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Trigonometric Functions

Trigonometric functions are used to model many phenomena, including sound waves, vibrations of strings, alternating electrical current, and the motion of pendulums. In fact, almost any repetitive, or cyclical, motion can be modeled by some combination of trigonometric functions. In this section, we define the six basic trigonometric functions and look at some of the main identities involving these functions.

Radian Measure

To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle \theta, let s be the length of the corresponding arc on the unit circle ((Figure)). We say the angle corresponding to the arc of length 1 has radian measure 1.

An image of a circle. At the exact center of the circle there is a point. From this point, there is one line segment that extends horizontally to the right a point on the edge of the circle and another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. These line segments have a length of 1 unit. The curved segment on the edge of the circle that connects the two points at the end of the line segments is labeled “s”. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta = s radians”.
Figure 1. The radian measure of an angle \theta  is the arc length s of the associated arc on the unit circle.

Since an angle of 360° corresponds to the circumference of a circle, or an arc of length 2\pi , we conclude that an angle with a degree measure of 360° has a radian measure of 2\pi . Similarly, we see that 180° is equivalent to \pi  radians. (Figure) shows the relationship between common degree and radian values.

DegreesRadiansDegreesRadians
001202\pi/3
30\pi/61353\pi/4
45\pi/41505\pi/6
60\pi/3180\pi
90\pi/2

Converting between Radians and Degrees

  1. Express 225° using radians.
  2. Express 5\pi/3 rad using degrees.

Solution

Use the fact that 180° is equivalent to \pi  radians as a conversion factor: 1=\frac{\pi \, \text{rad}}{180^{\circ}}=\frac{180^{\circ}}{\pi \, \text{rad}}.

  1. 225^{\circ}=225^{\circ}·\frac{\pi }{180^{\circ}}=\frac{5\pi }{4} rad
  2. \frac{5\pi }{3} rad = \frac{5\pi }{3}·\frac{180^{\circ}}{\pi }=300^{\circ}

Express 210° using radians. Express 11\pi/6 rad using degrees.

Solution

7\pi/6 rad; 330°

Hint

\pi  radians is equal to 180^{\circ}.

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point P=(x,y) on the unit circle. Let \theta  be an angle with an initial side that lies along the positive x-axis and with a terminal side that is the line segment OP. An angle in this position is said to be in standard position ((Figure)). We can then define the values of the six trigonometric functions for \theta  in terms of the coordinates x and y.

An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled “P = (x, y)”. These line segments have a length of 1 unit. From the point “P”, there is a dotted vertical line that extends downwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta”.
Figure 2. The angle \theta  is in standard position. The values of the trigonometric functions for \theta  are defined in terms of the coordinates x and y.

Definition

Let P=(x,y) be a point on the unit circle centered at the origin O. Let \theta  be an angle with an initial side along the positive x-axis and a terminal side given by the line segment OP. The trigonometric functions are then defined as

\begin{array}{cccc}\sin \theta =y & & & \csc \theta =\large{\frac{1}{y}} \normalsize \\ \cos \theta =x & & & \sec \theta =\large{\frac{1}{x}} \normalsize \\ \tan \theta =\large{\frac{y}{x}} \normalsize & & & \cot \theta =\large{\frac{x}{y}} \end{array}

If x=0, then \sec \theta  and \tan \theta  are undefined. If y=0, then \cot \theta  and \csc \theta  are undefined.

We can see that for a point P=(x,y) on a circle of radius r with a corresponding angle \theta, the coordinates x and y satisfy\begin{array}{c} \cos \theta =\large{\frac{x}{r}} \\ x=r \cos \theta \\ \sin \theta =\large{\frac{y}{r}} \\ y=r \sin \theta \end{array}.

The values of the other trigonometric functions can be expressed in terms of x, \, y, and r ((Figure)).

An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one blue line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another blue line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled “P = (x, y)”. These line segments have a length of “r” units. Between these line segments within the circle is the label “theta”, representing the angle between the segments. From the point “P”, there is a blue vertical line that extends downwards until it hits the x axis and thus hits the horizontal line segment, at a point labeled “x”. At the intersection horizontal line segment and vertical line segment at the point x, there is a right triangle symbol. From the point “P”, there is a dotted horizontal line segment that extends left until it hits the y axis at a point labeled “y”.
Figure 3. For a point P=(x,y) on a circle of radius r, the coordinates x and y satisfy x=r \cos \theta  and y=r \sin \theta.

(Figure) shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of \sin \theta  and \cos \theta.

\theta \sin \theta \cos \theta
001
\large{\frac{\pi}{6}}\large{\frac{1}{2}}\large{\frac{\sqrt{3}}{2}}
\large{\frac{\pi}{4}}\large{\frac{\sqrt{2}}{2}}\large{\frac{\sqrt{2}}{2}}
\large{\frac{\pi}{3}}\large{\frac{\sqrt{3}}{2}}\large{\frac{1}{2}}
\large{\frac{\pi}{2}}10

Evaluating Trigonometric Functions

Evaluate each of the following expressions.

  1. \sin \Big(\large\frac{2\pi}{3}\Big)
  2. \cos \Big(-\large\frac{5\pi}{6}\Big)
  3. \tan \Big(\large\frac{15\pi}{4}\Big)

Solution

  1. On the unit circle, the angle \theta =\large\frac{2\pi}{3} corresponds to the point \Big(-\large\frac{1}{2}, \frac{\sqrt{3}}{2}\Big). Therefore, \sin \Big(\large\frac{2\pi}{3}\Big) \normalsize = y = \large\frac{\sqrt{3}}{2}.
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the left to another point on the edge of the circle. This point is labeled “(-(1/2), ((square root of 3)/2))”. These line segments have a length of 1 unit. From the point “(-(1/2), ((square root of 3)/2))”, there is a vertical line that extends downwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise until it hits the diagonal line segment. This arrow has the label “theta = (2 pi)/3”.
  2. An angle \theta =-\large\frac{5\pi}{6} corresponds to a revolution in the negative direction, as shown. Therefore, \cos \Big(-\large\frac{5\pi}{6}\Big) =x=-\large\frac{\sqrt{3}}{2}.
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the left to another point on the edge of the circle. This point is labeled “(-((square root of 3)/2)), -(1/2))”. These line segments have a length of 1 unit. From the point “(-((square root of 3)/2)), -(1/2))”, there is a vertical line that extends upwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels clockwise until it hits the diagonal line segment. This arrow has the label “theta = -(5 pi)/6”.
  3. An angle \theta =\large\frac{15\pi}{4} \normalsize = 2\pi +\large\frac{7\pi}{4}. Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of \large\frac{7\pi}{4} corresponds to the point \Big(\large\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\Big), we can conclude that \tan \Big(\large\frac{15\pi}{4}\Big)\normalsize =\large\frac{y}{x} =-1.
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the right to another point on the edge of the circle. This point is labeled “(((square root of 2)/2), -((square root of 2)/2))”. These line segments have a length of 1 unit. From the point “(((square root of 2)/2), -((square root of 2)/2))”, there is a vertical line that extends upwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise. The arrow makes one full rotation around the circle and then keeps traveling until it hits the diagonal line segment. This arrow has the label “theta = (15 pi)/4”.

Evaluate \cos (3\pi/4) and \sin (−\pi/6).

Solution

\cos (3\pi/4)=−\sqrt{2}/2; \, \sin(−\pi/6)=-1/2

Hint

Look at angles on the unit circle.

As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let \theta  be one of the acute angles. Let A be the length of the adjacent leg, O be the length of the opposite leg, and H be the length of the hypotenuse. By inscribing the triangle into a circle of radius H, as shown in (Figure), we see that A, \, H, and O satisfy the following relationships with \theta:

\begin{array}{cccc}\sin \theta =\large \frac{O}{H} & & & \normalsize \csc \theta =\large \frac{H}{O} \\ \cos \theta =\large \frac{A}{H} & & & \sec \theta =\large \frac{H}{A} \\ \tan \theta =\large \frac{O}{A} & & & \cot \theta =\large \frac{A}{O} \end{array}
An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment with length labeled “H” that extends diagonally upwards and to the right to another point on the edge of the circle. From the point, there is vertical line with a length labeled “O” that extends downwards until it hits the x axis and thus the horizontal line segment at a point with a right triangle symbol. The distance from this point to the center of the circle is labeled “A”. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta”.
Figure 4. By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at \theta.

Constructing a Wooden Ramp

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be 10^{\circ}, how long does the ramp need to be?

Solution

Let x denote the length of the ramp. In the following image, we see that x needs to satisfy the equation \sin(10^{\circ})=4/x. Solving this equation for x, we see that x=4/ \sin(10^{\circ}) \approx 23.035 ft.

An image of a ramp and a staircase. The ramp starts at a point and increases diagonally upwards and to the right at an angle of 10 degrees for x feet. At the end of the ramp, which is 4 feet off the ground, a staircase descends downwards and to the right.

A house painter wants to lean a 20-ft ladder against a house. If the angle between the base of the ladder and the ground is to be 60^{\circ}, how far from the house should she place the base of the ladder?

Solution

10 ft

Hint

Draw a right triangle with hypotenuse 20 ft.

Trigonometric Identities

trigonometric identity is an equation involving trigonometric functions that is true for all angles \theta  for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

Rule: Trigonometric Identities

Reciprocal identities

\begin{array}{cccc}\tan \theta =\large \frac{\sin \theta}{\cos \theta} & & & \cot \theta =\large \frac{\cos \theta}{\sin \theta} \\ \csc \theta =\large \frac{1}{\sin \theta} & & & \sec \theta =\large \frac{1}{\cos \theta} \end{array}

Pythagorean identities

\sin^2 \theta +\cos^2 \theta =1\phantom{\rule{2em}{0ex}}1+\tan^2 \theta =\sec^2 \theta \phantom{\rule{2em}{0ex}}1+\cot^2 \theta =\csc^2 \theta

Addition and subtraction formulas

\sin(\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta
\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta

Double-angle formulas

\sin(2\theta)=2\sin \theta \cos \theta
\cos(2\theta)=2\cos^2 \theta -1=1-2\sin^2 \theta =\cos^2 \theta -\sin^2 \theta

Solving Trigonometric Equations

For each of the following equations, use a trigonometric identity to find all solutions.

  1. 1+\cos(2\theta)=\cos \theta
  2. \sin(2\theta)=\tan \theta

[reveal-answer q=”501288″]Show Answer[/reveal-answer][hidden-answer a=”501288″]a. Using the double-angle formula for \cos(2\theta), we see that \theta  is a solution of

1+\cos(2\theta)=\cos \theta

if and only if

1+2\cos^2 \theta -1=\cos \theta,

which is true if and only if

2\cos^2 \theta -\cos \theta =0.

To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by \cos \theta. The problem with dividing by \cos \theta is that it is possible that \cos \theta is zero. In fact, if we did divide both sides of the equation by \cos \theta, we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that \theta  is a solution of this equation if and only if

\cos \theta (2\cos \theta -1)=0.

Since \cos \theta =0 when

\theta =\large \frac{\pi }{2}, \, \frac{\pi}{2} \normalsize \pm \pi, \, \large \frac{\pi}{2} \normalsize \pm 2\pi, \cdots,

and \cos \theta =1/2 when

\theta =\large \frac{\pi}{3}, \, \frac{\pi}{3} \normalsize \pm 2\pi, \cdots, or \theta =-\large \frac{\pi}{3}, \, -\frac{\pi}{3} \normalsize \pm 2\pi, \cdots,

we conclude that the set of solutions to this equation is

\theta =\large \frac{\pi}{2} \normalsize +n\pi, \, \theta =\large \frac{\pi}{3} \normalsize +2n\pi, and \theta =-\large \frac{\pi}{3}\normalsize +2n\pi, \, n=0, \pm 1, \pm 2,\cdots.

b. Using the double-angle formula for \sin(2\theta) and the reciprocal identity for \tan(\theta), the equation can be written as

2\sin \theta \cos \theta =\large \frac{\sin\theta}{\cos \theta}.

To solve this equation, we multiply both sides by \cos \theta  to eliminate the denominator, and say that if \theta  satisfies this equation, then \theta  satisfies the equation

2\sin \theta \cos^2 \theta -\sin \theta =0.

However, we need to be a little careful here. Even if \theta  satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by \cos \theta. However, if \cos \theta =0, we cannot divide both sides of the equation by \cos \theta. Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor \sin \theta  out of both terms on the left-hand side instead of dividing both sides of the equation by \sin \theta. Factoring the left-hand side of the equation, we can rewrite this equation as

\sin \theta (2\cos^2 \theta -1)=0.

Therefore, the solutions are given by the angles \theta  such that \sin \theta =0 or \cos^2 \theta =1/2. The solutions of the first equation are \theta =0, \pm \pi, \pm 2\pi, \cdots. The solutions of the second equation are \theta =\pi /4, \, (\pi/4) \pm (\pi/2), \, (\pi/4) \pm \pi, \cdots. After checking for extraneous solutions, the set of solutions to the equation is

\theta =n\pi and \theta =\large \frac{\pi}{4}+\frac{n\pi}{2}, \, \normalsize n=0, \pm 1, \pm 2, \cdots.

Find all solutions to the equation \cos(2\theta)=\sin \theta.

Solution

\theta =\large \frac{3\pi}{2} \normalsize +2n\pi, \, \large \frac{\pi}{6} \normalsize +2n\pi, \, \large \frac{5\pi}{6} \normalsize +2n\pi for n=0, \pm 1, \pm 2, \cdots

Hint

Use the double-angle formula for cosine.

Proving a Trigonometric Identity

Prove the trigonometric identity 1+\tan^2 \theta =\sec^2 \theta.

Solution

We start with the identity\sin^2 \theta +\cos^2 \theta =1.

Dividing both sides of this equation by \cos^2 \theta, we obtain\frac{\sin^2 \theta}{\cos^2 \theta}+1=\frac{1}{\cos^2 \theta}.

Since \sin \theta / \cos \theta =\tan \theta and 1 / \cos \theta =\sec \theta, we conclude that\tan^2 \theta +1=\sec^2 \theta.

Prove the trigonometric identity 1+\cot^2 \theta =\csc^2 \theta.

Hint

Divide both sides of the identity \sin^2 \theta + \cos^2 \theta =1 by \sin^2 \theta.

Graphs and Periods of the Trigonometric Functions

We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let P=(x,y) be a point on the unit circle and let \theta  be the corresponding angle.  Since the angle \theta  and \theta +2\pi  correspond to the same point P, the values of the trigonometric functions at \theta  and at \theta +2\pi  are the same. Consequently, the trigonometric functions are periodic functions. The period of a function f is defined to be the smallest positive value p such that f(x+p)=f(x) for all values x in the domain of f. The sine, cosine, secant, and cosecant functions have a period of 2\pi. Since the tangent and cotangent functions repeat on an interval of length \pi, their period is \pi  ((Figure)).

An image of six graphs. Each graph has an x axis that runs from -2 pi to 2 pi and a y axis that runs from -2 to 2. The first graph is of the function “f(x) = sin(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 0) and increases until the point (-((3 pi)/2), 1). After this point, the function decreases until the point (-(pi/2), -1). After this point, the function increases until the point ((pi/2), 1). After this point, the function decreases until the point (((3 pi)/2), -1). After this point, the function begins to increase again. The x intercepts shown on the graph are at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The second graph is of the function “f(x) = cos(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 1) and decreases until the point (-pi, -1). After this point, the function increases until the point (0, 1). After this point, the function decreases until the point (pi, -1). After this point, the function increases again. The x intercepts shown on the graph are at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0). The y intercept is at the point (0, 1). The graph of cos(x) is the same as the graph of sin(x), except it is shifted to the left by a distance of (pi/2). On the next four graphs there are dotted vertical lines which are not a part of the function, but act as boundaries for the function, boundaries the function will never touch. They are known as vertical asymptotes. There are infinite vertical asymptotes for all of these functions, but these graphs only show a few. The third graph is of the function “f(x) = csc(x)”. The vertical asymptotes for “f(x) = csc(x)” on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. Between the “x = -2 pi” and “x = -pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (-((3 pi)/2), 1). Between the “x = -pi” and “x = 0” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-(pi/2), -1). Between the “x = 0” and “x = pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point ((pi/2), 1). Between the “x = pi” and “x = 2 pi” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (((3 pi)/2), -1). The fourth graph is of the function “f(x) = sec(x)”. The vertical asymptotes for this function on this graph are at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. Between the “x = -((3 pi)/2)” and “x = -(pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-pi, -1). Between the “x = -(pi/2)” and “x = (pi/2)” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (0, 1). Between the “x = (pi/2)” and “x = (3pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (pi, -1). The graph of sec(x) is the same as the graph of csc(x), except it is shifted to the left by a distance of (pi/2). The fifth graph is of the function “f(x) = tan(x)”. The vertical asymptotes of this function on this graph occur at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. In between all of the vertical asymptotes, the function is always increasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The sixth graph is of the function “f(x) = cot(x)”. The vertical asymptotes of this function on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. In between all of the vertical asymptotes, the function is always decreasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0) and there is no y intercept.
Figure 5. The six trigonometric functions are periodic.

Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the following function:f(x)=A \sin(B(x-\alpha))+C.

In (Figure), the constant \alpha  causes a horizontal or phase shift. The factor B changes the period. This transformed sine function will have a period 2\pi / |B|. The factor A results in a vertical stretch by a factor of |A|. We say |A| is the “amplitude of f.” The constant C causes a vertical shift.

An image of a graph. The graph is of the function “f(x) = Asin(B(x - alpha)) + C”. Along the y axis, there are 3 hash marks: starting from the bottom and moving up, the hash marks are at the values “C - A”, “C”, and “C + A”. The distance from the origin to “C” is labeled “vertical shift”. The distance from “C - A” to “A” and the distance from “A” to “C + A” is “A”, which is labeled “amplitude”. On the x axis is a hash mark at the value “alpha” and the distance between the origin and “alpha” is labeled “horizontal shift”. The distance between two successive minimum values of the function (in other words, the distance between two bottom parts of the wave that are next to each other) is “(2 pi)/(absolute value of B)” is labeled the period. The period is also the distance between two successive maximum values of the function.
Figure 6. A graph of a general sine function.

Notice in (Figure) that the graph of y=\cos x is the graph of y=\sin x shifted to the left \pi /2 units. Therefore, we can write \cos x=\sin(x+\pi /2). Similarly, we can view the graph of y=\sin x as the graph of y=\cos x shifted right \pi /2 units, and state that \sin x=\cos(x-\pi /2).

A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.7 hours and December 21 is the shortest day of the year with 8.3 hours. It can be shown that the function

h(t)=3.7\sin(\frac{2\pi}{365}(t-80.5))+12

is a model for the number of hours of daylight h as a function of day of the year t ((Figure)).

An image of a graph. The x axis runs from 0 to 365 and is labeled “t, day of the year”. The y axis runs from 0 to 20 and is labeled “h, number of daylight hours”. The graph is of the function “h(t) = 3.7sin(((2 pi)/365)(t - 80.5)) + 12”, which is a curved wave function. The function starts at the approximate point (0, 8.4) and begins increasing until the approximate point (171.8, 15.7). After this point, the function decreases until the approximate point (354.3, 8.3). After this point, the function begins increasing again.
Figure 7. The hours of daylight as a function of day of the year can be modeled by a shifted sine curve.

Sketching the Graph of a Transformed Sine Curve

Sketch a graph of f(x)=3\sin(2(x-\frac{\pi}{4}))+1.

Solution

This graph is a phase shift of y=\sin x to the right by \pi /4 units, followed by a horizontal compression by a factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of f is \pi.

An image of a graph. The x axis runs from -((3 pi)/2) to 2 pi and the y axis runs from -3 to 5. The graph is of the function “f(x) = 3sin(2(x-(pi/4))) + 1”, which is a curved wave function. The function starts decreasing from the point (-((3 pi)/2), 4) until it hits the point (-pi, -2). At this point, the function begins increasing until it hits the point (-(pi/2), 4). After this point, the function begins decreasing until it hits the point (0, -2). After this point, the function increases until it hits the point ((pi/2), 4). After this point, the function decreases until it hits the point (pi, -2). After this point, the function increases until it hits the point (((3 pi)/2), 4). After this point, the function decreases again.

Describe the relationship between the graph of f(x)=3\sin(4x)-5 and the graph of y=\sin x.

Solution

To graph f(x)=3\sin(4x)-5, the graph of y=\sin x needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function f will have a period of \pi /2 and an amplitude of 3.

Hint

The graph of f can be sketched using the graph of y=\sin x and a sequence of three transformations.

Key Concepts

  • Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian measure 1. An angle with a degree measure of 180° has a radian measure of \pi  rad.
  • For acute angles \theta, the values of the trigonometric functions are defined as ratios of two sides of a right triangle in which one of the acute angles is \theta.
  • For a general angle \theta, let (x,y) be a point on a circle of radius r corresponding to this angle \theta. The trigonometric functions can be written as ratios involving x, \, y, and r.
  • The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period 2\pi. The tangent and cotangent functions have period \pi.

Key Equations

  • Generalized sine function
    f(x)=A\sin(B(x-\alpha))+C

An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.

Existence of an Inverse Function

We begin with an example. Given a function f and an output y=f(x), we are often interested in finding what value or values x were mapped to y by f. For example, consider the function f(x)=x^3+4. Since any output y=x^3+4, we can solve this equation for x to find that the input is x=\sqrt[3]{y-4}. This equation defines x as a function of y. Denoting this function as f^{-1}, and writing x=f^{-1}(y)=\sqrt[3]{y-4}, we see that for any x in the domain of f, \, f^{-1}(f(x))=f^{-1}(x^3+4)=x. Thus, this new function, f^{-1}, “undid” what the original function f did. A function with this property is called the inverse function of the original function.

Definition

Given a function f with domain D and range R, its inverse function (if it exists) is the function f^{-1} with domain R and range D such that f^{-1}(y)=x if f(x)=y. In other words, for a function f and its inverse f^{-1},f^{-1}(f(x))=x for all x in D, and f(f^{-1}(y))=y for all y in R.

Note that f^{-1} is read as “f inverse.” Here, the -1 is not used as an exponent and f^{-1}(x) \ne 1/f(x)(Figure) shows the relationship between the domain and range of f and the domain and range of f^{-1}.

An image of two bubbles. The first bubble is orange and has two labels: the top label is “Domain of f” and the bottom label is “Range of f inverse”. Within this bubble is the variable “x”. An orange arrow with the label “f” points from this bubble to the second bubble. The second bubble is blue and has two labels: the top label is “range of f” and the bottom label is “domain of f inverse”. Within this bubble is the variable “y”. A blue arrow with the label “f inverse” points from this bubble to the first bubble.
Figure 1. Given a function f and its inverse f^{-1}, \, f^{-1}(y)=x if and only if f(x)=y. The range of f becomes the domain of f^{-1} and the domain of f becomes the range of f^{-1}.

Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for f(x)=x^2. Solving the equation y=x^2 for x, we arrive at the equation x= \pm \sqrt{y}. This equation does not describe x as a function of y because there are two solutions to this equation for every y>0″ height=”16″ width=”42″>. The problem with trying to find an inverse function for <img loading= is that two inputs are sent to the same output for each output y>0″ height=”16″ width=”42″>. The function <img loading= discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a different output is called a one-to-one function.

Definition

We say a f is a one-to-one function if f(x_1) \ne f(x_2) when x_1 \ne x_2.

One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the xy-plane, according to the horizontal line test, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one ((Figure)).

Rule: Horizontal Line Test

A function f is one-to-one if and only if every horizontal line intersects the graph of f no more than once.

An image of two graphs. Both graphs have an x axis that runs from -3 to 3 and a y axis that runs from -3 to 4. The first graph is of the function “f(x) = x squared”, which is a parabola. The function decreases until it hits the origin, where it begins to increase. The x intercept and y intercept are both at the origin. There are two orange horizontal lines also plotted on the graph, both of which run through the function at two points each. The second graph is of the function “f(x) = x cubed”, which is an increasing curved function. The x intercept and y intercept are both at the origin. There are three orange lines also plotted on the graph, each of which only intersects the function at one point.
Figure 2. (a) The function f(x)=x^2 is not one-to-one because it fails the horizontal line test. (b) The function f(x)=x^3 is one-to-one because it passes the horizontal line test.

Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

  1. An image of a graph. The x axis runs from -3 to 11 and the y axis runs from -3 to 11. The graph is of a step function which contains 10 horizontal steps. Each steps starts with a closed circle and ends with an open circle. The first step starts at the origin and ends at the point (1, 0). The second step starts at the point (1, 1) and ends at the point (1, 2). Each of the following 8 steps starts 1 unit higher in the y direction than where the previous step ended. The tenth and final step starts at the point (9, 9) and ends at the point (10, 9)
  2. An image of a graph. The x axis runs from -3 to 6 and the y axis runs from -3 to 6. The graph is of the function “f(x) = (1/x)”, a curved decreasing function. The graph of the function starts right below the x axis in the 4th quadrant and begins to decreases until it comes close to the y axis. The graph keeps decreasing as it gets closer and closer to the y axis, but never touches it due to the vertical asymptote. In the first quadrant, the graph of the function starts close to the y axis and keeps decreasing until it gets close to the x axis. As the function continues to decreases it gets closer and closer to the x axis without touching it, where there is a horizontal asymptote.

Solution

  1. Since the horizontal line y=n for any integer n\ge 0 intersects the graph more than once, this function is not one-to-one.
    An image of a graph. The x axis runs from -3 to 11 and the y axis runs from -3 to 11. The graph is of a step function which contains 10 horizontal steps. Each steps starts with a closed circle and ends with an open circle. The first step starts at the origin and ends at the point (1, 0). The second step starts at the point (1, 1) and ends at the point (1, 2). Each of the following 8 steps starts 1 unit higher in the y direction than where the previous step ended. The tenth and final step starts at the point (9, 9) and ends at the point (10, 9). There are also two horizontal orange lines plotted on the graph, each of which run through an entire step of the function.
  2. Since every horizontal line intersects the graph once (at most), this function is one-to-one.
    An image of a graph. The x axis runs from -3 to 6 and the y axis runs from -3 to 6. The graph is of the function “f(x) = (1/x)”, a curved decreasing function. The graph of the function starts right below the x axis in the 4th quadrant and begins to decreases until it comes close to the y axis. The graph keeps decreasing as it gets closer and closer to the y axis, but never touches it due to the vertical asymptote. In the first quadrant, the graph of the function starts close to the y axis and keeps decreasing until it gets close to the x axis. As the function continues to decreases it gets closer and closer to the x axis without touching it, where there is a horizontal asymptote. There are also three horizontal orange lines plotted on the graph, each of which only runs through the function at one point.

Is the function f graphed in the following image one-to-one?

An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of the function “f(x) = (x cubed) - x” which is a curved function. The function increases, decreases, then increases again. The x intercepts are at the points (-1, 0), (0,0), and (1, 0). The y intercept is at the origin.

Solution

No.

Hint

Use the horizontal line test.

Finding a Function’s Inverse

We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of f to elements in the range of f. The inverse function maps each element from the range of f back to its corresponding element from the domain of f. Therefore, to find the inverse function of a one-to-one function f, given any y in the range of f, we need to determine which x in the domain of f satisfies f(x)=y. Since f is one-to-one, there is exactly one such value x. We can find that value x by solving the equation f(x)=y for x. Doing so, we are able to write x as a function of y where the domain of this function is the range of f and the range of this new function is the domain of f. Consequently, this function is the inverse of f, and we write x=f^{-1}(y). Since we typically use the variable x to denote the independent variable and y to denote the dependent variable, we often interchange the roles of x and y, and write y=f^{-1}(x). Representing the inverse function in this way is also helpful later when we graph a function f and its inverse f^{-1} on the same axes.

Problem-Solving Strategy: Finding an Inverse Function

  1. Solve the equation y=f(x) for x.
  2. Interchange the variables x and y and write y=f^{-1}(x).

Finding an Inverse Function

Find the inverse for the function f(x)=3x-4. State the domain and range of the inverse function. Verify that f^{-1}(f(x))=x.

Solution

Follow the steps outlined in the strategy.

Step 1. If y=3x-4, then 3x=y+4 and x=\frac{1}{3}y+\frac{4}{3}.

Step 2. Rewrite as y=\frac{1}{3}x+\frac{4}{3} and let y=f^{-1}(x).

Therefore, f^{-1}(x)=\frac{1}{3}x+\frac{4}{3}.

Since the domain of f is (−\infty ,\infty), the range of f^{-1} is (−\infty ,\infty). Since the range of f is (−\infty ,\infty), the domain of f^{-1} is (−\infty ,\infty).

You can verify that f^{-1}(f(x))=x by writingf^{-1}(f(x))=f^{-1}(3x-4)=\frac{1}{3}(3x-4)+\frac{4}{3}=x-\frac{4}{3}+\frac{4}{3}=x.

Note that for f^{-1}(x) to be the inverse of f(x), both f^{-1}(f(x))=x and f(f^{-1}(x))=x for all x in the domain of the inside function.

Find the inverse of the function f(x)=3x/(x-2). State the domain and range of the inverse function.

Solution

f^{-1}(x)=\frac{2x}{x-3}. The domain of f^{-1} is \{x|x \ne 3\}. The range of f^{-1} is \{y|y \ne 2\}.

Hint

Use the (Note) for finding inverse functions.

Graphing Inverse Functions

Let’s consider the relationship between the graph of a function f and the graph of its inverse. Consider the graph of f shown in (Figure) and a point (a,b) on the graph. Since b=f(a), then f^{-1}(b)=a. Therefore, when we graph f^{-1}, the point (b,a) is on the graph. As a result, the graph of f^{-1} is a reflection of the graph of f about the line y=x.

An image of two graphs. The first graph is of “y = f(x)”, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs “y = f(x)” with the point (a, b), but also graphs the function “y = f inverse (x)”, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
Figure 3. (a) The graph of this function f shows point (a,b) on the graph of f. (b) Since (a,b) is on the graph of f, the point (b,a) is on the graph of f^{-1}. The graph of f^{-1} is a reflection of the graph of f about the line y=x.

Sketching Graphs of Inverse Functions

For the graph of f in the following image, sketch a graph of f^{-1} by sketching the line y=x and using symmetry. Identify the domain and range of f^{-1}.

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4).

Solution

Reflect the graph about the line y=x. The domain of f^{-1} is [0,\infty). The range of f^{-1} is [-2,\infty). By using the preceding strategy for finding inverse functions, we can verify that the inverse function is f^{-1}(x)=x^2-2, as shown in the graph.

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from -2 to 2. The graph is of two functions. The first function is “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). The second function is “f inverse (x) = (x squared) -2”, an increasing curved function that starts at the point (0, -2). The x intercept is at the approximate point (1.4, 0) and the y intercept is at the point (0, -2). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.

Sketch the graph of f(x)=2x+3 and the graph of its inverse using the symmetry property of inverse functions.

Solution

An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of two functions. The first function is “f(x) = 2x +3”, an increasing straight line function. The function has an x intercept at (-1.5, 0) and a y intercept at (0, 3). The second function is “f inverse (x) = (x - 3)/2”, an increasing straight line function, which increases at a slower rate than the first function. The function has an x intercept at (3, 0) and a y intercept at (0, -1.5). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.

Hint

The graphs are symmetric about the line y=x.

Restricting Domains

As we have seen, f(x)=x^2 does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of f such that the function is one-to-one. This subset is called a restricted domain. By restricting the domain of f, we can define a new function g such that the domain of g is the restricted domain of f and g(x)=f(x) for all x in the domain of g. Then we can define an inverse function for g on that domain. For example, since f(x)=x^2 is one-to-one on the interval [0,\infty), we can define a new function g such that the domain of g is [0,\infty) and g(x)=x^2 for all x in its domain. Since g is a one-to-one function, it has an inverse function, given by the formula g^{-1}(x)=\sqrt{x}. On the other hand, the function f(x)=x^2 is also one-to-one on the domain (−\infty,0]. Therefore, we could also define a new function h such that the domain of h is (−\infty,0] and h(x)=x^2 for all x in the domain of h. Then h is a one-to-one function and must also have an inverse. Its inverse is given by the formula h^{-1}(x)=−\sqrt{x} ((Figure)).

An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is “g(x) = x squared”, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is “g inverse (x) = square root of x”, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is “h(x) = x squared”, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is “h inverse (x) = -(square root of x)”, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
Figure 4. (a) For g(x)=x^2 restricted to [0,\infty), \, g^{-1}(x)=\sqrt{x}. (b) For h(x)=x^2 restricted to (−\infty,0], \, h^{-1}(x)=−\sqrt{x}.

Restricting the Domain

Consider the function f(x)=(x+1)^2.

  1. Sketch the graph of f and use the horizontal line test to show that f is not one-to-one.
  2. Show that f is one-to-one on the restricted domain [-1,\infty). Determine the domain and range for the inverse of f on this restricted domain and find a formula for f^{-1}.

Solution

  1. The graph of f is the graph of y=x^2 shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, f is not one-to-one.
    An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function “f(x) = (x+ 1) squared”, which is a parabola. The function decreases until the point (-1, 0), where it begins it increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1). There is also a horizontal dotted line plotted on the graph, which crosses through the function at two points.
  2. On the interval [-1,\infty), \, f is one-to-one.
    An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function “f(x) = (x+ 1) squared”, on the interval [1, infinity). The function starts from the point (-1, 0) and increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1).
    The domain and range of f^{-1} are given by the range and domain of f, respectively. Therefore, the domain of f^{-1} is [0,\infty) and the range of f^{-1} is [-1,\infty). To find a formula for f^{-1}, solve the equation y=(x+1)^2 for x. If y=(x+1)^2, then x=-1 \pm \sqrt{y}. Since we are restricting the domain to the interval where x \ge -1, we need \pm \sqrt{y} \ge 0. Therefore, x=-1+\sqrt{y}. Interchanging x and y, we write y=-1+\sqrt{x} and conclude that f^{-1}(x)=-1+\sqrt{x}.

Consider f(x)=1/x^2 restricted to the domain (−\infty ,0). Verify that f is one-to-one on this domain. Determine the domain and range of the inverse of f and find a formula for f^{-1}.

Solution

The domain of f^{-1} is (0,\infty). The range of f^{-1} is (−\infty ,0). The inverse function is given by the formula f^{-1}(x)=-1/\sqrt{x}.

Hint

The domain and range of f^{-1} is given by the range and domain of f, respectively. To find f^{-1}, solve y=1/x^2 for x.

Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function ((Figure)). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [-\frac{\pi}{2},\frac{\pi}{2}]. By doing so, we define the inverse sine function on the domain [-1,1] such that for any x in the interval [-1,1], the inverse sine function tells us which angle \theta  in the interval [-\frac{\pi}{2},\frac{\pi}{2}] satisfies  \sin \theta =x. Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Definition

The inverse sine function, denoted  \sin^{-1} or arcsin, and the inverse cosine function, denoted \cos^{-1} or arccos, are defined on the domain D=\{x|-1 \le x \le 1\} as follows:

\begin{array}{c}\sin^{-1}(x)=y \,\, \text{if and only if} \, \sin (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2};\hfill \\ \cos^{-1}(x)=y \,\, \text{if and only if} \, \cos (y)=x \, \text{and} \, 0 \le y \le \pi \hfill \end{array}

The inverse tangent function, denoted \tan^{-1} or arctan, and inverse cotangent function, denoted \cot^{-1} or arccot, are defined on the domain D=\{x|-\infty <x<\infty \} as follows:

\begin{array}{c}\tan^{-1}(x)=y \,\, \text{if and only if} \, \tan (y)=x \, \text{and} \, -\frac{\pi}{2}<y<\frac{\pi}{2};\hfill \\ \cot^{-1}(x)=y \,\, \text{if and only if} \, \cot (y)=x \, \text{and} \, 0<y<\pi \hfill \end{array}

The inverse cosecant function, denoted \csc^{-1} or arccsc, and inverse secant function, denoted \sec^{-1} or arcsec, are defined on the domain D=\{x| \, |x| \ge 1\} as follows:

\begin{array}{c}\csc^{-1}(x)=y \,\, \text{if and only if} \, \csc (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2}, \, y\ne 0;\hfill \\ \sec^{-1}(x)=y \,\, \text{if and only if} \, \sec (y)=x \, \text{and} \, 0 \le y \le \pi, \, y \ne \pi/2\hfill \end{array}

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line y=x ((Figure)).

An image of six graphs. The first graph is of the function “f(x) = sin inverse(x)”, which is an increasing curve function. The function starts at the point (-1, -(pi/2)) and increases until it ends at the point (1, (pi/2)). The x intercept and y intercept are at the origin. The second graph is of the function “f(x) = cos inverse (x)”, which is a decreasing curved function. The function starts at the point (-1, pi) and decreases until it ends at the point (1, 0). The x intercept is at the point (1, 0). The y intercept is at the point (0, (pi/2)). The third graph is of the function f(x) = tan inverse (x)”, which is an increasing curve function. The function starts close to the horizontal line “y = -(pi/2)” and increases until it comes close the “y = (pi/2)”. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The x intercept and y intercept are both at the origin. The fourth graph is of the function “f(x) = cot inverse (x)”, which is a decreasing curved function. The function starts slightly below the horizontal line “y = pi” and decreases until it gets close the x axis. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The fifth graph is of the function “f(x) = csc inverse (x)”, a decreasing curved function. The function starts slightly below the x axis, then decreases until it hits a closed circle point at (-1, -(pi/2)). The function then picks up again at the point (1, (pi/2)), where is begins to decrease and approach the x axis, without ever touching the x axis. There is a horizontal asymptote at the x axis. The sixth graph is of the function “f(x) = sec inverse (x)”, an increasing curved function. The function starts slightly above the horizontal line “y = (pi/2)”, then increases until it hits a closed circle point at (-1, pi). The function then picks up again at the point (1, 0), where is begins to increase and approach the horizontal line “y = (pi/2)”, without ever touching the line. There is a horizontal asymptote at the “y = (pi/2)”.
Figure 5. The graph of each of the inverse trigonometric functions is a reflection about the line y=x of the corresponding restricted trigonometric function.

Go to the following site for more comparisons of functions and their inverses.

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate \cos^{-1}(\frac{1}{2}), we need to find an angle \theta  such that  \cos \theta =\frac{1}{2}. Clearly, many angles have this property. However, given the definition of \cos^{-1}, we need the angle \theta  that not only solves this equation, but also lies in the interval [0,\pi]. We conclude that \cos^{-1}(\frac{1}{2})=\frac{\pi}{3}.

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions  \sin (\sin^{-1}(\frac{\sqrt{2}}{2})) and \sin^{-1}(\sin(\pi)). For the first one, we simplify as follows: \sin (\sin^{-1}(\frac{\sqrt{2}}{2}))= \sin (\frac{\pi}{4})=\frac{\sqrt{2}}{2}.

For the second one, we have\sin^{-1}( \sin (\pi))=\sin^{-1}(0)=0.

The inverse function is supposed to “undo” the original function, so why isn’t \sin^{-1}(\sin (\pi))=\pi ? Recalling our definition of inverse functions, a function f and its inverse f^{-1} satisfy the conditions f(f^{-1}(y))=y for all y in the domain of f^{-1} and f^{-1}(f(x))=x for all x in the domain of f, so what happened here? The issue is that the inverse sine function, \sin^{-1}, is the inverse of the restricted sine function defined on the domain [-\frac{\pi}{2},\frac{\pi}{2}]. Therefore, for x in the interval [-\frac{\pi}{2},\frac{\pi}{2}], it is true that \sin^{-1}(\sin x)=x. However, for values of x outside this interval, the equation does not hold, even though \sin^{-1}(\sin x) is defined for all real numbers x.

What about  \sin (\sin^{-1}y)? Does that have a similar issue? The answer is no. Since the domain of \sin^{-1} is the interval [-1,1], we conclude that  \sin (\sin^{-1}y)=y if -1 \le y \le 1 and the expression is not defined for other values of y. To summarize,

 \sin (\sin^{-1}y)=y \, \text{if} \, -1 \le y \le 1

and\sin^{-1}( \sin x)=x \, \text{if} \, -\frac{\pi}{2} \le x \le \frac{\pi}{2}.

Similarly, for the cosine function,

 \cos (\cos^{-1}y)=y \, \text{if} \, -1 \le y \le 1

and\cos^{-1}( \cos x)=x \, \text{if} 0 \le x \le \pi.

Similar properties hold for the other trigonometric functions and their inverses.

Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

  1. \sin^{-1}(-\frac{\sqrt{3}}{2})
  2.  \tan (\tan^{-1}(-\frac{1}{\sqrt{3}}))
  3. \cos^{-1}( \cos (\frac{5\pi}{4}))
  4. \sin^{-1}( \cos (\frac{2\pi}{3}))

Solution

  1. Evaluating \sin^{-1}(−\sqrt{3}/2) is equivalent to finding the angle \theta  such that  \sin \theta =−\sqrt{3}/2 and −\pi/2 \le \theta \le \pi/2. The angle \theta =−\pi/3 satisfies these two conditions. Therefore, \sin^{-1}(−\sqrt{3}/2)=−\pi/3.
  2. First we use the fact that \tan^{-1}(-1/\sqrt{3})=−\pi/6. Then  \tan (\pi/6)=-1/\sqrt{3}. Therefore,  \tan (\tan^{-1}(-1/\sqrt{3}))=-1/\sqrt{3}.
  3. To evaluate \cos^{-1}( \cos (5\pi/4)), first use the fact that  \cos (5\pi/4)=−\sqrt{2}/2. Then we need to find the angle \theta  such that  \cos (\theta )=−\sqrt{2}/2 and 0 \le \theta \le \pi. Since 3\pi/4 satisfies both these conditions, we have  \cos (\cos^{-1}(5\pi/4))= \cos (\cos^{-1}(−\sqrt{2}/2))=3\pi/4.
  4. Since  \cos (2\pi/3)=-1/2, we need to evaluate \sin^{-1}(-1/2). That is, we need to find the angle \theta  such that  \sin (\theta )=-1/2 and −\pi/2 \le \theta \le \pi/2. Since −\pi/6 satisfies both these conditions, we can conclude that \sin^{-1}( \cos (2\pi/3))=\sin^{-1}(-1/2)=−\pi/6.

The Maximum Value of a Function

In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.

This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable x.

  1. Consider the graph in (Figure) of the function y= \sin x + \cos x. Describe its overall shape. Is it periodic? How do you know?An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of the function “y = sin(x) + cos(x)”, a curved wave function. The graph of the function decreases until it hits the approximate point (-(3pi/4), -1.4), where it increases until the approximate point ((pi/4), 1.4), where it begins to decrease again. The x intercepts shown on this graph of the function are at (-(5pi/4), 0), (-(pi/4), 0), and ((3pi/4), 0). The y intercept is at (0, 1).Figure 6. The graph of y= \sin x + \cos x.Using a graphing calculator or other graphing device, estimate the x– and y-values of the maximum point for the graph (the first such point where x>0″ height=”12″ width=”43″>). It may be helpful to express the <img loading=-value as a multiple of \pi.
  2. Now consider other graphs of the form y=A \sin x + B \cos x for various values of A and B. Sketch the graph when A = 2 and B = 1, and find the x– and y-values for the maximum point. (Remember to express the x-value as a multiple of \pi, if possible.) Has it moved?
  3. Repeat for A = 1, \, B = 2. Is there any relationship to what you found in part (2)?
  4. Complete the following table, adding a few choices of your own for A and B:
    ABxyABxy01\sqrt{3}1101\sqrt{3}111251251221223443
  5. Try to figure out the formula for the y-values.
  6. The formula for the x-values is a little harder. The most helpful points from the table are (1,1), \, (1,\sqrt{3}), \, (\sqrt{3},1). (HintConsider inverse trigonometric functions.)
  7. If you found formulas for parts (5) and (6), show that they work together. That is, substitute the x-value formula you found into y=A \sin x + B \cos x and simplify it to arrive at the y-value formula you found.

Key Concepts

  • For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
  • If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
  • For a function f and its inverse f^{-1}, \, f(f^{-1}(x))=x for all x in the domain of f^{-1} and f^{-1}(f(x))=x for all x in the domain of f.
  • Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
  • The graph of a function f and its inverse f^{-1} are symmetric about the line y=x.

Key Equations

  • Inverse functions
    f^{-1}(f(x))=x for all x in D, and f(f^{-1}(y))=y for all y in R.[/latex]