Chapter 2: Graphs and Inverse Functions

Author

Pablo Chalmeta, New River Community College

2.1 Graphs of Sine and Cosine

Basic Sine and Cosine Graphs

We can graph trigonometric functions the same as we can graph any other function. We will graph the trigonometric functions on the \(xy\)-plane and the \(x\) coordinate will always be in radians. We will demonstrate two ways to look at the graph of \(y= \sin x\). First we will plot points by selecting angle values for \(x\) and calculating the \(y\) values. Second we will use the unit circle.

The following table (Table 2.1) is a list of common angles and their trigonometric function values.

Table 1: Table 2.1 Table of Common Trigonometric Function Values
\(\theta\) (radians) \(y = \sin \theta\) \(y = \cos \theta\) \(y = \tan \theta\)
\(0\) \(0\) \(1\) \(0\)
\(\frac{\pi}{6}\) \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{3}}\)
\(\frac{\pi}{4}\) \(\frac{1}{\sqrt{2}}\) \(\frac{1}{\sqrt{2}}\) \(1\)
\(\frac{\pi}{3}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(\sqrt{3}\)
\(\frac{\pi}{2}\) \(1\) \(0\) undefined
\(\frac{2\pi}{3}\) \(\frac{\sqrt{3}}{2}\) \(-\frac{1}{2}\) \(-\sqrt{3}\)
\(\frac{3\pi}{4}\) \(\frac{1}{\sqrt{2}}\) \(-\frac{1}{\sqrt{2}}\) \(-1\)
\(\frac{5\pi}{6}\) \(\frac{1}{2}\) \(-\frac{\sqrt{3}}{2}\) \(-\frac{1}{\sqrt{3}}\)
\(\pi\) \(0\) \(-1\) \(0\)
\(\frac{7\pi}{6}\) \(-\frac{1}{2}\) \(-\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{3}}\)
\(\frac{5\pi}{4}\) \(-\frac{1}{\sqrt{2}}\) \(-\frac{1}{\sqrt{2}}\) \(1\)
\(\frac{4\pi}{3}\) \(-\frac{\sqrt{3}}{2}\) \(-\frac{1}{2}\) \(\sqrt{3}\)
\(\frac{3\pi}{2}\) \(-1\) \(0\) undefined
\(\frac{5\pi}{3}\) \(-\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(-\sqrt{3}\)
\(\frac{7\pi}{4}\) \(-\frac{1}{\sqrt{2}}\) \(\frac{1}{\sqrt{2}}\) \(-1\)
\(\frac{11\pi}{6}\) \(-\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(-\frac{1}{\sqrt{3}}\)

Using the numbers in Table 2.1 we can plot the sine function from \(0 \leq x \leq 2\pi\). In Figure 2.1 the points are indicated on the graph and some have been labeled. We saw in Section 1.4 that the trigonometric functions are periodic. This means that the values repeat at regular intervals. The sine repeats every \(2\pi\) radians so this graph repeats forever in both directions as seen in Figure 2.3.

Graph of y equals sine x from 0 to 2 pi. The curve starts at origin, rises to maximum of 1 at pi over 2, returns to 0 at pi, reaches minimum of negative 1 at 3 pi over 2, and returns to 0 at 2 pi. Key points are labeled on the curve.
Figure 1

Figure 2.1: Graph of \(y=\sin x\) for \(0 \leq x \leq 2\pi\)

Another way to consider the graph of the sine is to remember that every point on the unit circle (circle of radius 1) is \((x, y) = (\cos \theta, \sin \theta)\) on the terminal side of \(\theta\). Here you can see how for each angle, we use the \(y\) value of the point on the circle to determine the output value of the sine function. The correspondence is shown in Figure 2.2.

Two panels showing the unit circle on the left and the sine graph on the right. Arrows connect points on the unit circle to corresponding points on the sine wave, illustrating how the y-coordinate of the circle produces the height of the sine curve.
Figure 2

Figure 2.2: Graph of sine function based on \(y\)-coordinate of points on unit circle

It is most common to use the variable \(x\) and \(y\) to represent the horizontal and vertical axes so we will relabel the axes when we draw the graphs of the trigonometric functions from now on. In our graph in {} we have plotted both positive an negative angles. You will notice that if you pick any starting \(x\) value and move \(2\pi\) units in either direction the values of the function are the same because the period of the sine function in \(2\pi\).

WARNING: Be careful because we reuse variables. \(x\) and \(y\) are used to represent the cosine and sine on the unit circle but here \(x\) is the angle and \(y\) is the trigonometric value of that angle.

Complete graph of y equals sine x showing multiple periods extending from negative 2 pi to 2 pi. The wave oscillates between negative 1 and 1 with period 2 pi.
Figure 3

Figure 2.3: Graph of \(y = \sin x\)

Similarly we can construct a graph for the cosine function. Note that the cosine function has the same shape as the sine function but it is shifted \(\dfrac{\pi}{2}\) units to the left. From algebra you may recall that a \(\dfrac{\pi}{2}\) shift to the left can be represented \(f\left(x + \frac{\pi}{2}\right) = \sin \left(x + \frac{\pi}{2}\right) = \cos(x)\). This is the same cofunction identity presented in Section 1.4.

Both the sine and cosine functions alternate between \(+1\) and \(-1\) passing through zero at regular points. When we label the axes of the graphs we want to make sure we label the angles where the functions are \(0\), \(1\) or \(-1\) on the \(x\)-axis and the values for the maximum, minimum and center line for the \(y\)-axis. You can certainly include more labels but this would generally be the minimum amount of information for a graph. Notice that all the multiples of \(\frac{\pi}{2}\) have been labeled on the graphs in Figures 2.3 and Figure 2.4.

Complete graph of y equals cosine x showing multiple periods. The curve starts at 1 when x equals 0, descends to negative 1 at pi, and returns to 1 at 2 pi. The wave repeats with period 2 pi.
Figure 4

Figure 2.4: Graph of \(y = \cos x\)

Algebraic Transformations

The graphs can be altered by standard algebraic transformations.A function may be stretched or compressed vertically by multiplying by a number.

NoteStretching the function \(f(x)\) vertically

\(h(x) = A \cdot f(x)\) stretches or compresses \(f(x)\) vertically by a factor of \(A\).

If \(A>1\) the function is stretched vertically and if \(0 < A < 1\) the function is compressed vertically.

In the case of the sine and cosine this has the effect of making the {} of the function larger or smaller. The amplitude of the function is the distance from the center line to the maximum height. It can be calculated using the formula:

\[\text{amplitude of } f(x) = \frac{(\text{maximum of } f(x)) - (\text{minimum of } f(x)) }{2}\]

Since \(-1 \leq \sin x \leq 1\) and \(-1 \leq \cos x \leq 1\) then for any \(A>0\)

\[-A \leq A\sin x \leq A ~~ \text{and} ~~ -A \leq A\cos x \leq A\]

Notice that the \(x\)-axis is labeled at the maximums, minimums and zeros of the function in Figure 2.5.

Diagram showing sine wave with amplitude A marked as the distance from midline to maximum, period marked as the distance for one complete cycle, and vertical shift D marked as the distance the midline is shifted from the x-axis.
Figure 5

Figure 2.5: \(y = 2\cos x\)

TipExample 2.1.1

Sketch the graph of \(y = 2\cos x\) for two complete periods.

Solution: The coefficient 2 in front of the cosine means the amplitude is 2. The graph will oscillate between -2 and 2 instead of between -1 and 1. The period remains \(2\pi\).

Graph of y equals 2 cosine x showing two complete periods. The curve oscillates between 2 and negative 2 with period 2 pi.
Figure 6

Figure 2.6: \(y = 2\cos x\)

A function may be shifted up or down by adding or subtracting a number on the outside. This is called a vertical shift.

NoteShifting the function \(f(x)\) vertically

\(h(x) = f(x) + D\) moves \(f(x)\) up “\(D\)” units.

\(h(x) = f(x) - D\) moves \(f(x)\) down “\(D\)” units.

TipExample 2.1.2

Sketch the graph of \(y = 2\cos x + 3\) for two complete periods.

Solution: The amplitude is still 2, but now the entire graph is shifted up 3 units. The midline of the function is now \(y = 3\) instead of \(y = 0\). The function oscillates between 1 and 5.

Graph of y equals 2 cosine x plus 3 showing two complete periods. The curve oscillates between 5 and 1 with midline at y equals 3.
Figure 7

Figure 2.7: \(y = 2\cos x + 3\)

A function may be stretched or compressed horizontally by multiplying the variable by a number.

NoteStretching the function \(f(x)\) horizontally

\(h(x) = f(B \cdot x)\) stretches or compresses \(f(x)\) horizontally by a factor of \(\frac{1}{B}\).

If \(B>1\) the function is compressed horizontally and if \(0 < B < 1\) the function is stretched horizontally.

In the case of the sine and cosine multiplying the variable by a number \(B\) changes the period. The period of \(y=\sin(Bx)\) and of \(y = \cos(Bx)\) is

\[\text{period of } y=\sin(Bx) \text{ is } \frac{2\pi}{B}\]

\[\text{period of } y=\cos(Bx) \text{ is } \frac{2\pi}{B}\]

TipExample 2.1.3

Sketch the graph of \(y = \cos (2x)\) and \(y = \cos x\) on the same set of axes.

Solution: Since we have a \(2x\) inside the cosine it goes around the circle twice as fast which is why in the space of \(2\pi\) the graph will repeat twice. We will graph both \(y=\cos x\) and \(y=\cos (2x)\) on the same set of axes.

Two cosine curves on the same axes. The curve y equals cosine x completes one period from 0 to 2 pi. The curve y equals cosine 2x completes two periods in the same interval, with period pi.
Figure 8

Figure 2.8: \(y = \cos (2x)\) and \(y = \cos x\)

A function may be reflected across the \(x\)-axis by multiplying by \((-1)\). (Making it negative.)

NoteReflecting the function \(f(x)\) over the \(x\)-axis

\(h(x) = -f(x)\) reflects \(f(x)\) across the \(x\)-axis.

TipExample 2.1.4

Sketch the graphs of \(y=-\cos\left(\frac{x}{2}\right)\) and \(y=-\cos\left(\frac{x}{2}\right)+3\) on the same set of axes. Draw two complete periods for each function.

Solution: Here we will have to adjust the period using the period formula period \(= \frac{2\pi}{B}\). Since we have \(\cos\left(\frac{x}{2}\right) =\cos\left(\frac{1}{2} x\right)\) we can see that \(B= \frac{1}{2}\) and the period is \(= \frac{2\pi}{1/2} = 4\pi\). The function will repeat every \(4\pi\) units. Since the horizontal axis is divided into 4 pieces for each period those divisions are all of size \(\pi\). See Figure 2.9.

Two curves shown. First curve y equals negative cosine of x over 2 is reflected across x-axis with period 4 pi. Second curve is the first shifted up 3 units with midline at y equals 3.
Figure 9

Figure 2.9: \(y=-\cos\left(\frac{x}{2}\right)\) and \(y=-\cos\left(\frac{x}{2}\right)+3\)

So far we’ve looked at amplitude, period, and vertical shifts. There is one more transformation we need to consider: the phase shift or horizontal shift. This occurs when we add or subtract a constant inside the function argument.

NoteShifting the function \(f(x)\) left and right

\(h(x) = f(x + C)\) shifts \(f(x)\) to the left by \(C\) units. \(h(x) = f(x - C)\) shifts \(f(x)\) to the right by \(C\) units.

TipExample 2.1.5

Graph \(y = 3\cos \left(2x - \pi \right)\) for two complete cycles.

Solution: Since we have added \(\frac{\pi}{4}\) inside the function the graph will be the same as the graph of \(y = \sin x\) but shifted to the left \(\frac{\pi}{4}\) units. Rather than having zeros at \(0\), \(\pm \pi\) and \(\pm 2\pi\) the zeros are now at \(-\frac{5\pi}{4}\), \(-\frac{\pi}{4}\), \(\frac{3\pi}{4}\) and \(\frac{7\pi}{4}\). The graphs of both \(y = \sin \left(x + \frac{\pi}{4}\right)\) and \(y = \sin x\) are presented in Figure 2.10.

Two curves shown. First curve y equals sine of x. Second curve is the first shifted left by pi over 4.
Figure 10
TipExample 2.1.6

Graph \(y = 3\cos \left(2x - \pi \right)\) for two complete cycles.

Solution: Here we have to be careful because there are three of our transformations in the same problem. First we need to identify the amplitude. That is given to us by the number multiplied in front of the function so \(A = 3\). The period is determined by the number multiplied by the \(x\), in this case \(B = 2\). The period of the function is \(\frac{2\pi}{B} = \pi\).

The phase shift is a bit more difficult because our original definition of phase shift was written as \(f(x+C)\) but we don’t have that, we have \(f(2x+\phi)\). That \(2\) multiplied by the \(x\) is going to influence our shift. We have to write the function as \(f(2(x+C))\) to find the correct value of the phase shift. To see why this is true let’s consider that the cosine function goes through an entire cycle when its angle goes from \(0\) to \(2\pi\). In this case our angle is represented by \(2x - \pi\) so that cycle starts when \[2x - \pi = 0 ~~\implies~~ x = \frac{\pi}{2}\] and ends when \[2x - \pi = 2\pi ~~\implies~~ x = \frac{2\pi}{2} +\frac{\pi}{2} = \pi + \frac{\pi}{2}\] Our phase shift is \(\frac{\pi}{2}\) and the period is \(\pi\) which is exactly what we see when we write the function as \(y = 3\cos \left[2\left(x - \frac{\pi}{2} \right)\right]\).

Graph of y equals 3 cosine of 2x minus pi. The curve shows two complete cycles from x equals 0 to x equals 2 pi. The amplitude is 3, with maximum values at 3 and minimum values at negative 3. The first maximum occurs at x equals pi over 2, showing a phase shift of pi over 2 to the right. The period is pi, so the curve completes one full cycle from x equals pi over 2 to x equals 3 pi over 2, and another cycle from x equals 3 pi over 2 to x equals 5 pi over 2.

Figure 2.11: \(y = 3\cos \left(2x - \pi \right)\)

NoteSummary of trigonometric transformations for sine and cosine

Given the functions \[y = A\sin \left(Bx + C \right) + D= A\sin \left(B\left(x + \frac{C}{B}\right) \right) + D\] or \[y = A\cos \left(Bx + C \right) + D= A\cos \left(B\left(x + \frac{C}{B}\right) \right) + D\] the following transformations occur:

  1. The amplitude of the function is \(|A|\).
  2. The period of the function is \(\dfrac{2\pi}{B}\)
  3. The phase shift of the function is \(-\dfrac{C}{B}\). The shift is to the left for \(\left(x + \frac{C}{B}\right)\) and to the right for \(\left(x - \frac{C}{B}\right)\)
  4. The vertical shift is \(D\)

A negative sign in front of the function will reflect it over the \(x\)-axis.

TipExample 2.1.7

Find the amplitude, period and phase shift of \(y = -2\sin\left( 3x + \frac{\pi}{2} \right)\)

Solution: The amplitude is \(2\), the period is \(\frac{2\pi}{3}\), and the phase shift is \(-\frac{\frac{\pi}{2}}{3} = -\frac{\pi}{6}\). Since the phase shift is negative we move the graph to the left. Or if you write the function as \[y = -2\sin\left( 3\left(x + \frac{\pi}{6}\right) \right)\] we are adding \(\frac{\pi}{6}\) inside the sine function which is a shift to the left. Also note the negative in front of the sine, this reflects the graph over the \(x\)-axis.

Graph of y equals negative 2 sine of 3x plus pi over 2. The curve shows multiple complete cycles. The amplitude is 2, with the curve oscillating between 2 and negative 2. The graph is reflected over the x-axis due to the negative sign in front. The period is 2 pi over 3. The phase shift of negative pi over 6 (or pi over 6 to the left) means the curve starts its cycle earlier than the standard sine function. The curve crosses the x-axis at x equals negative pi over 6, then reaches a minimum, crosses again, reaches a maximum, and repeats.

Figure 2.12: \(y = -2\sin\left( 3x + \frac{\pi}{2} \right) = -2\sin \left[ 3 \left( x + \frac{\pi}{6} \right)\right]\)

2.1 Exercises

For Exercises 1-12, determine the amplitude, period, vertical shift, horizontal shift, and draw the graph of the given function for two complete periods.

  1. \(y=3 \sin x\)

  2. \(f(x)=-3\sin x\)

  3. \(y=-3\sin(2x)\)

  4. \(f(x) = -3 \sin(2x)+4\)

  5. \(y = \dfrac{\cos x}{4}\)

  6. \(y=\cos\left(\frac{x}{4}\right)\)

  7. \(f(x) = \frac{1}{2}\cos x -4\)

  8. \(y = 2\cos\left(x - \frac{\pi}{4}\right)\)

  9. \(g(x) =-3+ 2\cos\left(x - \frac{\pi}{4}\right)\)

  10. \(y=2\sin\left(2x + \frac{\pi}{2}\right)\)

  11. \(y=\frac{1}{2}\sin\left(2x + \frac{\pi}{2}\right)+1\)

  12. \(y=3\sin \frac{\pi t}{3}\)

For Exercises 13-14, sketch \(f(x)\) and \(g(x)\) on the same set of axes for \(0\leq x \leq 2\pi\).

  1. \(f(x) = 2\sin x\), \(g(x)=\sin(2x)\)

  2. \(f(x) = 3\cos (2x)\), \(g(x)=3\cos (2x)-2\)

For Exercises 15-19, determine the amplitude, period and vertical shift, then find a formula for the function.

Graph of a periodic function that resembles a sine or cosine curve. The curve oscillates between a maximum value of 3 and a minimum value of negative 1, giving an amplitude of 2 and a vertical shift of 1 (midline at y equals 1). The curve appears to complete one full cycle over an interval of 2 pi, indicating a period of 2 pi. The curve starts at the midline at x equals 0, rises to a maximum, returns to the midline at x equals pi, reaches a minimum, and returns to the midline at x equals 2 pi.

Graph of a periodic function that resembles a sine or cosine curve. The curve oscillates between a maximum value of 2 and a minimum value of negative 2, giving an amplitude of 2 and no vertical shift (midline at y equals 0). The curve appears to complete one full cycle over an interval of pi, indicating a period of pi. The curve starts at a maximum value of 2 at x equals 0, decreases to the midline, reaches a minimum of negative 2, and returns to the midline.

Graph of a periodic function that resembles a sine or cosine curve. The curve oscillates between a maximum value of 4 and a minimum value of 0, giving an amplitude of 2 and a vertical shift of 2 (midline at y equals 2). The curve appears to complete one full cycle over an interval of 2 pi, indicating a period of 2 pi. The curve starts at the midline at x equals 0, decreases to a minimum of 0 at x equals pi over 2, returns to the midline at x equals pi, rises to a maximum of 4 at x equals 3 pi over 2, and returns to the midline at x equals 2 pi.

Graph of a periodic function that resembles a sine or cosine curve. The curve oscillates between a maximum value of approximately 1.5 and a minimum value of approximately negative 1.5, giving an amplitude of approximately 1.5 and no vertical shift (midline at y equals 0). The curve appears to complete one full cycle over an interval of 4 pi, indicating a period of 4 pi. The curve starts at the maximum value at x equals 0, decreases through the midline, reaches a minimum, and returns through the midline to the maximum.

Graph of a periodic function that resembles a sine or cosine curve. The curve oscillates between a maximum value of 4 and a minimum value of negative 4, giving an amplitude of 4 and no vertical shift (midline at y equals 0). The curve appears to complete one full cycle over an interval of 4 pi, indicating a period of 4 pi. The curve starts at y equals 0 at x equals 0, rises to a maximum of 4 at x equals pi, returns to 0 at x equals 2 pi, reaches a minimum of negative 4 at x equals 3 pi, and returns to 0 at x equals 4 pi.

  1. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 50 degrees at midnight and the high and low temperature during the day are 57 and 43 degrees, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t.

  2. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68 degrees at midnight and the high and low temperature during the day are 80 and 56 degrees, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t.

  3. Consider the device shown in Figure 2.13 for converting rotary motion to linear motion (and vice versa). A nail on the edge of the wheel moves the arm back and forth. Relative to the coordinates shown, derive an expression for the position of point \(P\) as a function of the wheel radius \(R\), the bar length \(L\), and the angle \(\theta\).

Mechanical diagram showing a linear motion device. A circular wheel of radius R is positioned on the left side with its center at the origin of a coordinate system. An angle theta is marked from the positive x-axis to a radius of the wheel. At the edge of the wheel (at angle theta), a point is connected to a horizontal bar of length L that extends to the right. Point P is located at the right end of the bar on the positive x-axis. The bar can slide horizontally as the wheel rotates, converting the rotary motion of the wheel into linear motion of point P.

Figure 2.13: Linear motion device

2.2 Graphs of tan(x), cot(x), csc(x) and sec(x)

Tangent and Cotangent Graphs

The graph of the tangent can be constructed by plotting points from Table 2.1 or by using the identity \(\tan x = \dfrac{\sin x}{\cos x}\). On the graph of the tangent notice that there are vertical asymptotes at multiples of \(\dfrac{\pi}{2}\). This is because \(\tan x = \dfrac{\sin x}{\cos x}\) and everywhere cosine is zero tangent is undefined. You can see from the cosine graph that it has zeros at \(x = \dfrac{\pi}{2} + n\pi\) where \(n \in \mathbb{Z}\). Also note that the period of the tangent function is \(\pi\). The graph repeats every \(\pi\) units, it is identical between any two asymptotes.

Graph of y equals tangent x showing multiple periods. The curve has vertical asymptotes at x equals negative 3 pi over 2, negative pi over 2, pi over 2, 3 pi over 2, and so on. Between each pair of consecutive asymptotes, the tangent curve increases from negative infinity to positive infinity, passing through zero at multiples of pi. The period is pi, meaning the pattern repeats every pi units. The curve passes through the origin with a slope of 1.

Figure 2.14: Graph of \(y=\tan x\)

We can perform similar transformations to what was done for the sine and cosine graphs. Those transformations are summarized here:

NoteSummary of trigonometric transformations for tangent

Given the function \[y = A\tan \left(Bx + C \right) + D\] the following transformations occur:

  1. The amplitude of the function is undefined.
  2. The period of the function is \(\dfrac{\pi}{B}\)
  3. The phase shift of the function is \(\dfrac{C}{B}\).
  4. The vertical shift is \(D\)

A negative sign in front of the function will reflect it over the \(x\)-axis.

TipExample 2.2.1

Find the amplitude, period, phase shift, and vertical shift for the function \(y=\frac{1}{2}\tan \left(2x \right)-3\)

Solution: The amplitude is undefined, the period is \(\frac{\pi}{2}\), there is no phase shift, and the vertical shift is down 3 units.

Graph of y equals one half tangent of 2x minus 3. The curve shows multiple periods of a transformed tangent function. Vertical asymptotes occur at x equals negative 3 pi over 4, negative pi over 4, pi over 4, 3 pi over 4, and so on, spaced pi over 2 apart (the period). The entire graph is shifted down 3 units, so the curves pass through y equals negative 3 at multiples of pi over 2. Between asymptotes, the curve increases from negative infinity to positive infinity, but more gradually than the standard tangent due to the coefficient of one half.

Figure 2.15: \(y=\frac{1}{2}\tan \left(2x \right)-3\)

The graph of the cotangent Figure 2.16 can be constructed by using the identity \(\cot x = \dfrac{\cos x}{\sin x}\) or by using the relation \(\cot x = -\tan \left(x+\frac{\pi}{2}\right)\). On the graph of the cotangent notice that there are vertical asymptotes at multiples of \(\pi\). This is because \(\cot x = \dfrac{\cos x}{\sin x}\) and everywhere sine is zero the cotangent is undefined. \(y = \sin x\) has zeros at \(x = \pi + n\pi\) where \(n \in \mathbb{Z}\) so \(y = \cot x\) has vertical asymptotes at \(x = \pi + n\pi\). Also note that the period of the cotangent function is \(\pi\). The graph repeats every \(\pi\) units, it is identical between any two asymptotes.

Graph of y equals cotangent x showing multiple periods. The curve has vertical asymptotes at x equals negative 2 pi, negative pi, 0, pi, 2 pi, and so on, at multiples of pi. Between each pair of consecutive asymptotes, the cotangent curve decreases from positive infinity to negative infinity. The period is pi. At x equals pi over 2 plus n pi (where n is an integer), the cotangent passes through zero. The curve is decreasing throughout each period, in contrast to tangent which is increasing.

Figure 2.16: Graph of \(y=\cot x\)

Cosecant and Secant Graphs

The graph of the cosecant can be constructed by using the identity \(\csc x = \dfrac{1}{\sin x}\). On the graph of the cosecant notice that there are vertical asymptotes at multiples of \(\pi\). This is because \(\csc x = \dfrac{1}{\sin x}\) and everywhere sine is zero the cosecant is undefined. The period of the cosecant function is \(2\pi\) which is the same as the sine function. The graph repeats every \(2\pi\) units. Figure 2.17 shows the graph of \(y = \csc x\), with the graph of \(y = \sin x\) (the dashed curve) for reference.

Graph of y equals cosecant x in blue with y equals sine x shown as a dashed reference curve. The cosecant has vertical asymptotes at x equals negative 2 pi, negative pi, 0, pi, 2 pi, and so on, wherever sine equals zero. Between asymptotes, the cosecant curve has U-shaped or inverted U-shaped branches. Where sine reaches a maximum of 1, cosecant also equals 1 at that point. Where sine reaches a minimum of negative 1, cosecant also equals negative 1. The cosecant curves approach positive or negative infinity near the asymptotes. The period is 2 pi.

Figure 2.17: Graph of \(y=\csc x\) in blue and \(y=\sin x\) (dashed line)

The graph of the secant can be constructed by using the identity \(\sec x = \dfrac{1}{\cos x}\). On the graph of the secant notice that there are vertical asymptotes at multiples of \(\dfrac{\pi}{2}\) because the graph of \(y = \cos x\) has zeros at \(x = \dfrac{\pi}{2} + n\pi\) where \(n \in \mathbb{Z}\). The period of the secant function is \(2\pi\) which is the same as the cosine function. The graph repeats every \(2\pi\) units. Figure 2.18 shows the graph of \(y = \sec x\), with the graph of \(y = \cos x\) (the dashed curve) for reference.

Graph of y equals secant x with y equals cosine x shown as a dashed reference curve. The secant has vertical asymptotes at x equals negative 3 pi over 2, negative pi over 2, pi over 2, 3 pi over 2, and so on, wherever cosine equals zero. Between asymptotes, the secant curve has U-shaped or inverted U-shaped branches. Where cosine reaches a maximum of 1, secant also equals 1 at that point (at x equals 0, plus or minus 2 pi, etc.). Where cosine reaches a minimum of negative 1, secant also equals negative 1 (at x equals plus or minus pi, etc.). The secant curves approach positive or negative infinity near the asymptotes. The period is 2 pi.

Figure 2.18: Graph of \(y=\sec x\)

All the same transformations that were done to the sine, cosine and tangent can be done to the other functions.

NoteSummary of trigonometric transformations for cosecant, secant and cotangent

\(y = A\csc \left(Bx + C \right)\) has undefined amplitude, period \(\dfrac{2\pi}{B}\) and phase shift \(\dfrac{C}{B}\)

\(y = A\sec \left(Bx + C \right)\) has undefined amplitude, period \(\dfrac{2\pi}{B}\) and phase shift \(\dfrac{C}{B}\)

\(y = A\cot \left(Bx + C \right)\) has undefined amplitude, period \(\dfrac{\pi}{B}\) and phase shift \(\dfrac{C}{B}\)

A negative sign in front of the function will reflect it over the \(x\)-axis.

2.2 Exercises

For Exercises 1-9, determine the amplitude, period, vertical shift, horizontal shift, and draw the graph of the given function for two complete periods.

  1. \(y=3 \tan x\)

  2. \(f(x)=-3\csc x\)

  3. \(y=-3\sec(2x)\)

  4. \(f(x) = -3 \sec(\pi x)\)

  5. \(y = \dfrac{\cot x}{4}\)

  6. \(y=\cot\left(\frac{x}{4}\right)\)

  7. \(y=\tan\left(x + \frac{\pi}{4}\right)\)

  8. \(y=\frac{1}{2}\cot\left(x - \frac{\pi}{4}\right)\)

  9. \(y=\sec (t)+2\)

2.3 Inverse Trigonometric Functions

Review of Functions and Inverse Functions

NoteDefinition 2.1

A function is a rule that establishes a correspondence between two sets of elements (called the domain and range) so that for every element in the domain there corresponds EXACTLY ONE element in the range.

Often the domain is \(x\) and the range is \(y\) but any symbols can be used. With trigonometric functions frequently \(\theta\) or another Greek letter is used for the domain. For a function we can have repeated range elements but all the domain elements are unique. For example with \(f(x) = x^2\) both \(x=2\) and \(x=-2\) are mapped to \(y=4\) when put into the function.

There is a special type of function known as a one-to-one (sometimes written \(1-1\)) function where all the range values are unique as well. In other words if \(x_1 \neq x_2\) then \(f(x_1) \neq f(x_2)\). The example above of \(f(x) =x^2\) is not a \(1-1\) function because two different \(x\) values give the same \(y\) value. Much like there was a vertical line test for functions we have a horizontal line test for \(1-1\) functions.

The vertical line test says that \(f(x)\) is a function if and only if every vertical line intersects the graph of \(f(x)\) at most once. Similarly the horizontal line tests says that a function \(f(x)\) is \(1-1\) if every horizontal line intersects the graph at most once.

This idea of a \(1-1\) function is important when discussing inverse functions. An inverse function is a function \(f^{-1}(x)\) such that \[f(f^{-1}(x)) = x ~ \mbox{and}~ f^{-1}(f(x)) = x.\] In other words if \(f\) is a function that takes \(x\) to \(y\) then the inverse function \(f^{-1}\) takes \(y\) back to \(x\). We need the original function to be \(1-1\) because when we reverse the operation we want to make sure we get a unique answer. In the \(f(x) = x^2\) example we can’t have an inverse function because reversing the operation results in two \(x\)-values because \(f(2) = f(-2) = 4\).

None of the trigonometric functions are \(1-1\). Consider the sine function \(y = \sin x\). There are an infinite number of \(x\)-values that will produce every \(y\) value since the sine repeats every \(2\pi\) radians. If we want to reverse the operation of the sine function with an inverse sine function we will have to restrict the domain so that the original sine produces one set of range values. We will make sure that this restriction includes the angle zero. In Figure 2.19 the extended dotted line is to show that the sine function would fail the horizontal line test and sine is \(1-1\) on the domain \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\). It also shows that we have one complete set of range values (\(-1 \leq \sin x \leq 1\)) for the sine function.

Graph of y equals sine x with a restricted domain highlighted. The full sine curve is shown as a dotted line extending beyond the restricted region. The solid portion of the curve shows the restricted domain from x equals negative pi over 2 to x equals pi over 2. This portion passes through the origin, increases from a minimum of negative 1 at x equals negative pi over 2 to a maximum of 1 at x equals pi over 2. A horizontal line is shown at y equals 0.5 to demonstrate that within this restricted domain, the horizontal line intersects the curve only once, showing the function is one-to-one on this interval.

Figure 2.19: Restricted domain for sine

We will do the same for the cosine and tangent. Figure 2.20 shows the domain restrictions.

Graph of y equals cosine x with a restricted domain highlighted. The full cosine curve is shown as a dotted line extending beyond the restricted region. The solid portion shows the restricted domain from x equals 0 to x equals pi. This portion starts at the maximum value of 1 at x equals 0, decreases through 0 at x equals pi over 2, and reaches the minimum value of negative 1 at x equals pi. Within this restricted domain, any horizontal line intersects the curve only once, showing the function is one-to-one on this interval. Graph of y equals tangent x with a restricted domain highlighted. The full tangent curve is shown as a dotted line with vertical asymptotes. The solid portion shows the restricted domain from x equals negative pi over 2 to x equals pi over 2 (excluding the endpoints where asymptotes occur). The curve increases continuously from negative infinity to positive infinity within this interval, passing through the origin. Within this restricted domain, any horizontal line intersects the curve only once, showing the function is one-to-one on this interval.

Figure 2.20: Restricted domains for cosine and tangent

Recall that there are two ways to find the inverse of a function. The graphical way to find the inverse is to look at the graph and reflect it across the line \(y = x\). The algebraic way to solve for the inverse of a function is to switch the \(x\) and \(y\) coordinates and solve for \(y\).

We can find inverse functions of the sine, cosine and tangent using the graphing method. The graph of \(y = \sin^{-1} x\) (sometimes called the arcsine and denoted \(y=\arcsin x\)) is shown in Figure 2.21. Notice the symmetry about the line \(y = x\) with the graph of \(y = \sin x\).

Graph showing both y equals sine x (dotted curve) with restricted domain from negative pi over 2 to pi over 2, and y equals inverse sine x (solid curve). The graph also shows the line y equals x as a dashed diagonal line. The inverse sine curve is the reflection of the restricted sine curve across the line y equals x. The inverse sine has domain from negative 1 to 1 on the x-axis and range from negative pi over 2 to pi over 2 on the y-axis. The curve increases continuously from the point (negative 1, negative pi over 2) through the origin to the point (1, pi over 2).

Figure 2.21: Graph of \(y = \sin^{-1} x\)

The sine function gives you the ratio \(\frac{\text{opposite}}{\text{hypotenuse}}\) for some angle \(\theta\). The inverse sine function give you the angle \(\theta\) if you know the ratio \(\frac{\text{opposite}}{\text{hypotenuse}}\). It is the reverse of the sine. It is often good to think of \(y = \sin^{-1} x\) as “the inverse sine of \(x\) is the angle whose sine is \(x\).”

On your calculator these functions are not displayed as arc functions. Your calculator probably has keys that look like: \(\sin^{-1}\), \(\cos^{-1}\) and \(\tan^{-1}\). These features are often found just above the regular trigonometric function, but different models of calculator have it in different places.

NoteThe inverse sine function \(y = \sin^{-1} x = \arcsin x\)

The sine has restricted domain \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\) and range \(-1 \leq \sin x \leq 1\). The inverse sine is the function whose domain is \(-1 \leq x \leq 1\) and whose range is \(-\frac{\pi}{2} \leq \sin^{-1} x \leq \frac{\pi}{2}\) such that

\[ \sin \left( \sin^{-1} x \right) = x ~~\mbox{for} ~~ -1 \leq x \leq 1\]

and

\[ \sin^{-1} \left( \sin x \right) = x ~~\mbox{for} ~~ -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\]

There are a couple of important things to remember here.

Note 1: With the restriction we have put on the inverse sine, it is ONLY defined in quadrants I and IV so all your answers for arcsine must lie between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

Note 2: The notation for inverse functions is to have an exponent of \(-1\) on the function. This should not be confused with the reciprocal of the function. If we want the reciprocal of the sine we would write it one of the following ways: \[ \frac{1}{\sin x} = \left( \sin x \right)^{-1} = \csc x.\] For this reason some prefer to write \(y = \arcsin x\) and both are often used interchangeably without warning.

The cosine is similar but in this case we restrict the domain to \(0 \leq x \leq \pi\) because this also gives us all the \(y\) values between \(-1\) and \(1\). Figure 2.22 shows the graph. It is often good to think of \(y=\cos^{-1} x\) as “the inverse cosine of \(x\) is the angle whose cosine is \(x\).”

Graph of y equals inverse cosine x (arccosine). The curve has domain from negative 1 to 1 on the x-axis and range from 0 to pi on the y-axis. The curve decreases continuously from the point (negative 1, pi) through the point (0, pi over 2) to the point (1, 0). This is the reflection of the restricted cosine function across the line y equals x.

Figure 2.22: Graph of \(y = \cos^{-1} x = \arccos x\)

Again we reflect this dotted curve across the line \(y = x\) to get the inverse cosine function.

Note: With the restriction we have put on the arccosine, it is ONLY defined in quadrants I and II so all your answers for arccosine must lie between \(0\) and \(\pi\).

The tangent has the same restrictions as the sine but in this case we have vertical asymptotes. When we reflect across the line \(y=x\) the vertical asymptotes become horizontal asymptotes. It is often good to think of \(y = \tan^{-1} x\) as “the inverse tangent of \(x\) is the angle whose tangent is \(x\).” See Figure 2.23.

Graph of y equals inverse tangent x (arctangent). The curve has domain of all real numbers on the x-axis and range from negative pi over 2 to pi over 2 on the y-axis. The curve has horizontal asymptotes at y equals negative pi over 2 and y equals pi over 2. The curve increases continuously from the lower asymptote through the origin (where it has a slope of 1) toward the upper asymptote. This is the reflection of the restricted tangent function across the line y equals x, with the vertical asymptotes of tangent becoming horizontal asymptotes for arctangent.

Figure 2.23: Graph of \(y = \tan^{-1} x = \arctan x\)

Note: With the restriction we have put on the arctangent, it is ONLY defined in quadrants I and IV so all your answers for arctangent must lie between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

NoteSummary of inverse trigonometric functions
Function Definition In Words Range
\(\sin^{-1} x = y\) \(x = \sin y\) \(y\) is the angle whose sine is \(x\) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
\(\cos^{-1} x = y\) \(x = \cos y\) \(y\) is the angle whose cosine is \(x\) \(0 \leq y \leq \pi\)
\(\tan^{-1} x = y\) \(x = \tan y\) \(y\) is the angle whose tangent is \(x\) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
TipExample 2.3.1

Find \(y\) when \(y = \cos^{-1} \left( \dfrac{\sqrt{3}}{2} \right)\)

Solution:

Step 1: Draw a triangle in the appropriate quadrant and label the sides. Since \(\cos^{-1}\) is defined in quadrant I and II and \(\dfrac{\sqrt{3}}{2}\) is positive, we draw the triangle in quadrant I (See Figure 2.24). Since we have the arccosine here we know that \(\dfrac{\sqrt{3}}{2} = \dfrac{\text{adjacent}}{\text{hypotenuse}}\) and we can use the Pythagorean theorem to find the missing side.

Step 2: Identify the angle in the triangle. Very often it will be one of the special triangles. In this case we have a \(30 - 60 - 90\) triangle so our angle is

\[\boxed{y =30^\circ} \text{ or } \boxed{y =\frac{\pi}{6}}\]

Right triangle drawn in Quadrant I with angle y at the origin. The hypotenuse has length 1, the adjacent side (along the positive x-axis) has length square root of 3 over 2, and the opposite side (vertical) has length 1 over 2. This forms a 30-60-90 degree triangle. The angle y is marked at the origin between the positive x-axis and the hypotenuse.

TipExample 2.3.2

Find \(y\) when \(y = \cos^{-1} \left( -\dfrac{\sqrt{3}}{2} \right)\)

Solution:

Step 1: Draw a triangle in the appropriate quadrant and label the sides. Since \(\cos^{-1}\) is defined in quadrant I and II and \(-\dfrac{\sqrt{3}}{2}\) is negative, we draw the triangle in quadrant II (See figure at right). Since we have the arccosine here we know that \(-\dfrac{\sqrt{3}}{2} = \dfrac{\text{adjacent}}{\text{hypotenuse}}\) and we can use the Pythagorean theorem to find the missing side.

Step 2: Identify the angle in the triangle. In this case we have a \(30 - 60 - 90\) triangle again so our reference angle is \(\frac{\pi}{6}\) and the answer is

\[\boxed{y=\frac{5\pi}{6}}\]

Notice that this falls in the range we want for answers to arccosine problems: \(0 \leq y \leq \pi\).

Right triangle drawn in Quadrant II with angle y at the origin. The hypotenuse has length 1, the adjacent side (along the negative x-axis) has length negative square root of 3 over 2, and the opposite side (vertical, pointing up) has length 1 over 2. This forms a 30-60-90 degree triangle in Quadrant II. The angle y is marked at the origin, measured from the positive x-axis counterclockwise to the hypotenuse.

TipExample 2.3.3

Find \(y\) when \(y = \sin^{-1} \left( -\dfrac{\sqrt{2}}{2} \right)\)

Solution:

Step 1: Draw a triangle in the appropriate quadrant and label the sides. Since \(\sin^{-1}\) is defined in quadrant I and IV and \(-\dfrac{\sqrt{2}}{2}\) is negative, we draw the triangle in quadrant IV (See Figure at right). Since we have the arcsine here we know that \(-\dfrac{\sqrt{2}}{2} = \dfrac{\text{opposite}}{\text{hypotenuse}}\) and we can use the Pythagorean theorem to find the missing side.

Step 2: Identify the angle in the triangle. In this case we have a \(45-45 - 90\) triangle so our reference angle is \(\frac{\pi}{4}\) and the answer is

\[\boxed{y=-\dfrac{\pi}{4}}\]

Notice that this falls in the range we want for answers to arcsine problems: \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).

Right triangle drawn in Quadrant IV with angle y at the origin. The hypotenuse has length 1, the adjacent side (along the positive x-axis) has length square root of 2 over 2, and the opposite side (vertical, pointing down) has length negative square root of 2 over 2. This forms a 45-45-90 degree triangle in Quadrant IV. The angle y is marked at the origin, measured as a negative angle from the positive x-axis clockwise to the hypotenuse.

TipExample 2.3.4

Evaluate \(\sin^{-1}(0.97)\) using your calculator.

Solution: Since the output of the inverse function is an angle, your calculator will give you a degree value if in degree mode, and a radian value if in radian mode.

In radian mode, \(\boxed{\sin^{-1}(0.97) \approx 1.3252}\)

In degree mode, \(\boxed{\sin^{-1}(0.97)\approx 75.93^\circ}\)

TipExample 2.3.5

Evaluate \(\cos^{-1} \left( \cos \left( \dfrac{13\pi}{6}\right) \right)\)

Solution: Here we want to be careful. The cosine and arccosine are direct inverses of each other only between \(0\) and \(\pi\) so our answer can’t be \(\dfrac{13\pi}{6}\). What we need to do is to first find the value of \(\cos \left( \dfrac{13\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\). Once we know this we are now looking for \(\cos^{-1} \left( \dfrac{\sqrt{3}}{2} \right)\) which we found in Example 2.3.1. So \[\cos^{-1} \left( \cos \left( \dfrac{13\pi}{6}\right) \right) =\cos^{-1} \left( \dfrac{\sqrt{3}}{2} \right)= \boxed{\dfrac{\pi}{6}}\]

TipExample 2.3.6

Find \(\tan^{-1} \left( \tan \pi \right)\).

Solution: Since \(\pi > \frac{\pi}{2}\), tangent and arctangent are not direct inverses. But we know that \(\tan \pi = 0\). Thus, \(\tan^{-1} \left( \tan \pi \right) = \tan^{-1} 0\) is, by definition, the angle \(y\) such that \(\tan y = 0\) where \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\). That angle is \(y = 0\). Thus,

\[\boxed{\tan^{-1} \left( \tan \pi \right) = \tan^{-1} (0) = 0}\]

TipExample 2.3.7

Evaluate \(\sin^{-1}0\)

Solution: We need to find an angle \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) such that \(\sin \theta = 0\). The only angle that satisfies this is \(\boxed{\theta = 0}\).

TipExample 2.3.8

Evaluate \(\tan^{-1}(-1)\)

Solution: We need to find an angle \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) such that \(\tan \theta = -1\). The answer will be in QIV: \(\boxed{\theta = -\frac{\pi}{4}}\).

TipExample 2.3.9

Evaluate \(\cos\left(\sin^{-1}\left( -\dfrac{3}{4} \right) \right)\)

Solution: We could do this problem the way we did the earlier examples where we drew a triangle but another solution is to use one of our pythagorean identities from Section 1.2. \[\cos^2 \theta + \sin^2 \theta = 1\]

Let \(\theta = \sin^{-1}\left( -\frac{3}{4} \right)\). Since \(\sin(\theta) = -\frac{3}{4}\) we know that \(\theta\) is in QIV so \(\cos \theta >0\) (positive). Using our identity we can now calculate \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left( -\frac{3}{4} \right)^2 = \frac{7}{16} ~~ \implies~~\cos \theta = \frac{\sqrt{7}}{4}.\]

Note that we took the positive square root since \(\cos \theta >0\). Thus our answer is

\[\boxed{\cos\left(\sin^{-1}\left( -\dfrac{3}{4} \right) \right) = \dfrac{\sqrt{7}}{4}}\]

TipExample 2.3.10

Evaluate \(\csc\left( \tan^{-1} \left(-\dfrac{5}{12} \right) \right)\)

Solution: This problem is similar to Example 2.3.9 but for this one we will construct a triangle (see figure at right) to show a different way to arrive at the solution.

Let \(\theta = \tan^{-1} \left(-\frac{5}{12} \right)\). Then we can draw a triangle for \(\theta\) in QIV since \[\tan \theta = -\dfrac{5}{12} =\dfrac{\text{opposite}}{\text{adjacent}}.\]

Using the Pythagorean theorem we can find the hypotenuse length of \(13\). Now we can read the cosecant off the triangle.

\[\csc \theta = \dfrac{\text{hypotenuse}}{\text{opposite}}= \boxed{\dfrac{13}{-5}}\]

Right triangle drawn in Quadrant IV with angle theta at the origin. The adjacent side (along the positive x-axis) has length 12, the opposite side (vertical, pointing down) has length negative 5, and the hypotenuse has length 13 (found by the Pythagorean theorem). The angle theta is marked at the origin as a negative angle from the positive x-axis.

TipExample 2.3.11

Find a simplified expression for \(\tan \left( \sin^{-1} x \right)\) for \(-1< x<1\)

Solution: Let \(\theta = \sin^{-1} x\). Then we can draw a triangle for \(\theta\) since we know that \[\sin \theta = \dfrac{x}{1} =\dfrac{\text{opposite}}{\text{hypotenuse}}.\]

There are two triangles we can draw, one in QI for \(0<x<1\) and one in QIV for \(-1<x<0\) but the adjacent side length is the same for both so only the one in QI is presented in Figure 2.24. The adjacent side length is calculated using the Pythagorean theorem and is \(\sqrt{1-x^2}\). Notice that since the \(x\) is squared it is always positive no matter the sign of \(x\). Then we can read the tangent right off the graph and

\[\boxed{\tan \left( \sin^{-1} x \right) = \dfrac{x}{\sqrt{1-x^2}} \text{ for } -1< x<1}\]

Right triangle drawn in Quadrant I with angle theta at the origin. The hypotenuse has length 1, the opposite side (vertical) has length x, and the adjacent side (horizontal) has length square root of 1 minus x squared (found by the Pythagorean theorem). The angle theta is marked at the origin between the positive x-axis and the hypotenuse.

Figure 2.24

TipExample 2.3.12

A cellular telephone tower that is 50 meters tall is placed on top of a mountain that is 1200 meters above sea level. What is the angle of depression to two decimal places from the top of the tower to a cell phone user who is 5 horizontal kilometers away and 400 meters above sea level?

Diagram showing a mountain with a cellular tower on top. The mountain peak is at 1200 meters above sea level with a 50 meter tower on top, making the total height 1250 meters. A cell phone user is shown 5 kilometers (5000 meters) horizontally away at 400 meters above sea level. A right triangle is formed with the vertical height difference h equals 1250 minus 400 equals 850 meters, and the horizontal distance of 5000 meters. The angle theta is marked at the top of the tower, representing the angle of depression from the tower to the user.

Figure 2.25

Solution: Figure 2.25 above describes the situation. We need to measure the distance from the top of the hill to the top of the cellular tower marked \(h\). Thus \(h = 1200 + 50 - 400 = 850\) m. We also need to convert the horizontal distance to meters, \(5\) km = \(5000\) m and we can use the tangent function to write an equation relating the height and the adjacent side:

\[\frac{850}{5000} ~=~ \tan\;\theta \quad\Rightarrow\quad \theta ~=~ \tan^{-1} \left( \frac{850}{5000} \right) ~=~ \boxed{9.65^\circ}\]

We can calculate the inverse function by using a calculator, the inverse button looks something like: \(\tan^{-1}\). Again, be careful that your calculator is in degree mode.

2.3 Exercises

For Exercises 1-28, find the exact value of the given expression. If an answer is an angle answer in radians.

  1. \(\tan^{-1} 1\)

  2. \(\tan^{-1} (-1)\)

  3. \(\tan^{-1} 0\)

  4. \(\cos^{-1} 1\)

  5. \(\cos^{-1} (-1)\)

  6. \(\cos^{-1} 0\)

  7. \(\sin^{-1} 1\)

  8. \(\sin^{-1} (-1)\)

  9. \(\cos^{-1} \left( \frac{\sqrt{3}}{2}\right)\)

  10. \(\cos^{-1}\left( \frac{-\sqrt{3}}{2}\right)\)

  11. \(\sin^{-1} \left( \frac{\sqrt{2}}{2}\right)\)

  12. \(\sin^{-1} \left( \frac{-\sqrt{2}}{2}\right)\)

  13. \(\sin^{-1} 0\)

  14. \(\sin^{-1}\left(\sin \frac{\pi}{3}\right)\)

  15. \(\sin^{-1}\left(\sin \frac{4\pi}{3}\right)\)

  16. \(\sin^{-1}\left(\sin \left(-\frac{4\pi}{3}\right)\right)\)

  17. \(\cos^{-1}\left(\cos \frac{\pi}{5}\right)\)

  18. \(\cos^{-1}\left(\cos \frac{6\pi}{5}\right)\)

  19. \(\cos^{-1}\left(\cos \left(-\frac{\pi}{5}\right)\right)\)

  20. \(\tan^{-1}\left(\tan \left(-\frac{5\pi}{6}\right)\right)\)

  21. \(\tan^{-1}\left(\tan \frac{5\pi}{6}\right)\)

  22. \(\cos^{-1}\left(\sin \frac{13\pi}{6}\right)\)

  23. \(\sin^{-1}\left(\cos \left(-\frac{\pi}{6}\right)\right)\)

  24. \(\csc^{-1}\left(\sec \left(-\frac{5\pi}{6}\right)\right)\)

  25. \(\tan \left(\sin^{-1} \frac{4}{3}\right)\)

  26. \(\sin\left(\tan^{-1} \frac{4}{3}\right)\)

  27. \(\sin \left(\cos^{-1} \left(-\frac{3}{5}\right)\right)\)

  28. \(\cos \left(\sin^{-1} \left(-\frac{4}{5}\right)\right)\)

  29. Find a simplified expression for \(\cos\left(\sin^{-1} x\right)\) for \(-1\leq x \leq 1\).

  30. Find a simplified expression for \(\cot\left(\sin^{-1} \left(\dfrac{x}{3}\right)\right)\) for \(-3\leq x \leq 3\).

  31. Find a simplified expression for \(\sin \left( \cos^{-1} \frac{x}{3} \right)\) for \(-3< x<3\).

  32. Find a simplified expression for \(\csc\left(\tan^{-1} \left(\dfrac{x}{2}\right)\right)\).

  33. The height of a playground basketball backboard is 12 feet 6 inches high. At 4:00 pm it casts a shadow 15 feet long. What is the angle of elevation of the sun at that time?

2.4 Solving Trigonometric Equations

To solve a trigonometric equation we use standard algebraic techniques such as combining like terms and factoring. The first goal for any trigonometric equation is to isolate the trigonometric function in the equation. We can’t algebraically solve for the variable from inside a trigonometric function.

TipExample 2.4.1

Solve the equation \(2\sin x + 1 = 0\)

Solution: We have to have the trigonometric function by itself on one side of the equation and the numbers on the other side. In this case we solve for \[\sin x = -\frac{1}{2}.\]

When we have it in this form we can then decide what values of \(x\) will work here. In the section on inverse functions (Section 2.3) we saw that we could ask our calculator for the value \[x = \sin^{-1} \left(-\frac{1}{2}\right) = -\frac{\pi}{6}.\]

While it is true that this value of \(x\) satisfies the equation, it is not the complete solution. If we plot the graph of \(y = \sin x\) and \(y = -\frac{1}{2}\) on the same set of axes we can find where they intersect.

Graph showing a blue sine curve intersecting a horizontal dashed line at y = -1/2, with solutions marked at x = -5π/6, -π/6, 7π/6, and 11π/6
Figure 11

Figure 2.26: Intersections of \(y = \sin x\) and \(y = -\frac{1}{2}\)

Four of the solutions \((x, y)\) are labeled \(\left(\frac{7\pi}{6}, -\frac{1}{2}\right)\), \(\left(\frac{11\pi}{6}, -\frac{1}{2}\right)\), \(\left(-\frac{\pi}{6}, -\frac{1}{2}\right)\), \(\left(-\frac{5\pi}{6}, -\frac{1}{2}\right)\) and four more are indicated as multiples of \(2\pi\). There are an infinite number of solutions because the graph of sine continues indefinitely in both directions and the line \(y = -\frac{1}{2}\) will intersect it an infinite number of times.

We need to have a way to describe all the solutions. Since the sine is periodic we know that it repeats every \(2\pi\) so our solutions will repeat every \(2\pi\). We find all the positive solutions on one time around the circle \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\) and then add multiples of \(2\pi\) to it. Our two solutions can be written:

\[\boxed{x = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2n\pi, \quad \text{where} \ n \in \mathbb{Z}}\]

Figure 2.27: Reference triangles for Example 2.4.1
Figure 12

Figure 2.27: Reference triangles for Example 2.4.1

The graph of sine and cosine are not convenient for finding the \(x\)-values that satisfy the equation. Most often reference triangles or the unit circle are used. In the previous example we wanted the solutions to \(\sin x = -\frac{1}{2} = \frac{\text{opposite}}{\text{hypotenuse}}\) and we can draw two reference triangles that satisfy this angle. We need two because the sine is negative in both QIII and QIV. See Figure 2.27.

These two triangles are recognizable as our 30-60-90 triangle and as such we can find the reference angle \(\frac{\pi}{6}\) and the two basic solutions \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\). From there the complete solution can be written as above.

\[\boxed{x = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2n\pi, \quad \text{where} \ n \in \mathbb{Z}}\]

Recall that the integers are represented by the symbol \(\mathbb{Z} = \{0, \pm 1, \pm 2, \pm 3, \ldots\}\)

TipExample 2.4.2

Solve the equation \(3\cot^2 x - 1 = 0\)

Solution: Here we need to isolate the \(\cot x\) on one side of the equation.

\[\begin{align*} 3\cot^2 x - 1 &= 0\\ 3\cot^2 x &= 1\\ \cot^2 x &= \frac{1}{3}\\ \cot x &= \pm \frac{1}{\sqrt{3}} \end{align*}\]

There are two solutions because you always need to take into account both the positive and negative answers when taking a square root. We can draw reference triangles for these two solutions. There are 4 we could draw for \(0 \leq x < 2\pi\), one in each quadrant. Figure 2.28 shows the solutions in QI and QIII for \(\cot x = \frac{1}{\sqrt{3}}\) and solutions in QII and QIV for the negative.

Cotangent has a period of \(\pi\) so we can start with two basic solutions \(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\). Then add multiples of \(\pi\) to each of these to get the general form:

\[\boxed{x = \frac{\pi}{3} + n\pi \quad \text{and} \quad x = \frac{2\pi}{3} + n\pi, \quad \text{where} \ n \in \mathbb{Z}}\]

Blue reference triangles on left for $\cot x = \frac{1}{\sqrt{3}}$, red for $\cot x = -\frac{1}{\sqrt{3}}$
Figure 13

Figure 2.28: Reference triangles for Example 2.4.2

TipExample 2.4.3

Solve the equation \(2\cos^2 \theta - 1 = 0\).

Solution: Isolating \(\cos^2 \theta\) gives us

\[\cos^2 \theta = \frac{1}{2} \quad\Rightarrow\quad \cos \theta = \pm \frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4},\]

and since the period of cosine is \(2\pi\), we would add \(2n\pi\) to each of those angles to get the general solution. But notice that the above angles differ by multiples of \(\frac{\pi}{2}\). Since every multiple of \(2\pi\) is also a multiple of \(\frac{\pi}{2}\), we can combine those four separate answers into one:

\[\boxed{\theta = \frac{\pi}{4} + \frac{\pi}{2}n \quad \text{for } n \in \mathbb{Z}}\]

TipExample 2.4.4

Solve the equation \(4\cos^3 x - 3\cos x = 0\)

Solution: This equation will require some factoring. In our previous examples we were able to isolate a squared term and then take a square root. In this case that won’t be possible.

\[\begin{align*} 4\cos^3 x - 3\cos x &= 0\\ \cos x (4\cos^2 x - 3) &= 0 \end{align*}\]

Now we have two things multiplied together that equal zero so one of them must be zero. Set each factor equal to zero and find all the solutions between \(0 \leq x < 2\pi\).

\(\cos x = 0\) and \(4\cos^2 x - 3 = 0\)
\(\cos^2 x = \frac{3}{4}\)
\(\cos x = \pm \frac{\sqrt{3}}{2}\)
\(\cos x = \frac{\sqrt{3}}{2}\) and \(\cos x = -\frac{\sqrt{3}}{2}\)
\(x = \frac{\pi}{2}, \frac{3\pi}{2}\) and \(x = \frac{\pi}{6}, \frac{11\pi}{6}\) and \(x = \frac{5\pi}{6}, \frac{7\pi}{6}\)

Note that here \(\frac{\pi}{6}\) and \(\frac{7\pi}{6}\), \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\), as well as \(\frac{5\pi}{6}\) and \(\frac{11\pi}{6}\) are different by \(\pi\) so we can write our solutions as:

\[\boxed{x = \frac{\pi}{2} + n\pi, \quad x = \frac{\pi}{6} + n\pi, \quad \text{and} \quad x = \frac{5\pi}{6} + n\pi, \quad \text{where} \ n \in \mathbb{Z}}\]

TipExample 2.4.5

Solve the equation \(2\sin(5x) + 1 = 0\)

Solution: This problem is similar to Example 2.4.1 but now we have \((5x)\) in the sine. We need to isolate the \(\sin(5x)\) and solve for the values of \(5x\).

\[\begin{align*} 2\sin(5x) + 1 &= 0\\ 2\sin(5x) &= -1\\ \sin(5x) &= -\frac{1}{2} \end{align*}\]

In the interval \([0, 2\pi)\) we know that \(5x = \frac{7\pi}{6} + 2n\pi\) and \(5x = \frac{11\pi}{6} + 2n\pi\). We need to divide both sides by 5 to obtain the general solution:

\[\boxed{x = \frac{7\pi}{30} + \frac{2n\pi}{5}, \quad x = \frac{11\pi}{30} + \frac{2n\pi}{5}, \quad \text{where} \ n \in \mathbb{Z}}\]

TipExample 2.4.6

Find all solutions on \([0, 2\pi)\). \[2\sin^2 x + 5\sin x + 3 = 0\]

Solution: Here we need to factor the equation because it is a quadratic. Also, since we only want solutions on \([0, 2\pi)\) and the angle is just \(x\) then we don’t need to write the solution with the \(+2n\pi\).

\[\begin{align*} 2\sin^2 x + 5\sin x + 3 &= 0\\ (2\sin x + 3)(\sin x + 1) &= 0 \end{align*}\]

Now we have two things multiplied together that equal zero so one of them must be zero. Set each factor equal to zero and find all the solutions between \(0 \leq x < 2\pi\).

\(2\sin x + 3 = 0\) and \(\sin x + 1 = 0\)
\(\sin x = -\frac{3}{2}\) \(\sin x = -1\)
No solution because \(-1 \leq \sin x \leq 1\) and \(x = \frac{3\pi}{2}\)

The only solution here is \(\boxed{x = \frac{3\pi}{2}}\).

TipExample 2.4.7

Find all solutions on \([0, 2\pi)\). \[2\sin^2(2x) = 1\]

Solution: As in Example 2.4.5 we need to first solve for the value of \(2x\) and then divide by two. We don’t want the general solution but we do need to start with it to find all values of \(x\) on \([0, 2\pi)\).

We need to isolate the \(\sin(2x)\) and solve for the values of \(2x\).

\[\begin{align*} 2\sin^2(2x) &= 1\\ \sin^2(2x) &= \frac{1}{2}\\ \sin(2x) &= \pm \sqrt{\frac{1}{2}} \end{align*}\]

There are two equations to solve: \(\sin(2x) = \frac{1}{\sqrt{2}}\) and \(\sin(2x) = -\frac{1}{\sqrt{2}}\) so we have 4 general solutions for \(2x\), one in each quadrant:

General solution for \(2x\) Solution for \(x\)
\(2x = \frac{\pi}{4} + 2n\pi\) \(\implies\) \(x = \frac{\pi}{8} + n\pi\)
\(2x = \frac{3\pi}{4} + 2n\pi\) \(\implies\) \(x = \frac{3\pi}{8} + n\pi\)
\(2x = \frac{5\pi}{4} + 2n\pi\) \(\implies\) \(x = \frac{5\pi}{8} + n\pi\)
\(2x = \frac{7\pi}{4} + 2n\pi\) \(\implies\) \(x = \frac{7\pi}{8} + n\pi\)

To find all solutions on \([0, 2\pi)\) we will substitute values for \(n\) until we find all the solutions starting with \(n=0\):

\(n=0\): \(n=1\):
\(x = \frac{\pi}{8} + (0)\pi = \frac{\pi}{8}\) \(x = \frac{\pi}{8} + (1)\pi = \frac{9\pi}{8}\)
\(x = \frac{3\pi}{8} + (0)\pi = \frac{3\pi}{8}\) \(x = \frac{3\pi}{8} + (1)\pi = \frac{11\pi}{8}\)
\(x = \frac{5\pi}{8} + (0)\pi = \frac{5\pi}{8}\) \(x = \frac{5\pi}{8} + (1)\pi = \frac{13\pi}{8}\)
\(x = \frac{7\pi}{8} + (0)\pi = \frac{7\pi}{8}\) \(x = \frac{7\pi}{8} + (1)\pi = \frac{15\pi}{8}\)

We don’t need to go any further because any other answers will be larger than \(2\pi\). There are 8 possible solutions.

\[\boxed{x = \frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{9\pi}{8}, \ \frac{11\pi}{8}, \ \frac{13\pi}{8}, \ \frac{15\pi}{8}}\]

TipExample 2.4.8

There has been a murder at the Toronto docks. The coroner places the time of death around 8 AM. The main suspect claims she was on her boat fishing in Lake Ontario at the time and that she was waiting for the tide in order to tie up her boat. Detective Murdoch knows that the depth of water at the docks rises and falls with the tide, following the equation \[f(t) = 4\sin\left(\frac{\pi}{12}t\right) + 7,\] where \(t\) is measured in hours after midnight. The suspect’s boat requires a depth of 9 feet to tie up at the dock. Between what times will the depth be 9 feet? Is the suspect lying?

Solution: To find when the depth is 9 feet, we need to solve \(f(t) = 9 = 4\sin\left(\frac{\pi}{12}t\right) + 7\). We start by isolating the sine.

\[\begin{align*} 4\sin\left(\frac{\pi}{12}t\right) + 7 &= 9\\ 4\sin\left(\frac{\pi}{12}t\right) &= 2\\ \sin\left(\frac{\pi}{12}t\right) &= \frac{1}{2} \end{align*}\]

We know that \(\sin \theta = \frac{1}{2}\) when \(\theta = \frac{\pi}{6}\) or \(\theta = \frac{5\pi}{6}\) so the solutions to the equation \(\sin\left(\frac{\pi}{12}t\right) = \frac{1}{2}\) are \[\frac{\pi}{12}t = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad \frac{\pi}{12}t = \frac{5\pi}{6} + 2n\pi \quad n \in \mathbb{Z}\]

Multiply by \(\frac{12}{\pi}\) to find the solutions \(t = 2 + 24n\) and \(t = 10 + 24n\). The boat will be able to approach the dock between 2AM and 10AM. Notice that because we have \(+24n\) in each answer the cycle will repeat every day (24 hours). The suspect is lying about waiting for the tide at 8AM.

TipExample 2.4.9

Find all solutions to \(\sin \theta = 0.8\).

Unit circle with two reference angles marked, showing solutions in quadrants I and II where sine equals 0.8
Figure 14

Figure 2.29: Graph showing \(\sin \theta = 0.8\)

Solution: To find the solutions we will draw two reference angles. Since the sine is not one of the results for our special triangles we will use the inverse sine function here. When you ask your calculator for the inverse sine it will only give you one answer: \[\theta = \sin^{-1}(0.8) \approx 0.9273.\]

Recall that the inverse sine is answering the question: “What angle has sine 0.8?” We know that on the interval \(0 \leq \theta < 2\pi\) there are two answers. The second answer is in QII as shown in Figure 2.29. The second answer can be found with the reference angle and the size of the second angle is \[\theta \approx \pi - 0.9273 \approx 2.4130\]

To find all the solutions we add multiples of \(2\pi\).

\[\boxed{\theta = \sin^{-1}(0.8) + 2n\pi, \quad \theta = \pi - \sin^{-1}(0.8) + 2n\pi, \quad \text{where} \ n \in \mathbb{Z}}\]

\[\boxed{\theta \approx 0.9273 + 2n\pi, \quad \theta \approx 2.4130 + 2n\pi, \quad \text{where} \ n \in \mathbb{Z}}\]

2.4 Exercises

For Exercises 1-6, find all solutions on the interval \([0, 2\pi)\). Leave exact answers in radians.

  1. \(2\sin x = \sqrt{2}\)

  2. \(2\sin x + \sqrt{3} = 0\)

  3. \(\csc x = -2\)

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

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

  3. \(\tan(\theta) - \sqrt{3} = 0\)

For Exercises 7-12, find the general solution for each equation. Leave exact answers in radians.

  1. \(\tan \theta + 1 = 0\)

  2. \(2\sin x - 1 = 0\)

  3. \(2\cos x = \sqrt{3}\)

  1. \(\sqrt{3}\sec x = 2\)

  2. \(\sin \theta = 0\)

  3. \(\sqrt{3}\cot(x) - 1 = 0\)

For Exercises 13-18, find all solutions on the interval \([0, 2\pi)\). Leave exact answers in radians.

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

  2. \(\tan(2x) = -1\)

  3. \(\sqrt{3}\csc\left(\frac{x}{2}\right) = -2\)

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

  2. \(2\cos^2(2\theta) = 1\)

  3. \(\cos(3x) = \frac{\sqrt{2}}{2}\)

For Exercises 19-28, find all solutions on the interval \([0, 2\pi)\). Leave exact answers in radians.

  1. \(\tan \theta(\tan \theta + 1) = 0\)

  2. \(\cot^2 x = 3\)

  1. \(\tan x \sin x - \sin x = 0\)

  2. \(2\cos^2 x + 3\cos x + 1 = 0\)

  1. \((4\sin^2 x - 3)(\sqrt{2}\cos x + 1) = 0\)

  2. \(\sin x(\sec x + 2) = 0\)

  1. \(2\sin^2 x + \sin x - 1 = 0\)

  2. \(2\sin^3 x = \sin x\)

  1. \(\tan^5 x = \tan x\)

  2. \(2\cos^2 x - \sin x = 1\)

For Exercises 29-34, use a calculator to find all solutions on the interval \([0, 2\pi)\). Round answers to 4 decimal places.

  1. \(7\sin \theta = 2\)

  2. \(\cos x = -0.27\)

  3. \(\tan x = 9.27\)

  1. \(7\sin(2\theta) = 2\)

  2. \(\sec^2 x = 7\)

  3. \(\tan(\pi x) = 9.27\)

  1. An observer views a rocket take off from a distance of 7 km from the launch pad, and tracks the angle of elevation. Express the height of the rocket as a function of the angle of elevation, \(\theta\). Express the angle of elevation \(\theta\) as a function of the height, h, of the rocket. When the height of the rocket is 22 km what is the angle of elevation?

  2. The height of a rider on the London Eye Ferris wheel can be determined by the equation \(h(t) = -67.5\cos\left(\frac{\pi}{15}t\right) + 69.5\). How long is the rider more than 100 meters above ground?