unit circle, noting theta equals t. The point P where theta's terminal side meets the unit circle is labeled. — Real Number Relationships to Angles

In practice this means in expressions like ` $\cos(\pi)$ ‘ and ` $\sin(2)$ ,’ the inputs can be thought of as either angles in radian measure or real numbers, whichever is more convenient.

Suppose, as in the Exercises, we are asked to find all real number solutions to the equation such as $\sin(t) = -\frac{1}{2}$ . The discussion above allows us to find the real number solutions to this equation by thinking in angles. Indeed, we would solve this equation in the exact way we solved $\sin(\theta) = -\frac{1}{2}$ in Example 7.2.4 number 2. Our solution is only cosmetically different in that the variable used is $t$ rather than $\theta$ : $t = \frac{7\pi}{6} + 2\pi k$ or $t = \frac{11\pi}{6} + 2\pi k$ for integers, $k$ .

We will study the sine and cosine functions in greater detail in Section 7.3. Until then, keep in mind that any properties of the sine and cosine functions developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers.

7.2.3 Beyond the Unit Circle

In Definition 7.3, we define the sine and cosine functions using the Unit Circle, $x^2+y^2=1$ . It turns out that we can use any circle centered at the origin to determine the sine and cosine values of angles. To show this, we essentially recycle the same similarity arguments used in Section 7.2.1 to show the trigonometric ratios described in Definition 7.2 are independent of the choice of right triangle used.^[16]

Consider for the moment the acute angle $\theta$ drawn below in standard position. Let $Q(x,y)$ be the point on the terminal side of $\theta$ which lies on the circle $x^2+y^2 = r^2$ , and let $P(x',y')$ be the point on the terminal side of $\theta$ which lies on the Unit Circle. Now consider dropping perpendiculars from $P$ and $Q$ to create two right triangles, $\Delta OPA$ and $\Delta OQB$ . These triangles are similar,^[17] thus it follows that $\frac{x}{x'} = \frac{r}{1} = r$ , so $x = r x'$ and, similarly, we find $y = r y'$ . Because, by definition, $x' = \cos(\theta)$ and $y' = \sin(\theta)$ , we get the coordinates of $Q$ to be $x = r \cos(\theta)$ and $y = r \sin(\theta)$ . By reflecting these points through the $x$ -axis, $y$ -axis and origin, we obtain the result for all non-quadrantal angles $\theta$ , and we leave it to the reader to verify these formulas hold for the quadrantal angles as well.

Two sets of circles side by side. On the first graph is a unit circle with a circle of radius r around it. There is an angle, theta, with terminal side in the first quadrant. The point P is labeled where the terminal side touches the unit circle. The point Q is where the terminal side touches the circle of radius r. The second diagram has all of the same information with vertical lines from the each point down to the x-axis. — Graphical Representations of Trigonometry Beyond the Unit Circle

Not only can we describe the coordinates of $Q$ in terms of $\cos(\theta)$ and $\sin(\theta)$ but due to the fact that the radius of the circle is $r = \sqrt{x^2 + y^2}$ , we can also express $\cos(\theta)$ and $\sin(\theta)$ in terms of the coordinates of $Q$ . These results are summarized in the following theorem.

Theorem 7.4

If $Q(x,y)$ is the point on the terminal side of an angle $\theta$ , plotted in standard position, which lies on the circle $x^2+y^2 = r^2$ then $x = r \cos(\theta)$ and $y = r \sin(\theta)$ . Moreover,

$\begin{array}{ccc} \cos(\theta)= \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2+y^2}} & \text{and} & \sin(\theta) = \dfrac{y}{r} = \dfrac{y}{\sqrt{x^2+y^2}} \\ \end{array}$

Note that in the case of the Unit Circle we have $r = \sqrt{x^2+y^2} = 1$ , so Theorem 7.4 reduces to our definitions of $\cos(\theta)$ and $\sin(\theta)$ in Definition 7.3. Our next example makes good use of Theorem 7.4.

Example 7.2.5

Example 7.2.5.1

Suppose that the terminal side of an angle $\theta$ , when plotted in standard position, contains the point $Q(4,-2)$ . Compute $\sin(\theta)$ and $\cos(\theta)$ .

Solution:

Suppose that the terminal side of an angle $\theta$ , when plotted in standard position, contains the point $Q(4,-2)$ . Compute $\sin(\theta)$ and $\cos(\theta)$ .

We are given both the $x$ and $y$ coordinates of a point on the terminal side of this angle, so we can use Theorem7.4 directly. First, we find

$\begin{array}{rcl} r &=& \sqrt{x^2+y^2} \\ &=& \sqrt{(-2)^2+4^2} \\ &=& \sqrt{20} \\ &=& 2\sqrt{5} \end{array}$

This means the point $Q$ lies on a circle of radius $2\sqrt{5}$ units.

Hence, $\cos(\theta) = \frac{x}{r} = \frac{4}{2 \sqrt{5}} = \frac{2 \sqrt{5}}{5}$ and $\sin(\theta) = \frac{y}{r} = \frac{-2}{2 \sqrt{5}} = -\frac{\sqrt{5}}{5}$ .

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Example 7.2.5.2

Suppose $\frac{\pi}{2} < \theta < \pi$ with $\sin(\theta) = \frac{8}{17}$ . Compute $\cos(\theta)$ .

Solution:

Suppose $\frac{\pi}{2} < \theta < \pi$ with $\sin(\theta) = \frac{8}{17}$ . Compute $\cos(\theta)$ .

We are told $\frac{\pi}{2} < \theta < \pi$ , so, in particular, $\theta$ is a Quadrant II angle.

Per Theorem 7.4, $\sin(\theta) = \frac{8}{17} = \frac{y}{r}$ where $y$ is the $y$ -coordinate of the intersection point of the circle $x^2+y^2 = r^2$ and the terminal side of $\theta$ (when plotted in standard position, of course!) For convenience, we choose $r=17$ so that $y = 8$ , and we get the diagram.

Given $x^2+y^2 = r^2$ , we get $x^2 + 8^2 = 17^2$ . We find $x = \pm 15$ , and because $\theta$ is a Quadrant II angle, we get $x = -15$ .

Hence, $\cos(\theta) = -\frac{15}{17}$ .

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Example 7.2.5.3

In Example 7.1.5 in Section 7.1.2, we approximated the radius of the earth at $41.628^{\circ}$ north latitude to be $2960$ miles. Justify this approximation if the spherical radius of the Earth is $3960$ miles.

Solution:

In Example 7.1.5 in Section 7.1.2, we approximated the radius of the earth at $41.628^{\circ}$ north latitude to be $2960$ miles. Justify this approximation if the spherical radius of the Earth is $3960$ miles.

Recall the diagram below on the left indicating the circles which are the parallels of latitude.^[18]

A sketch of the global with latitude lines marked, along with the north and south pole and equator. Parts of North and South America as well as Europe and Africa are visible in the sketch. — Sketch of Earth

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Assuming the Earth is a sphere of radius $3960$ miles, a cross-section through the poles produces a circle of radius $3960$ miles. Viewing the Equator as the $x$ -axis, the value we seek is the $x$ -coordinate of the point $Q(x,y)$ indicated in the figure above on the right. Using Theorem 7.4, we get $x = 3960 \cos\left(41.628^{\circ}\right) \approx 2960$ . Hence, the radius of the Earth at North Latitude $41.628^{\circ}$ is approximately $2960$ miles.

Theorem 7.4 gives us what we need to `circle back’ to the question posed at the the beginning of the section: how to describe the position of an object traveling in a circular path of radius $r$ with constant angular velocity $\omega$ . Suppose that at time $t$ , the object has swept out an angle measuring $\theta$ radians. If we assume that the object is at the point $(r,0)$ when $t=0$ , the angle $\theta$ is in standard position. By definition, $\omega = \frac{\theta}{t}$ which we rewrite as $\theta = \omega t$ . According to Theorem 7.4, the location of the object $Q(x,y)$ on the circle is found using the equations $x = r \cos(\theta) = r \cos(\omega t)$ and $y = r \sin(\theta) = r \sin(\omega t)$ . Hence, at time $t$ , the object is at the point $(r \cos(\omega t), r \sin(\omega t))$ , as seen in the diagram below.

A circle of radius r centered at the origin. The angle is given as theta equals omega times t, with a terminal side in the first quadrant. The point Q where the terminal side meets the circle is labeled as well, — Equations for Circular Motion

We have just argued the following.

Equation 7.3

Suppose an object is traveling in a circular path of radius $r$ centered at the origin with constant angular velocity $\omega$ . If $t=0$ corresponds to the point $(r,0)$ , then the $x$ and $y$ coordinates of the object are functions of $t$ and are given by $x = r \cos(\omega t)$ and $y = r \sin(\omega t)$ . Here, $\omega > 0$ indicates a counter-clockwise direction and $\omega < 0$ indicates a clockwise direction.

Example 7.2.6

Suppose we are in the situation of Example 7.1.5. Determine the equations of motion of Lakeland Community College as the earth rotates.

Solution:

From Example 7.1.5, we take $r = 2960$ miles and and $\omega = \frac{\pi}{12 \, \text{hours}}$ .

Hence, the equations of motion are

$x = r \cos(\omega t) = 2960 \cos\left(\frac{\pi}{12} t\right) \text{ and } y = r \sin(\omega t) = 2960 \sin\left(\frac{\pi}{12} t\right),$

where $x$ and $y$ are measured in miles and $t$ is measured in hours.

7.2.4 Section Exercises

In Exercises 1 – 4, compute the requested quantities.

Determine $\theta$ , $a$ , and $c$ .
Determine $\alpha$ , $b$ , and $c$ .
Find $\theta$ , $a$ , and $c$ .
Find $\beta$ , $b$ , and $c$ .

In Exercises 5 – 7, answer the following questions assuming $\theta$ is an angle in a right triangle.

If $\theta = 15^{\circ}$ and the hypotenuse has length $10$ , how long is the side opposite $\theta$ ?
If $\theta = 38.2^{\circ}$ and the side opposite $\theta$ has length $14$ , how long is the hypotenuse?
If $\theta = 2.05^{\circ}$ and the hypotenuse has length $3.98$ , how long is the side adjacent to $\theta$ ?

In Exercises 8 – 27, determine the exact value of the cosine and sine of the given angle.

$\theta = 0$
$\theta = \dfrac{\pi}{4}$
$\theta = \dfrac{\pi}{3}$
$\theta = \dfrac{\pi}{2}$
$\theta = \dfrac{2\pi}{3}$
$\theta = \dfrac{3\pi}{4}$
$\theta = \pi$
$\theta = \dfrac{7\pi}{6}$
$\theta = \dfrac{5\pi}{4}$
$\theta = \dfrac{4\pi}{3}$
$\theta = \dfrac{3\pi}{2}$
$\theta = \dfrac{5\pi}{3}$
$\theta = \dfrac{7\pi}{4}$
$\theta = \dfrac{23\pi}{6}$
$\theta = -\dfrac{13\pi}{2}$
$\theta = -\dfrac{43\pi}{6}$
$\theta = -\dfrac{3\pi}{4}$
$\theta = -\dfrac{\pi}{6}$
$\theta = \dfrac{10\pi}{3}$
$\theta = 117\pi$

In Exercises 28 – 36, compute all of the angles which satisfy the given equation.

$\sin(\theta) = \dfrac{1}{2}$
$\cos(\theta) = -\dfrac{\sqrt{3}}{2}$
$\sin(\theta) = 0$
$\cos(\theta) = \dfrac{\sqrt{2}}{2}$
$\sin(\theta) = \dfrac{\sqrt{3}}{2}$
$\cos(\theta) = -1$
$\sin(\theta) = -1$
$\cos(\theta) = \dfrac{\sqrt{3}}{2}$
$\cos(\theta) = -1.001$

In Exercises 37 – 45, solve the equation for $t$ . (See the remarks in 7.2.2.)

$\cos(t) = 0$
$\sin(t) = -\dfrac{\sqrt{2}}{2}$
$\cos(t) = 3$
$\sin(t) = -\dfrac{1}{2}$
$\cos(t) = \dfrac{1}{2}$
$\sin(t) = -2$
$\cos(t) = 1$
$\sin(t) = 1$
$\cos(t) = -\dfrac{\sqrt{2}}{2}$

In Exercises 46 – 49, let $\theta$ be the angle in standard position whose terminal side contains the given point then compute $\cos(\theta)$ and $\sin(\theta)$ .

$P(-7, 24)$
$Q(3, 4)$
$R(5, -9)$
$T(-2, -11)$

In Exercises 50 – 59, use the results developed throughout the section to find the requested value.

If $\sin(\theta) = -\dfrac{7}{25}$ with $\theta$ in Quadrant IV, what is $\cos(\theta)$ ?
If $\cos(\theta) = \dfrac{4}{9}$ with $\theta$ in Quadrant I, what is $\sin(\theta)$ ?
If $\sin(\theta) = \dfrac{5}{13}$ with $\theta$ in Quadrant II, what is $\cos(\theta)$ ?
If $\cos(\theta) = -\dfrac{2}{11}$ with $\theta$ in Quadrant III, what is $\sin(\theta)$ ?
If $\sin(\theta) = -\dfrac{2}{3}$ with $\theta$ in Quadrant III, what is $\cos(\theta)$ ?
If $\cos(\theta) = \dfrac{28}{53}$ with $\theta$ in Quadrant IV, what is $\sin(\theta)$ ?
If $\sin(\theta) = \dfrac{2\sqrt{5}}{5}$ and $\dfrac{\pi}{2} < \theta < \pi$ , what is $\cos(\theta)$ ?
If $\cos(\theta) = \dfrac{\sqrt{10}}{10}$ and $2\pi < \theta < \dfrac{5\pi}{2}$ , what is $\sin(\theta)$ ?
If $\sin(\theta) = -0.42$ and $\pi < \theta < \dfrac{3\pi}{2}$ , what is $\cos(\theta)$ ?
If $\cos(\theta) = -0.98$ and $\dfrac{\pi}{2} < \theta < \pi$ , what is $\sin(\theta)$ ?

In Exercises 60 – 64, write the given function as a nontrivial decomposition of functions as directed.

For $f(t) = 3t + \sin(2t)$ , find functions $g$ and $h$ so that $f=g+h$ .
For $f(\theta) = 3\cos(\theta) - \sin(4\theta)$ , find functions $g$ and $h$ so that $f=g-h$ .
For $f(t) = e^{-0.1t} \sin(3t)$ , find functions $g$ and $h$ so that $f=gh$ .
For $r(t) = \dfrac{\sin(t)}{t}$ , find functions $f$ and $g$ so $r = \dfrac{f}{g}$ .
For $r(\theta) =\sqrt{3 \cos(\theta)}$ , find functions $f$ and $g$ so $r = g \circ f$ .
For each function $S(t)$ listed below, compute the average rate of change over the indicated interval.^[19] What trends do you notice? Be sure your calculator is in radian mode! $\begin{array}{|r||c|c|c|} \hline S(t) & [-0.1, 0.1] & [-0.01, 0.01] &[-0.001, 0.001] \\ \hline \sin(t) &&& \\ \hline \sin(2t) &&& \\ \hline \sin(3t) &&& \\ \hline \sin(4t) &&& \\ \hline \end{array}$

In Exercises 66 – 69, find the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that $t = 0$ corresponds to a position along the positive $x$ -axis. (See Equation 7.3 and Example 7.1.5.)

A point on the edge of the spinning yo-yo in Exercise 50 from Section 7.1.2.Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 revolutions per minute.
The yo-yo in exercise 52 from Section 7.1.2.Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds.
A point on the edge of the hard drive in Exercise 53 from Section 7.1.2.Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute.
A passenger on the Big Wheel in Exercise 55 from Section 7.1.2.Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds.
Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide each by 2. The resulting numbers should look hauntingly familiar.
In this section, we saw that the sine and cosine functions of angles can be considered functions of real numbers. With help from your classmates, discuss the domains and ranges of $f(t) = \sin(t)$ and $g(t) = \cos(t)$ . Write your answers using interval notation.
Another way to establish Theorem 7.4 is to use transformations. Transform the Unit Circle, $x^2+y^2 = 1$ , to $x^2 + y^2 = r^2$ using horizontal and vertical stretches. Show if the coordinates on the Unit Circle are $(\cos(\theta), \sin(\theta))$ , then the corresponding coordinates on $x^2+y^2 = r^2$ are $(r \cos(\theta), r \sin(\theta))$ .
In the scenario of Equation 7.3, we assumed that at $t=0$ , the object was at the point $(r,0)$ . If this is not the case, we can adjust the equations of motion by introducing a `time delay.’ If $t_{0} > 0$ is the first time the object passes through the point $(r,0)$ , show, with the help of your classmates, the equations of motion are $x = r \cos(\omega (t - t_{0}))$ and $y = r \sin(\omega (t-t_{0}))$ .

Section 7.2 Exercise Answers can be found in the Appendix … Coming soon

Including one by Mentor, Ohio native President James Garfield. ↵
We will prove this in Section 8.5 by generalizing the Pythagorean Theorem to a formula that works for all triangles. ↵
That is, a triangle with the same `shape' - that is, the same angles. ↵
We will do a little of this in Section 8.2. ↵
See Definition 1.3 in Section 1.2. ↵
See Section 1.1. ↵
The etymology of the name `sine' is quite colorful, and the interested reader is invited to research it; the `co' in `cosine' is related to the concept of `co'mplementary angles (see Sections 7.1 and 7.2.1) and is explained in detail in Section 8.2. ↵
For instance, Definition 7.2 in Section 7.2.1. ↵
Do you see why? ↵
Recall we say they are `coterminal.' ↵
For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a `natural' way to match oriented angles with real numbers! ↵
Because $\pi + \alpha = \alpha + \pi$ , $\theta$ may be plotted by reversing the order of rotations given here. You should do this. ↵
See Section 7.2.1 for more examples of Pythagorean Triples. ↵
We will study equations in more detail in Section 8.3.2. ↵
This ensures we keep building the `fluency with radians' which is so necessary for later work. ↵
Another approach uses transformations. See Exercise 72. ↵
Do you remember why? ↵
Diagram credit: Pearson Scott Foresman [Public domain], via Wikimedia Commons. ↵
See Definition 1.11 in Section 1.3.4 for a review of this concept, as needed. ↵

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Functions, Trigonometry, and Systems of Equations Copyright © by Vanessa Coffelt is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

7.2.1 Right Triangle Definitions

Definition 7.2

7.2.2 Unit Circle Definitions

Definition 7.3

Example 7.2.1

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Example 7.2.2

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Example 7.2.3

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Example 7.2.4

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7.2.3 Beyond the Unit Circle

Example 7.2.5

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Equation 7.3

Example 7.2.6

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