Measurement of Triangles

In this section on trigonometric functions, we will discover how ideas from compound angles can be used to rewrite expressions in harmonic form. This skill makes it easier to find the highest and lowest points of a vibrating system, such as the motion of a string on a guitar or the swing of a pendulum. We will also explore the shapes of sine, cosine, and tangent graphs, noticing their repeating patterns and how they rise and fall. By sketching these graphs by hand and experimenting with digital tools like GeoGebra, we can see how mathematics describes waves, rhythms, and cycles that appear all around us. Linking these graphs to the unit circle will give us a clear picture of why trigonometric functions repeat and how their values change as we move around the circle. In the end, you will gain not only the ability to analyse trig graphs and find their maximum and minimum values but also an appreciation of how these ideas explain real-life phenomena such as sound waves, tides, and even satellite motion.

KEY IDEAS

• The maximum, minimum, and periodic behaviour of trigonometric graphs come directly from their definitions on the unit circle. • Trigonometric expressions of the form $a\cos\theta + b\sin\theta$ can be rewritten in harmonic form using compound angle identities. • The amplitude in harmonic form shows the maximum value, and the minimum is its negative (with vertical shifts considered). • The graphs of $y = \sin x$ and $y = \cos x$ oscillate between maximum and minimum values and are periodic. • The graph of $y = \tan x$ repeats every $180^\circ$ and has vertical asymptotes where $\cos x = 0$ .

In Year 1 Section 7, we discovered about quadrantal angles and their special properties. We established that they are angles that terminate on the $x$ and $y$ axes. The angles are $0, 90^\circ, 180^\circ, 270^\circ$ and $360^\circ$ . In Year 2 Section 9, we learnt that a unit circle is a circle with a radius of 1 unit. If the circle is centred at the origin $(0, 0)$ it has the equation $x^2 + y^2 = 1$ .

Regardless of the scale you choose, a unit circle will always intersect the $x$ and $y$ axes at the points $(1, 0), (0, 1), (-1, 0)$ and $(0, -1)$ as represented in Figure 4.1. [Figure] Figure 4.1: Representation of a Unit Circle Now, let us explore the relationship between the coordinates of a point on a unit circle and the cosine/sine value of the angle between the line that joins the point to the origin and the $x$ or $y$ axis using a diagram. [Figure] Figure 4.2: Unit circle with a point From Figure 4.2, $\sin(\alpha) = y$ and $\cos(\alpha) = x$ . This means that for any point on the unit circle, the $x -$ coordinate is the value of the cosine value of the angle between the line that joins the point to the origin and the $x\text{-axis}$ . The $y -$ coordinate of a point on the unit circle is the value of the sine value of the angle between line that joins the point to the origin and the $x\text{-axis}$ .

Combining what we know about unit circles and quadrantal angles we may have a representation as seen in Figure 4.3. [Figure] Figure 4.3: Unit Circle with Quadrantal angles Table 4.1: Functional values of quadrantal angles for sine and cosine Quadrantal Angles $0^\circ$ $90^\circ$ $180^\circ$ $270^\circ$ $360^\circ$ Point $(1, 0)$ $(0, 1)$ $(-1, 0)$ $(0, -1)$ $(1, 0)$ Sine $0$ $1$ $0$ $-1$ $0$ Cosine $1$ $0$ $-1$ $0$ $1$ Table 4.1 contains details of the sine and cosine values of the quadrantal angles in relation to the positioning of the points on the unit circle that intersect with the $x$ and $y$ axes.

Graphs of basic trigonometric functions

In Year 2 Section 9 while learning about verification of identities, we explored how to use technological tools such as GeoGebra and Desmos to graph given trigonometric functions and determine their equivalence. Now, we are going to explore how to graph the basic trigonometric functions (sine, cosine and tangent) and identify their distinctive properties.

Activity 4.1 Graphs of basic trigonometric functions

1. Refer to Table 4.1, use the quadrantal angles as $x$ and represent the values of $\sin x.$

$\boldsymbol{x}$ $\boldsymbol{0^\circ}$ $\boldsymbol{90^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{360^\circ}$ $\boldsymbol{\sin x}$ $0$ $1$ $0$ $-1$ $0$

2. Using the same quadrantal angles as in Step 1, represent the values of $\cos x$ . $\boldsymbol{x}$ $\boldsymbol{0^\circ}$ $\boldsymbol{90^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{360^\circ}$ $\boldsymbol{\cos x}$ $1$ $0$ $-1$ $0$ $1$

3. Using a scale of $2\mathrm{cm} : 90^\circ$ on the $x -$ axis and $2\mathrm{cm}:$ 0.5 units on the $y -$ axis, represent $y = \sin x$ and $y = \cos x$ on the same graph sheet. Remember to draw the graphs using smooth curves. Your resulting graph should be similar to the one in Figure 4.4 below. [Figure] Figure 4.4: Graphical illustration of $\sin x$ and $\cos x$ where $0^\circ \leq x \leq 360^\circ$ .

4. Record your observations from the graph in your notebook and share with a classmate.

5. Recall the $\tan x = \frac{\sin x}{\cos x}$ , now, create a table of values for $\tan x$ using $30^\circ, 45^\circ, 60^\circ$ and the quadrantal angles as $x$ .

$\boldsymbol{x}$ $\boldsymbol{0^\circ}$ $\boldsymbol{30^\circ}$ $\boldsymbol{45^\circ}$ $\boldsymbol{60^\circ}$ $\boldsymbol{90^\circ}$ $\boldsymbol{120^\circ}$ $\boldsymbol{150^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{240^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{330^\circ}$ $\boldsymbol{360^\circ}$ tan x $0$ $\frac{\sqrt{3}}{3}$ $1$ $\sqrt{3}$ unde- fined $-\sqrt{3}$ $-\frac{\sqrt{3}}{3}$ $0$ $\sqrt{3}$ unde- fined $-\frac{\sqrt{3}}{3}$ $0$

6. Represent $y = \tan x$ on a different graph sheet/book. [Figure]

Figure 4.5: Graphical illustration of $\tan x$ , where $0^\circ \leq x \leq 360^\circ$ .

Generalisations

1. For the sine graph you should arrive at the conclusion that: a. The range of the graph is $[-1, 1]$ . The graph has a maximum value of $1$ and minimum value of $-1$ . b. The graph increases before decreasing when starting from the origin. c. The graph cuts the $x -$ axis at $0, 180^\circ$ and $360^\circ$ i.e., values of $x = 180n^\circ$ for integer values of $n$ . d. The graph is symmetric about the origin, confirming that it is an odd function. e. The $x -$ axis, with equation $y = 0$ is the midline. It divides the two halves horizontally.
2. For the cosine graph you should arrive at the conclusion that: a. The range of the graph is $[-1, 1]$ . The graph has a maximum value of $1$ and minimum value of $-1$ . b. The graph decreases before increasing when starting from the origin. c. The graph cuts the $x -$ axis at $90^\circ$ and $270^\circ$ i.e., values of $x = 90^\circ + 180n^\circ$ for integer values of $n$ .

d. The graph is symmetric about the $y -$ axis, confirming that it is an even function. e. The $x -$ axis, with equation $y = 0$ is the midline. It divides the two halves horizontally. 3. For the graph of the tangent function, you should arrive at the conclusion that: a. The range of the graph is $(-\infty, \infty)$ . b. The graph cuts the x- axis at $0, 180^\circ$ and $360^\circ$ i.e., values of $x = 180n^\circ$ for integer values of $n$ . c. There are asymptotes at $x = 90^\circ$ and $x = 270^\circ$ i.e., values of $x = 90^\circ + 180n^\circ$ for integer values of $n$ . (Remember that an asymptote is a line that a graph approaches but never meets.) d. The graph has no maximum or minimum values. e. The graph is symmetric about the origin, confirming that it is an odd function. f. The $x$ - axis, with equation $y = 0$ is the midline. It divides the two halves horizontally.

Transformation of graphs of trigonometric

functions

In Section 2, while learning about Matrices, we discovered various ways in which they can be transformed. From explorations and observations, we realised that transformations sometimes lead to a change in the size of the shape of a plane figure or even the direction. In the same way, transformations of trigonometric graphs bring about changes in the graphs. The original graph might express a shift up or down, a shift to the left or right, a stretch/compression vertically or horizontally. Sometimes, the graph can even experience a reflection across the $x -$ axis. To allow for transformation, basic trigonometric functions can be stated in a more general way. The sine function can be expressed as $f(x) = A\sin(Bx - C) + D$ and the cosine function can be expressed as $f(x) = A\cos(Bx - C) + D$ . Let us delve deeper into the various forms of transformation that may occur.

Transformation of the form A sin x or A cos x

Let us go through this activity to explore transformation of trigonometric functions.

Activity 4.2 Transformation of the form A sin x

1. Using the quadrantal angles as $x$ values, create a table of values for $\sin x$ , $3\sin x$ , $0.5\sin x$ , and $-2.5\sin x$ . Your table of values should be similar to this:

$\boldsymbol{x}$ $\boldsymbol{0^\circ}$ $\boldsymbol{90^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{360^\circ}$ sin x $0$ $1$ $0$ $-1$ $0$ 3 sin x $0$ $3$ $0$ $-3$ $0$ 0.5 sin x $0$ $0.5$ $0$ $-0.5$ $0$ -2.5 sin x $0$ $-2.5$ $0$ $2.5$ $0$

2. On the same graph sheet, draw the curves of points for $y = \sin x$ , $y = 3\sin x$ , $y = 0.5\sin x$ and $y = -2.5\sin x$ . Your graphical representation should be similar to Figure 4.6. [Figure]

Figure 4.6: Graphical representation of $y = \sin x$ , $y = 3\sin x$ , $y = 0.5\sin x$ and $y = -2.5\sin x$ 3. The functions you have can be represented generally as $\mathrm{A}\sin x$ where $\mathrm{A}$ assumes different values. 4. Post your graphical illustration on the walls of your class, move around and observe how your classmates did theirs. 5. Now, record your observation on the different functions as the value of $\mathrm{A}$ changes and share with your classmates.

Observations When:

1. $A > 1$ , the graph of the function stretches vertically.
2. $0 < A < 1$ , the graph of the function compresses or shrinks vertically.
3. $A < 0$ , the graph of the function reflects across the $x\text{-axis}$ . For these graphs, the amplitude, $|A|$ is the largest value the functions attain. $|A|$ is the distance between the midline and the highest or lowest point on the graph.
4. the graph of $y = A\sin(x)$ stretches or shrinks vertically as the value of $|A|$ changes.
5. when the value of $|A|$ gets larger, the graph is stretched more and as the value of $|A|$ gets smaller, it is compressed.
6. when $A > 0$ , the graph increases before decreasing.
7. when $A < 0$ , the graph decreases before increasing.
8. the $x -$ intercepts remain the same.
9. The width (domain of the function for a full revolution) does not change.
10. The midlines for all three graphs remain at $y = 0$ .

Activity 4.3 Transformation of the form A cos x

1. Using the quadrantal angles as $x$ values, create a table of values for $\cos x$ , $3\cos x$ , $0.5\cos x$ , and $-2.5\cos x$ . Your table of values should be similar to this: $\boldsymbol{x}$ $\boldsymbol{0^\circ}$ $\boldsymbol{90^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{360^\circ}$ $1$ $0$ $-1$ $0$ $1$ cos x $3$ $0$ $-3$ $0$ $3$ 3 cos x $0.5$ $0$ $-0.5$ $0$ $0.5$ 0.5 cos x $-2.5$ $0$ $2.5$ $0$ $-2.5$ -2.5 cos x

2. On the same graph sheet, draw the curves of points for $y = \cos x$ , $y = 3\cos x$ , $y = 0.5\cos x$ and $y = -2.5\cos x$ .

Your graphical representation should be similar to Figure 4.7. [Figure]

Figure 4.7: Graphical representation of $y = \cos x$ , $y = 3\cos x$ , $y = 0.5\cos x$ and $y = -2.5\cos x$ 3. The functions you have can be represented generally as $\mathrm{A}\cos x$ where $\mathrm{A}$ assumes different values. 4. Post your graphical illustration on the walls of your class, move around and observe how your classmates did theirs. 5. Now, record your observation on the different functions as the value of $\mathrm{A}$ changes and share with your classmates.

Example 4.1

Plot the graphs of $y = 2\sin x$ and $y = 4\sin x$ on the same axes and: a. Explain how the amplitude affects the graph without changing the period. b. Justify, using both algebraic reasoning and a diagram, why the maximum and minimum values change as they do.

Solution

$\boldsymbol{x}$ 0 $\boldsymbol{90^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{360^\circ}$ $2\sin(x)$ 0 $2$ 0 $-2$ 0 $4\sin(x)$ 0 $4$ 0 $-4$ 0

[Figure] Figure 4.8: Graphical representation of $y = 2\sin x$ and $y = 4\sin x$ a. Compare $y = 2$ sin $x$ and $y = 4$ sin $x$ For $y = A$ sin $x$ , amplitude $= |A|,$ period $= 2\pi = 360^\circ.$ So $2$ sin $x$ has amplitude 2; $4$ sin $x$ has amplitude 4; both have period $2\pi \ (or\ 360^\circ)$ . b. Max/min & range (algebraic justification): Since $-1 \leq$ sin $x \leq 1$ , multiplying by A gives $-|A| \leq A$ sin $x \leq |A|.$ Thus: $y = 2$ sin $x$ : range $[-2, 2];$ maxima $= 2$ at $x = \frac{\pi}{2} + 2k\pi$ ; minima $= -2$ at $x = \frac{3}{2}\pi + 2k\pi$ . $y = 4$ sin $x$ : range $[-4, 4];$ same $x$ -locations for peaks/troughs, but doubled height. Geometrically there is a vertical stretch by factor 2 from the first to the second as seen in the graphical illustration.

Example 4.2

A student claims that the graph of $y = -3\cos x$ is the same as $y = 3\cos(x + 180^\circ)$ a. Determine whether the claim is correct. b. Support your answer with a clear explanation, a sketch, and an example showing the correspondence between the two graphs.

Solution

a. Using the identity $\cos(x + \pi) = -\cos x$ , where $\pi \equiv 180^\circ$ . Then $3\cos(x + \pi) = 3(-\cos x) = -3\cos x,$ so the claim is correct. b. Point correspondence check: When $x = 0:$ LHS RHS $-3\cos 0 = -3$ $3\cos\pi = 3(-1) = -3$ When $x = \frac{\pi}{2}$ LHS RHS $-3\cos\left(\frac{\pi}{2}\right) = 0$ $3\cos\left(\frac{3\pi}{2}\right) = 0$ Using table of values $\boldsymbol{x}$ 0 $\boldsymbol{90^\circ}$ $\boldsymbol{180^\circ}$ $\boldsymbol{270^\circ}$ $\boldsymbol{360^\circ}$ $-3\cos x$ $-3$ $0$ $3$ $0$ $-3$ $3\cos(x + 180^\circ)$ $-3$ $0$ $3$ $0$ $-3$ The whole graph is a phase shift by $\pi$ (half-period) of $3\cos x$ , which exactly flips the sign. [Figure] [Figure] Figure 4.9: Graphical Illustration of $y = -3\cos x$ Figure 4.10: Graphical Illustration of $y = 3\cos(x + 180^\circ)$

Transformation of the form sin Bx or cos(Bx), A sin (Bx -

C) or A cos(Bx - C)

Recall in Year 1 Section 7 we learnt about radian measure. We explored how to convert radians to degrees and vice versa. Let us go through this example quickly as a refresher.

Example 4.3

Convert the following to degrees.

1. 1 radian
2. $\pi$ radians
3. 4 radians
4. 5 radians

Solution

1. $1 \times \frac{180^\circ}{\pi} = 57.30^\circ$
2. $\pi \times \frac{180^\circ}{\pi} = 180^\circ$
3. $4 \times \frac{180^\circ}{\pi} = 229.18^\circ$
4. $5 \times \frac{180^\circ}{\pi} = 286.48^\circ$

Example 4.4

Express the following in radians.

1. $180^\circ$
2. $75^\circ$
3. $32^\circ$

Solution

1. $180^\circ \times \frac{\pi}{180^\circ} = \pi$
2. $75^\circ \times \frac{\pi}{180^\circ} = \frac{5}{12}\pi$
3. $32^\circ \times \frac{\pi}{180^\circ} = \frac{8}{45}\pi$

Now that we remember how to change radians into degrees and vice versa, we shall make use of the two interchangeably while dealing with transformation of trigonometric functions. The graphs of the functions $\sin(\mathrm{B}x)$ or $\cos(\mathrm{B}x)$ have amplitude $|A|$ and period $\frac{2\pi}{B}$ or $\frac{360^\circ}{B}$ . The appropriate interval of one complete period is $\left[0, \frac{2\pi}{B}\right]$ or $[0, 360^\circ]$ . Note • $\sin(\mathrm{B}x) = \mathrm{A}\sin(\mathrm{B}x - C)$ where $\mathrm{A} = 1$ , $\mathrm{C} = 0$ • $\cos(\mathrm{B}x) = \mathrm{A}\cos(\mathrm{B}x - C)$ where $\mathrm{A} = 1$ , $\mathrm{C} = 0$

Activity 4.4 Transformation of the form A sin(Bx - C) or A cos(Bx - C)

1. Exploration on Amplitude $(\mathrm{A})$ : Watch the video https://youtu.be/Wvpa6He9qf0 and pay close attention to the changes happening to the functions.

a. What do you observe about the maximum and minimum values of the graph when you change the value of $\mathrm{A}$ ? b. If you set $\mathrm{A} = 1$ , then increase it to $\mathrm{A} = 3$ , what changes and what stays the same? c. Watch this video https://youtu.be/q7HS9tnPrQU . What happens to the graph when $\mathrm{A}$ becomes negative?

d. Describe the effect of a negative amplitude. e. Based on your observations, how would you define the amplitude of the function? 2. Exploration on Frequency ( $\mathrm{B}$ ) and Period. Watch this video https://youtu.be/ lQbxO7YDhiI and answer the questions that follow.

a. What happens to the distance between repeating patterns (periods) as you adjust the values for $\mathrm{B}$ ? b. Compare the graph of $\sin(x)$ and $\sin(2x)$ and identify the difference on how often each repeat. c. How would you calculate the period of the graph using $\mathrm{B}$ ? d. What do you notice about the graphs when $\mathrm{B} = 0$ ? 3. Exploration for Phase Shift a. Draw the graphs of $y = \sin x$ , $y = \sin(x - 2)$ and $y = \sin(x + 2)$ using the interval, $0 \leq x \leq 360^\circ$ . b. What happens to the position of the starting point of the sine wave when $\mathrm{C}$ changes? c. What is the horizontal shift from $C = 0$ to when $\mathrm{C} = 2$ ? d. What is the horizontal shift from $C = 0$ to when $\mathrm{C} = -2$ ? e. What is the horizontal shift when $\mathrm{C} = 0.5$ ? f. Set $\mathrm{B} = 1$ and make changes to $\mathrm{C}$ and explore how far the graph shifts with each change. g. Explore same using the cosine function. 4. Discuss your general observations about the transformation of the form $\mathrm{A}\sin(\mathrm{B}x - \mathrm{C})$ and $\mathrm{A}\cos(\mathrm{B}x - \mathrm{C})$ with your classmates.

Note Without graphing software, you would have to build a table of values for each function with different values for $\mathrm{A, B}$ and $\mathrm{C}$ and graph them to observe the changes in Amplitude, Period and Phase Shift. Table 4.2: Conclusions for transformations of the form $A\sin(Bx - C)$ or $A\cos(Bx - C)$ $\mathbf{A}\sin(\mathbf{B}x - \mathbf{C})$ or $\mathbf{A}\cos(\mathbf{B}x - \mathbf{C})$ Amplitude $= |\mathbf{A}|$ It has a period of $\frac{2\pi}{B}$ or $\frac{360^\circ}{B}$ The interval of one cycle is $\frac{C}{B} \leq x \leq \frac{C}{B} + \frac{360^\circ}{B}$ Note “ $Bx - C$ ” in the equation represents a shift/translation to the right while “ $Bx + C$ ” represents a shift/translation to the left.

Example 4.5

Describe the transformations performed on the graph of $y = 3\sin(x)$ to obtain the graph of one cycle of $y = 4\sin(2x - 60^\circ).$ Refer to Figure 4.11 for the graphs. The broken lines indicate a second period thus the first period ends at $180^\circ$ for $y = 4\sin(2x - 60^\circ)$ . [Figure] Figure 4.11: Graphical Illustration of $y = 4\sin(2x - 60^\circ)$ and $y = 3\sin(x)$ .

Solution

The graph of $y = 4\sin(2x - 60^\circ)$ has a period of $180^\circ$ , i.e., from $30^\circ$ to $210^\circ$ which is half the period of $y = 3\sin(x)$ . This implies a horizontal stretch of the graph of $y = 3\sin(x)$ by a scale factor of $\frac{1}{2}$ . The graph of $y = 4\sin(2x - 60^\circ)$ starts from $x = 30^\circ$ while the graph of $y = 3\sin(x)$ starts from the origin. This depicts a horizontal shift to the right by $30^\circ$ confirmed numerically by $\frac{C}{B} = \frac{60^\circ}{2} =$ $30^\circ$ . The maximum and minimum values of the graph of $y = 4\sin(2x - 60^\circ)$ are $4$ and $-4$ respectively, whereas those of $y = 3\sin x$ are $3$ and $-3$ . This implies a vertical stretch by a scale factor of $\frac{4}{3}$ . The midline remains unchanged at $y = 0$ , showing that there is no vertical translation.

Example 4.6

The up-and-down motion of a child on a rope swing can be modelled by $y = 3\sin\left(2x - \frac{\pi}{6}\right).$ a. Without graphing, describe when the child will be at the highest point, lowest point, and midline. b. Explain how amplitude and frequency affect the swing’s motion. c. Why would a parent need to know the maximum height of the swing?

Solution

a. For $y = 3\sin\left(2x - \frac{\pi}{6}\right)$ , $A = 3, B = 2, C = \frac{\pi}{6}$ . Period $(T) = \frac{2\pi}{B} = \frac{2\pi}{2} = \pi$ Phase shift $\left(\frac{C}{B}\right) = \left(\frac{\pi}{6}\right) \div 2 = \frac{\pi}{12}$ to the right. Since there is no vertical shift, the midline is $y = 0$ . Since $-1 \leq$ sin $x \leq 1$ , multiplying by A gives $-|A| \leq A$ sin $x \leq |A|.$ For $y = 3\sin\left(2x - \frac{\pi}{6}\right)$ : range $[-3, 3];$ Let $\theta = 2x - \frac{\pi}{6}$ then $y = 3\sin\theta$ For the midline, $\theta = k\pi, (k \in \mathbb{Z})$ $2x - \frac{\pi}{6} = k\pi$ $x = \frac{k\pi}{2} + \frac{\pi}{12}.$ So, the child is at midline when $x = \frac{\pi}{12}, \frac{7\pi}{12}, \frac{13\pi}{12}, \ldots$ For maximum height, the swing reaches its highest point when $\sin\theta = 1$ $\theta = \frac{\pi}{2} + 2k\pi$ $2x - \frac{\pi}{6} = \frac{\pi}{2} + 2k\pi$ $x = \frac{\pi}{3} + k\pi$ At such $x$ values, $y = 3$ For minimum height, the swing reaches its lowest point when $\sin\theta = -1$ $\theta = \frac{3\pi}{2} + 2k\pi$ $2x - \frac{\pi}{6} = \frac{3\pi}{2} + 2k\pi$

$x = \frac{5\pi}{6} + k\pi$ At such $x$ values, $y = -3$ b. For the amplitude, the child swings 3 units above and below the midline. Thus, the larger the amplitude, the higher the swing. For the frequency, since the period is $\pi$ , the swing completes one full cycle every $\pi$ units of time. A larger $B$ will make the swing faster. c. A parent needs to know the maximum height to decide if the swing is too high for the child’s age. It helps ensure the swing does not collide with the frame or nearby objects and aids in taking all necessary precautions.

Example 4.7

Draw the graph of $y = 2\cos\left(3x - \frac{\pi}{2}\right)$ and describe the transformations performed on the graph of $y = \cos x$ to obtain $y = 2\cos\left(3x - \frac{\pi}{2}\right)$ .

Solution

The broken lines indicate a second period. [Figure]

Figure 4.12: Graphical Illustration of $y = 2\cos\left(3x - \frac{\pi}{2}\right)$ and $y = \cos(x)$ .

Amplitude $= 2$ Period $= \frac{2\pi}{B} = \frac{2\pi}{3}$ Step $= \frac{2\pi}{3} \times \frac{1}{2} = \frac{\pi}{3}$ . Phase shift (Left endpoint): $\frac{C}{B} = \frac{\pi}{2} \div 3 = \frac{\pi}{6}$ Right endpoint: $\frac{C}{B} +$ period $= \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6}$ The interval is $\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$ Points are $\left(\frac{\pi}{6}, 2\right), \left(\frac{\pi}{3}, 0\right), \left(\frac{\pi}{2}, -2\right), \left(\frac{2\pi}{3}, 0\right)$ and $\left(\frac{5\pi}{6}, 2\right)$ . The graph of $y = 2\cos\left(3x - \frac{\pi}{2}\right)$ has period $\frac{2\pi}{3}$ , i.e., one cycle runs from $x = \frac{\pi}{6}$ , to $x = \frac{5\pi}{6}$ which is one third of the period of $y = \cos x$ . This corresponds to a horizontal shrink by a factor of 3. The graph is shifted right by $\frac{\pi}{6}$ . The maximum and minimum values are $2$ and $-2$ , so the graph is vertically stretched by a factor of 2. The midline remains $y = 0$ , so there is no vertical translation.

Transformation of the form A sin (Bx - C) + D or A cos

(Bx - C) + D

Activity 4. 5 Daylight hours modelling

1. Working in pairs, go through this scenario: In Accra, Ghana, the number of daylight hours varies slightly over the course of a year due to the tilt of the Earth's axis, but it remains within a narrow range, approximately 11.9 to 12.4 hours per day. The daylight hours over 12 months can be modelled by the function $\mathrm{M}(t) = \mathrm{A}\cos(\mathrm{B}t - \mathrm{C}) + \mathrm{D}$ and are presented in the table. Month Jan.(0) Feb.(1) Mar.(2) Apr. (3) May(4) Daylight 11.9 12 12.1 12.3 12.4 hours $(t)$

Jun.(5) Jul.(6) Aug. (7) Sep.(8) Oct. (9) Nov.(10) Dec. (11) 12.3 12.2 12.1 12.0 11.9 11.8 11.9

2. Using an appropriate scale, plot the data presented on your graph sheet.
3. Join the points to form a curve.
4. From your graph, estimate the: a. Average daylight $(\mathrm{D})$

b. Amplitude of the curve c. The period $\left(\frac{2\pi}{\mathrm{B}}\right)$ d. The phase shift $\left(\frac{\mathrm{C}}{\mathrm{B}}\right)$ 5. Substitute the values obtained to build your model. 6. According to your model, in which months will Accra experience the shortest and longest daylight hours? 7. How many hours of daylight are there in August? Use your model to estimate. 8. If you lived in a country near the poles, how would the amplitude and midline of the graph change? 9. Why is a cosine graph a better model than a straight line for this situation?

Example 4.8

Draw the graph of $y = -3\sin\left(2x + \frac{\pi}{3}\right) + 1$ and describe the transformations performed on the graph of $y = \sin(x)$ to obtain one cycle of the given graph.

Solution

Comparing $y = -3\sin\left(2x + \frac{\pi}{3}\right) + 1$ to $y = A\sin(Bx - C) + D$ , $A = -3, B = 2, C = -\frac{\pi}{3},$ $D = 1$ . Hence, the Amplitude $|A| = 3$ but reflects in the midline because $A < 0$ . Period $= \frac{2\pi}{B} = \frac{2\pi}{2} = \pi$ Phase shift (left endpoint) $= \frac{C}{B} = -\frac{\pi}{3} \div 2 = -\frac{\pi}{6}.$ Right endpoint: $\frac{C}{B} + period = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}.$ Points in one cycle are $\left(\frac{\pi}{6}, 1\right), \left(\frac{\pi}{12}, -2\right), \left(\frac{\pi}{3}, 1\right), \left(\frac{7\pi}{12}, 4\right), \left(\frac{5\pi}{6}, 1\right)$ . There is an upward vertical shift of 1 because $D = 1$ . Hence the midline $y = 1$ . The broken lines in Figure: 4.13 indicate a second period. [Figure] Figure: 4.13: Graphical Illustration of $y = -3\sin\left(2x + \frac{\pi}{3}\right) + 1$ and $y = \sin(x)\ y = \sin(x)$ .

The graph of $y = -3\sin\left(2x + \frac{\pi}{3}\right) + 1$ has period $\pi$ (one cycle from $x = -\frac{\pi}{6}$ to $\frac{5\pi}{6}$ ), which is half the period of $y = \sin x$ ; this corresponds to a horizontal shrink by factor 2. The “ $\frac{\pi}{3}$ ” inside causes a shift left by $\frac{\pi}{6}$ (since $\frac{C}{B} = -\frac{\pi}{6}$ ). The amplitude is 3 but $A$ is negative, so the graph is reflected in its midline and vertically stretched by a factor of 3. Finally, the whole graph is shifted up by 1 unit so the midline is $y = 1$ . Because $A$ is negative, it reverses the usual rising/falling pattern of $y = \sin x.$

Example 4.9

Compare the graphs of $y = \cos x$ and $y = -\frac{1}{2}\cos(4x - \pi) - 3$ . Explain how the negative sign, the coefficient of $x$ and the constants inside and outside the function affect the graph of $y = \cos x$ .

Solution

For $y = \cos x$ , Amplitude $=$ 1, Period $= 2\pi$ , Midline $y = 0$ , Range $=$ $[-1, 1]$ , Starts at $(0, 1)$ , symmetric about the $y$ -axis. Compared to $y = \cos x$ the graph of $y = -\frac{1}{2}\cos(4x - \pi) - 3$ is reflected in the $x$ -axis because of the negative sign of A. The coefficient of the cosine being $-\frac{1}{2}$ , implies an amplitude $|A|$ of $\frac{1}{2}$ thus vertically shrinking the graph by $\frac{1}{2}$ . Fo r the coefficient of $x$ the “4” inside $\cos(4x - \pi),$ the period becomes $\frac{2\pi}{4} = \frac{\pi}{2}$ . This means the cosine wave completes a cycle four times faster, a horizontal compression by factor 4. The constant inside $(-\pi)$ , causes a phase shift $\left(\frac{C}{B}\right) = \frac{\pi}{4}$ to the right. The constant outside $(-3)$ , translates the entire graph down by 3 units hence the midline moves from $y = 0$ to $y = -3$ . The range also shifts from $[-1, 1]$ to $[-3.5, -2.5].$

In Section 3, you learnt about Parabolas and how to determine the minimum and maximum points of such functions. We are going to apply that knowledge to determine minimum and maximum values of trigonometric graphs. When the graphs are sketched, we can use inspection to determine the maxima/minima values. Another way to find the maxima and minima of trigonometric expressions of the form $A\sin(B\theta \pm C) + D$ or $A\cos(B\theta \pm C) + D$ without sketching is to multiply $A$ by $1$ and $-1$ and add the $D$ .

Mathematically: Minimum or maximum value $= A + D$ or Minimum or maximum value $= -A + D$ Note

1. $D$ could be negative or positive.
2. The smaller values of “ $A + D$ ” and “ $-A + D$ ” is the minimum value of the expression.
3. The larger values of “ $A + D$ ” and “ $-A + D$ ” is the maximum value of the expression.

Example 4.10

If $y = 5\sin(\theta)$ , determine the maximum and minimum value of $y$ .

Solution

Rewriting the function in the form $A\sin(B\theta \pm C) + D$ , $A = 5$ and $D = 0$ Using $\mathrm{A}(-1) + \mathrm{D}$ , substitute the values of $\mathrm{A}$ and $\mathrm{D}$ Now, $\mathrm{A}(-1) + \mathrm{D} = 5(-1) + 0 = -5$ Using $A(1) + D$ , $A(1) + D = 5(1) + 0 = 5$ Since $-5 < 5$ , $-5$ is the minimum value while 5 is the maximum value of $y$

Example 4.11

Given the function $f(\phi) = 3\cos(\phi)$ , determine the maximum and minimum value of $f(\phi)$ .

Solution

Rewriting the function in the form $A\cos(B\theta \pm C) + D$ , $A = 3$ and $D = 0$ Using $\mathrm{A}(-1) + \mathrm{D}$ , substitute the values of $\mathrm{A}$ and $\mathrm{D}$ Now, $\mathrm{A}(-1) + \mathrm{D} = 3(-1) + 0 = -3$ Using $A(1) + D$ , $A(1) + D = 3(1) + 0 = 3$ Since $-3 < 3$ , $-3$ is the minimum value while 3 is the maximum value of $f(\phi)$ .

Example 4.12

Determine the minima and maxima values of: a. $\mathrm{h}(x) = -3\cos\left(\frac{1}{4}x\right) + 2$ b. $\mathrm{f}(\gamma) = 3\sin(2\gamma) + 4$

Solution

a. From the function, $A = -3$ and $D = 2$ For $\mathrm{A} + \mathrm{D}$ , $\mathrm{A} + \mathrm{D} = -3 + 2 = -1$ For $-\mathrm{A} + \mathrm{D}$ , $-\mathrm{A} + \mathrm{D} = 3 + 2 = 5$ For $\mathrm{h}(x)$ , t he maximum value is $5$ and the minimum value is $-1$ . b. From the function, $A = 3$ and $D = 4$ For $\mathrm{A} + \mathrm{D}$ , $\mathrm{A} + \mathrm{D} = 3 + 4 = 7$ For $-\mathrm{A} + \mathrm{D}$ , $-\mathrm{A} + \mathrm{D} = -3 + 4 = 1$ For $\mathrm{f}(\gamma)$ , t he maximum value is $7$ and the minimum value is $1$ .

In year 2 section 9, you learnt about compound angle identities, multiple and double angle identities. You engaged in activities where you researched the algebraic proof of the identities and solved examples to know when to apply these identities. The compound angle identities are listed below to serve as a reminder.

1. $\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$

2. $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$

3. $\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)$

4. $\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$

5. $\tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}$

6. $\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}$ The multiple angle identities are:

7. $\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$

8. $\cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha)$

9. $\tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)}$ We are going to make use of these identities in deriving harmonic functions.

Expression of a cos (0) + b sin(0) as R cos(0 - a)

We have $a\cos(\theta) + b\sin(\theta)$ as a trigonometric expression made up of two trigonometric functions. It can however, be expressed as $R\cos(\theta - \alpha)$ which comprises of only one trigonometric identity called a Harmonic Identity. There are several of such harmonic identities but we will go through an activity to generate the first one while you carry out personal research to prove the others.

Activity 4.6 Deriving Harmonic Identity R cos(0 - a)

1. Graph the function $y = 3\cos(\theta) + 4\sin(\theta)$ and analyse it.

2. Do you think the function can be rewritten as a single sine or cosine function?

3. Express $3\cos(\theta) + 4\sin(\theta)$ generally as $a\cos(\theta) + b\sin(\theta)$

4. Expand $R\cos(\theta - \alpha)$ using the appropriate compound angle identity. You should get $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\alpha\sin\theta$ . When rearranged, should be $R\cos\alpha\cos\theta + R\sin\alpha\sin\theta$

5. Compare the rearranged function to $a\cos(\theta) + b\sin(\theta)$ $R\cos\alpha\cos\theta + R\sin\alpha\sin\theta$ and $a\cos(\theta) + b\sin(\theta)$

6. Equate and match the coefficients of $\cos\theta$ and $\sin\theta$ to get; $a = R\cos\alpha$ , $b = R\sin\alpha$ This means that $a\cos(\theta) + b\sin(\theta) = R\cos(\theta - \alpha)$

7. From Step 5, square each of the terms on your LHS and RHS and add the two equations. $a^2 = R^2\cos^2\alpha$ , $b^2 = R^2\sin^2\alpha$ $a^2 + b^2 = R^2(\cos^2\alpha + \sin^2\alpha)$

8. Recall from Pythagorean identities that $\cos^2\alpha + \sin^2\alpha = 1$ . Thus $a^2 + b^2 = R^2$ .

9. Take the square root of both sides of the equation; $\sqrt{a^2 + b^2} = R$

10. Does the final equation look familiar?

11. Yes, it looks exactly like Pythagoras’ theorem.

12. From Step 5, taking a ratio of the two equations results in $\frac{R\sin\alpha}{R\cos\alpha} = \frac{b}{a}$

13. This can be simplified further to obtain $\tan\alpha = \frac{b}{a}$ and $\alpha = \tan^{-1}\left(\frac{b}{a}\right).$

14. Discuss your observations from this activity with a classmate and your teacher. Trigonometric Expression Harmonic identities $a\cos(\theta) + b\sin(\theta) + c$ $R\cos(\theta - \alpha) + c$ $a\cos(\theta) - b\sin(\theta) + c$ $R\cos(\theta + \alpha) + c$ $a\cos(\theta) + b\sin(\theta) + c$ $R\sin(\theta + \alpha) + c$ $a\cos(\theta) - b\sin(\theta) + c$ $R\sin(\theta - \alpha) + c$

Example 4.13

Express $3\cos(\theta) - 2\sin(\theta)$ in the form $R\cos(\theta + \alpha)$ .

Solution

Comparing $3\cos(\theta) - 2\sin(\theta)$ to $a\cos(\theta) - b\sin(\theta) + c$ , $a = 3$ , $b = 2$ and $c = 0$ . Recall that $R = \sqrt{a^2 + b^2}$ $R = \sqrt{3^2 + 2^2}$ $R = \sqrt{13}$ units Also, $\alpha = \tan^{-1}\left(\frac{b}{a}\right).$ Thus $\alpha = \tan^{-1}\left(\frac{2}{3}\right) = 33.6901^\circ$ Now, $R\cos(\theta + \alpha) = \sqrt{13}\cos(\alpha + 33.6901^\circ)$ $\therefore 3\cos(\theta) - 2\sin(\theta) = \sqrt{13}\cos(\alpha + 33.6901^\circ)$ .

Example 4.14

The height of a point on the surface of the sea due to a wave is given by $y = 7\cos(x) + 24\sin(x)$ . Calculate the maximum height of the wave above sea level.

Solution

Given functions in the form $A\sin(B\theta \pm C) + D$ or $A\cos(B\theta \pm C) + D$ , the maximum value is $A + D$ .

[Figure] Figure: 4.14: Wave Our function $y = 7\cos(x) + 24\sin(x)$ can be expressed as $A\sin(B\theta \pm C) + D$ or $A\cos(B\theta \pm C) + D$ when rewritten as a harmonic identity. Step 1 : Express the given function as a harmonic identity Comparing $7\cos(x) + 24\sin(x)$ to $a\cos(\theta) + b\sin(\theta) + c$ , $a = 7$ , $b = 24$ and $c = 0$ . Making use of a harmonic identity $R\sin(\theta + \alpha) + c$ Now, $7\cos(x) + 24\sin(x) = R\sin(\theta + \alpha) + c$ but $c = 0$ Step 2 : Compare the harmonic identity chosen to $\mathrm{A}\sin(B\theta \pm C) + D$ or $A\cos(B\theta \pm C) + D$ to determine the maximum value. Comparing $R\sin(\theta + \alpha) + c$ to $\mathrm{A}\sin(B\theta \pm C) + D$ , $A = R, B = 1, C = \alpha, D = c$ But maximum value is $A + D \equiv R + c$ Since $c = 0,$ the maximum value is $R + 0$ hence the maximum height will be; $R = \sqrt{a^2 + b^2}$ $R = \sqrt{7^2 + 24^2}$ $= \sqrt{576 + 49}$ $= 25$ $\therefore$ The maximum height of the wave above sea level is 25 units. Note In using any of the harmonic identities; • maximum value is $R + c$ • minimum value is $-R + c$

Example 4.15

The displacement of a pendulum bob from its rest position is modelled by: $y = 5\sin(x) - 12\cos(x).$ Find the greatest displacement of the pendulum from the equilibrium position.

Solution

Step 1 : Rearrange the function $5\sin(x) - 12\cos(x) = -12\cos(x) + 5\sin(x)$ Now, $y = -12\cos(x) + 5\sin(x)$ Step 2 : Express the given function as a harmonic function (identity) Now, $-12\cos(x) + 5\sin(x) = a\cos(\theta) + b\sin(\theta) + c$ , where $a = -12, b = 5, c = 0$ . Also, $a\cos(\theta) + b\sin(\theta) + c = R\cos(\theta - \alpha) + c$ Step 3: Determine the maximum value Since $c = 0,$ maximum value will be $R$ . $R = \sqrt{a^2 + b^2}$ $R = \sqrt{(-12)^2 + 5^2}$ $= \sqrt{169}$ $= 13$ $\therefore$ the greatest displacement of the pendulum from the equilibrium position is $13$ units.

Example 4.16

The vertical displacement of a vibrating spring is given by $y = 9\cos(\theta) - 40\sin(\theta) - 3$ . Calculate the maximum and minimum displacement of the spring.

Solution

Step 1 : Express the given function as a harmonic function (identity) $9\cos(\theta) - 40\sin(\theta) - 3 = a\cos(\theta) - b\sin(\theta) + c$ , where $a = 9, b = 40, c = -3$ . Also, $a\cos(\theta) - b\sin(\theta) + c = R\sin(\theta - \alpha) + c$ [Figure] Figure: 4.15: Displaced Spring

Step 2 : Determine the maximum value Since $c = -3,$ maximum value will be $R + c$ . $R + c = \sqrt{a^2 + b^2} + c$ $R + c = \sqrt{9^2 + 40^2} + (-3)$ $= \sqrt{1681} - 3$ $= 41 - 3$ $= 38$ Step 3 : Determine the minimum value Since $c = -3,$ and $R = 41,$ the minimum value will be $-R + c$ $-R + c = -41 - 3$ $= -44$ $\therefore$ the maximum and minimum displacement of the spring is $38$ units and $-41$ units respectively.