Degrees and Radians

Whilst we may be more familiar with degrees, radians are more convenient to work with when doing calculus.
One radian is the angle subtended at the centre of a unit circle by an arc of 1 unit in length. \[360^{\circ} = 2 \pi^{c}\]

Note: In general, we do not put any symbol to represent radians. Assume that all questions are in radians unless there is a degrees symbol. If a question is given to you in degrees, it should be answered in degrees.

Introduction

In this chapter we deal with three functions - sine, cosine and tangent. In previous years, you will have been introduced to these to find the angle or side lengths in right angle triangles.

\[ \text{ } \\ \begin{aligned} \sin(\theta) &= \frac{\text{Opposite}}{\text{Hypotenuse}} \\ \text{ }\\ \cos(\theta) &= \frac{\text{Adjacent}}{\text{Hypotenuse}} \\ \text{ }\\ \tan(\theta) &= \frac{\text{Opposite}}{\text{Adjacent}} \\ \end{aligned}\\ \]

Looking at the first letter of each equation in that order we have:
SOH CAH TOA
A mnemonic which can be used to remember this is: Some Old Hippy, Caught Another Hippy, Tripping On Acid.
While this knowledge is useful, we are now going to consider all angles, not only those in a right angled triangle.

Above we have a circle with radius one. This is called the unit circle. The x-coordinate of any point on this circle is equal to \(\cos(\theta)\) and the y-coordinate of any point on this circle is equal to \(\sin(\theta)\). The angle theta is the angle going anti - clockwise from the red dashed line and any line from the edge of the circle to the origin as shown in the above diagram.

The tangent function is defined as the sine function divided by the cosine function. Because of this, whenever the cosine function is equal to zero, the tangent function is not defined. So in summary we have, \[ \begin{aligned} \sin(\theta) &= y \\ \text{ }\\ \cos(\theta) &= x \\ \text{ }\\ \tan(\theta) &= \frac{\sin(\theta)}{\cos(\theta)} \\ \text{ }\\ &= \frac{y}{x} \end{aligned} \]

Using these definitions, sine is positive in quadrant one and two, cosine is positive in quadrant one and four, and tangent is positive in quadrants one and three.

The easiest way to remember in which quadrants each function is negative or positive is simply remembering which quadrants the functions are positive. If you read clockwise from quadrant one (1,4,3,2) we spell ACTS.
A key property of circular functions is that they are periodic. If we look at the unit circle, if we add another \(360^{\circ}\) or \(2\pi\) to \(\theta\) you will end up at the same point. Thus, adding or subtracting \(2\pi\) to the argument (inside the brackets) of the sine, cosine and tangent functions will not change their value.

Exact Values

Knowing all of the exact values is imperative. However, memorising them is very difficult and not recommended! Simply remember the two triangles below. The one on the left is an isosceles triangle with the two equal sides being of length 1. The one on the right is an equilateral triangle with side lengths of 2 which has been cut in half. You can then use pythagoras’ theorem in both of these cases to find any missing side lengths.

Supplementary Angle Formulae

Your teacher may have demonstrate how you can get all of the formulae below form the unit circle. We have found that the vast majority of students find this too complex and provides little help. Hence, this has not been explained.

Note that adding \(2\pi\) to the argument on the left hand side will mean that the equalities still stand as the sine, cosine and tangent functions are periodic. For example: \[\sin(3\pi-x) = \sin(\pi - x) = \sin(x)\] There are 9 formulae above. Memorising each individual one is very difficult. Supplementary angle formulae involve various multiples of \(\pi\). This is how we recommend you get around memorising all the formulae. Let’s always assume that \(x\) is an angle between \(\frac{\pi}{2}\) and \(0\). In this case, \(\sin(x)\) and \(\cos(x)\) are always positive.

Using \(\sin(\pi-x)\) as an example, \(\pi-x\) would be located in quadrant two. Sine is positive in quadrant two, so, \(\sin(\pi-x)\) = \(\sin(x)\).

Using \(\cos(\pi+x)\) as an example, \(\pi+x\) would be located in quadrant three. Cosine is negative in quadrant three, so, \(\cos(\pi+x)\) = \(-cos(x)\).

Using \(\tan(2\pi-x)\) as an example, \(2\pi-x\) would be located in quadrant four. The tangent function is negative in quadrant four, so, \(\tan(2\pi-x)\) = -\(\tan(x)\).

Complementary Angle Formulae

Again, your teacher may have demonstrate how you can get all of the formulae below form the unit circle. We have found that the vast majority of students find this too complex and provides little help. Hence, this has not been explained.

Note that adding \(2\pi\) to the argument on the left hand side will mean that the equalities still stand as the sine, cosine and tangent functions are periodic. For example: \[\sin(\frac{5\pi}{2}-x) = \sin(\frac{\pi}{2}-x) = \cos(x)\] There are 8 formulae above. Memorising each individual one is very difficult. Complementary angle formulae involve odd multiples of \(\frac{\pi}{2}\). This is how we recommend you get around memorising all the formulae. Again, let’s always assume that \(x\) is an angle between \(\frac{\pi}{2}\) and \(0\). In this case, \(\sin(x)\) and \(\cos(x)\) are always positive.

Using \(\sin(\frac{\pi}{2}-x)\) as an example, \(\frac{\pi}{2}\) would be located in quadrant one. Sine is positive in quadrant two, so, \(\sin(\frac{\pi}{2}-x) = \cos(x)\).

Using \(\frac{3\pi}{2}-x\) as an example, \(\frac{3\pi}{2}-x\) would be located in quadrant two. Cosine is negative in quadrant three, so, \(\cos(\frac{3\pi}{2}-x) =- \sin(x)\).

Note the difference between supplementary angle and complementary angle formulae. Supplementary angle formulae establish a relationship between the same circular functions and are relative to \(\pi\), while the complementary angle formulae establish a relationship between sine and cosine and are relative to \(\frac{\pi}{2}\).

Pythagorean Identity

\[\sin^2(\theta)+\cos^2(\theta)=1\]

\[ \text{Example 6.1: Given that } \sin(x)=-\frac{2}{5} \text{ where } x \in [\pi, \frac{3\pi}{2}), \text{ find } \cos(x) \text{ and }\tan(x).\\ \text{ }\\ \begin{aligned} \text{} \sin^2(x) + \cos^2(x) &= 1\\ \because \sin(x) &= -\frac{2}{5}\\ \therefore (-\frac{2}{5})^2 + \cos^2(x) &= 1\\ \cos^2(x) &= 1-\frac{4}{25}\\ \cos^2(x) &= \frac{21}{25}\\ \cos(x) &= \pm \sqrt{\frac{21}{25}}\\ \therefore \cos(x) &= -\frac{\sqrt{21}}{5}\quad \text{As }x \in [\pi, \frac{3\pi}{2})\\ \text{ }\\ \tan(x) &= \frac{\sin(x)}{\cos(x)}\\ \tan(x) &= -\frac{2}{5} \div -\frac{\sqrt{21}}{5}\\ \tan(x) &= \frac{2}{\sqrt{21}}\\ \end{aligned}\\ \]

Graphs of Circular Functions

The basic shape of circular functions are shown below.

The Amplitude and Period

The amplitude of a function is the distance from the centreline (the horizontal line which lies in between the peaks and troughs) to the peak or trough. That is half the distance from the trough to peak.
The period is the interval (horizontal distance) between each repetition.

A sine/cosine function is presented above. The length of the dotted vertical line is the amplitude. The length of the dotted horizontal line is the period.

Graphing Circular Functions

In general, unless told not to, all graphs must provide the coordinates and equations of any: x-intercepts, y-intercepts and asymptotes.

\[y = A\sin(nx+b)+c \text{ and } y = A\cos(nx+b)+c\]

\[y = A\tan(nx+b)+c\]

\[ \text{Example 6.2: Sketch the graph of } y = 2\sin(2x-\frac{\pi}{3})+1 \text{ for } x \in [-\pi, \pi].\\ \text{Label the coordinates of any y-intercept, x-intercept or endpoint.} \] \[ \text{1. y-intercept} \\ \begin{aligned} x&=0\\ y &= 2\sin(2\cdot 0-\frac{\pi}{3})+1 \\ y &= 2\sin(-\frac{\pi}{3}) + 1 \\ y&= 2 \cdot -\frac{\sqrt{3}}{2} + 1 \\ y&= -\sqrt{3} + 1 \\ \end{aligned}\\ \text{There is a y-intercept at } (0, -\sqrt{3} + 1) \\ \]

\[ \text{2. Range, period and centreline} \\ \text{Range : } [1,3] \\ \text{Period: } \frac{2\pi}{2} = \pi \\ \text{Centreline: } y = 1 \\ \] \[ \text{3. x-intercepts} \\ \text{For working, please view the “Solving circular functions” example, it is the exact same question} \\ x = -\frac{11\pi}{12}, \,-\frac{\pi}{4},\,\frac{\pi}{12}\,\frac{3\pi}{4} \\ \text{There are x-intercepts at: } (-\frac{11\pi}{12},0), \,(-\frac{\pi}{4},0), \,(\frac{\pi}{12},0),\,(\frac{3\pi}{4},0) \]

\[ 4. \, \sin(nx+b) = 1\\ \begin{aligned} x&\in[-\pi,\pi]\\ 2x&\in[-2\pi,2\pi] \\ \sin(2x-\frac{\pi}{3}) &=1 \\ \text{Basic Angle: }& \frac{\pi}{2} \\ \text{Let, } 2x-\frac{\pi}{3} &= \frac{\pi}{2} \\ 2x&=\frac{5\pi}{6} \\ x&=\frac{5\pi}{12} \\ \text{ }&\\ \sin(2x-\frac{\pi}{3})=1 \, &\rightarrow \, 2\sin(2x-\frac{\pi}{3}) + 1 = 3 \\ \end{aligned}\\ \text{so, the point } (\frac{5\pi}{12}, 3) \text{ lies on our graph} \]

\[ \text{5. Add or subtract the period from the x-coordinate in step 4} \\ \text{Period: } \pi \\ \text{Adding or subtracting } \pi \text{ from } \frac{5\pi}{12} \text{ we get:} \\ -\frac{7\pi}{12} \\ \text{All other values are outside of the domain} \\ \text{so, the point } (-\frac{7\pi}{12}, 3) \text{ lies on our graph} \]

\[ \text{6. Add or subtract half a period from the x-coordinate of one of the points in step 4 or 5} \\ \text{We will add half a period to the x-coordinate of the point found in step 5. \\ allowing the x-value to still be in the domain} \\ \text{Half a period: } \frac{\pi}{2} \\ -\frac{7\pi}{12} + \frac{\pi}{2} = -\frac{\pi}{12} \\ \text{At this point } \sin(2x-\frac{\pi}{3}) = -1 \\ \begin{aligned} \text{so, } y &= 2 \cdot (-1) + 1 \\ y &= -1 \\ \end{aligned}\\ \text{Thus, the point } (-\frac{\pi}{12}, -1) \text{ lies on our graph} \]

\[ \text{7. Add or subtract the period from the x-coordinate in step 6} \\ \text{Period: } \pi \\ \text{Adding or subtracting } \pi \text{ from } -\frac{\pi}{12} \text{ we get:} \\ \frac{11\pi}{12} \\ \text{All other values are outside of the domain} \\ \text{so, the point } (\frac{11\pi}{12}, -1) \text{ lies on our graph} \]

\[ \text{8. Endpoints} \\ \begin{aligned} x&=-\pi\\ y &= 2\sin(2\cdot (-\pi)-\frac{\pi}{3})+1 \\ y &= 2\sin(-\frac{7\pi}{3}) + 1 \\ y&= 2 \cdot -\frac{\sqrt{3}}{2} + 1 \\ y&= -\sqrt{3} + 1 \\ \text{ }& \\ x&=\pi\\ y &= 2\sin(2\cdot (-\pi)-\frac{\pi}{3})+1 \\ y &= 2\sin(\frac{5\pi}{3}) + 1 \\ y&= 2 \cdot -\frac{\sqrt{3}}{2} + 1 \\ y&= -\sqrt{3} + 1 \\ \end{aligned}\\ \text{The endpoints are at } (-\pi, -\sqrt{3}+1), \, (\pi, -\sqrt{3}+1) \]

\[ \text{9. Connect the dots and label everything} \]



\[ \text{Example 6.3: Sketch the graph of } y = \tan(x+\frac{\pi}{3})+1 \text{ for } x \in [-\pi, \pi]. \\ \text{Label the coordinates of any y-intercept, x-intercept, endpoints and the equations of any asymptotes.} \] \[ \text{1. y-intercept} \\ \begin{aligned} x&=0\\ y &= \tan(0+\frac{\pi}{3})+1 \\ y&= (\sqrt{3})+1\\ y &=\sqrt{3}+1 \\ \end{aligned}\\ \text{There is a y-intercept at } (0, \sqrt{3}+1) \]

\[ \text{2. X-intercepts} \\ \begin{aligned} x\in[-\pi,\pi]\\ x + \frac{\pi}{3} &\in [-\frac{2\pi}{3}, \frac{4\pi}{3}] \\ \text{ } \\ 0 &= \tan(x+\frac{\pi}{3})+1 \\ -1&=\tan(x+\frac{\pi}{3}) \\ \end{aligned}\\ \text{Basic Angle: } \frac{\pi}{4} \\ \text{Tan is negative in quadrant 2 and 4} \\ \text{ } \\ \begin{aligned} \tan(\pi-\frac{\pi}{4}) &= \tan(\frac{3\pi}{4}) = -1 \\ \tan(2\pi - \frac{\pi}{4} &= \tan(\frac{7\pi}{4}) = -1 \\ \end{aligned}\\ \text{ } \\ \begin{aligned} x + \frac{\pi}{3} &= \frac{3\pi}{4}, \frac{7\pi}{4}-2\pi \\ x&= \frac{9\pi}{12} - \frac{4\pi}{12}, -\frac{3\pi}{12} - \frac{4\pi}{12} \\ x&=\frac{5\pi}{12}, -\frac{7\pi}{12} \\ \end{aligned}\\ \text{There are x-intercepts at } (-\frac{7\pi}{12}, 0) \text{ and } (\frac{5\pi}{12}, 0) \]

\[ \text{3. Asymptotes} \\ \begin{aligned} x\in[-\pi,\pi]\\ x + \frac{\pi}{3} &\in [-\frac{2\pi}{3}, \frac{4\pi}{3}] \\ \text{ } \\ 0 &= \cos(x+\frac{\pi}{3}) \\ \end{aligned} \text{Basic Angle: } \frac{\pi}{2} \\ \begin{aligned} \cos(\frac{\pi}{2}) &= 0 \\ \cos(\frac{3\pi}{2}) &= 0 \\ \end{aligned}\\ \text{ } \\ \begin{aligned} x + \frac{\pi}{3} &= \frac{\pi}{2}, \frac{3\pi}{2}-2\pi \\ x&= \frac{3\pi}{6} - \frac{2\pi}{6},- \frac{3\pi}{6} - \frac{2\pi}{6} \\ x&=\frac{\pi}{6}, -\frac{5\pi}{6} \\ \end{aligned}\\ \text{The asymptotes have equations: } x = \frac{\pi}{6} \text{ and } x = -\frac{5\pi}{6} \]

\[ \text{4. Asymptote midpoints} \\ \text{The asymptotes have equations: } x = \frac{\pi}{6} \text{ and } x = -\frac{5\pi}{6} \\ \text{The only pair of asymptotes which are next to each other are: }\frac{\pi}{6} \text{ and } -\frac{5\pi}{6} \\ \begin{aligned} \frac{\frac{\pi}{6} + (-\frac{5\pi}{6})}{2} &= \frac{\frac{-4\pi}{6}}{2} \\ &= -\frac{2\pi}{6} \\ &= -\frac{\pi}{3} \\ \end{aligned}\\ \text{So, the point } (-\frac{\pi}{3}, 1) \text{ lies on the graph.} \]

\[ \text{5. Endpoints} \\ \begin{aligned} x&=-\pi\\ y &= \tan(-\pi+\frac{\pi}{3})+1\\ y &= \tan(\frac{-2\pi}{3})+1 \\ y&= \sqrt{3} + 1 \\ \text{ }& \\ x&=\pi\\ y &= \tan(\pi+\frac{\pi}{3})+1\\ y &= \tan(\frac{4\pi}{3})+1\\ y&= \sqrt{3} + 1 \\ \end{aligned}\\ \text{The endpoints are at } (-\pi, \sqrt{3}+1), \, (\pi, \sqrt{3}+1) \]

\[ \text{6. Connect the dots and label all necessary points} \]

Solving Trigonometric Equations

Solving trigonometric equations will seem quite difficult at first due to the large number of steps. It is important that you do it in a systematic manner. We have outlined the steps below and have provided a worked example.

Solving equations with sine and cosine functions in the one equation is slightly different. One example is provided below.

\[ \text{Example 6.4: Solve } 2\sin(2(x-\frac{\pi}{6}))+1 = 0, x \in [-\pi, \pi]\\ \text{ }\\ \text{ }\\ \text{1. Adjust the domain}\\ 2\sin(2x-\frac{\pi}{3}) + 1 =0\\ \begin{aligned} x &\in [-\pi, \pi]\\ 2x &\in [-2\pi, 2\pi]\\ 2x - \frac{\pi}{3} &\in [-2\pi-\frac{\pi}{3}, 2\pi-\frac{\pi}{3}]\\ 2x - \frac{\pi}{3} &\in [-\frac{7\pi}{3}, \frac{5\pi}{3}]\\ \end{aligned}\\ \text{ }\\ \text{ }\\ \text{2. Rearrange the equation}\\ \begin{aligned} 2\sin(2(x-\frac{\pi}{6}))+1 &= 0\\ 2\sin(2x-\frac{\pi}{3}) + 1 &=0\\ \sin(2x-\frac{\pi}{3})&=-\frac{1}{2}\\ \end{aligned}\\ \text{ }\\ \text{ }\\ \text{3. Find the basic angle}\\ \sin(\frac{\pi}{6}) = \frac{1}{2}\\ \text{So, the basic angle is } \frac{\pi}{6}\\ \text{ }\\ \text{ }\\ \text{4. Identify the quadrants}\\ \sin(2x-\frac{\pi}{3})=-\frac{1}{2}\\ \text{Sine is only negative in quadrant 3 and 4.}\\ \text{ }\\ \text{ }\\ \text{5. Use supplementary angle formulae, after you get use to this process try and merge steps 5 and 6}\\ \text{Our basic angle is } \frac{\pi}{6}\\ \begin{aligned} \sin(\pi + \frac{\pi}{6}) &= \sin(2\pi - \frac{\pi}{6}) = -\frac{1}{2}\\ \sin(\frac{7\pi}{6}) &= \sin(\frac{11\pi}{6}) = -\frac{1}{2}\\ \end{aligned}\\ \text{ }\\ \text{ }\\ \text{6. Add and Subtract } 2\pi\\ \begin{aligned} 2x-\frac{\pi}{3} &= \frac{7\pi}{6}, \frac{7\pi}{6}-2\pi , \frac{11\pi}{6}-2\pi, \frac{11\pi}{6}-4\pi\\ 2x-\frac{\pi}{3} &= \frac{7\pi}{6}, -\frac{5\pi}{6} , -\frac{1\pi}{6}, -\frac{13\pi}{6}\\ \end{aligned}\\ \text{ }\\ \text{ }\\ \text{7. Solve for } x\\ \begin{aligned} 2x &= \frac{7\pi}{6}+\frac{\pi}{3}, -\frac{5\pi}{6}+\frac{\pi}{3} , -\frac{1\pi}{6}+\frac{\pi}{3}, -\frac{13\pi}{6}+\frac{\pi}{3}\\ 2x &= \frac{9\pi}{6}, -\frac{3\pi}{6}, \frac{1\pi}{6}, -\frac{11\pi}{6}\\ x &= \frac{9\pi}{12}, -\frac{3\pi}{12}, \frac{1\pi}{12}, -\frac{11\pi}{12}\\ x &= \frac{3\pi}{4}, -\frac{1\pi}{4}, \frac{1\pi}{12}, -\frac{11\pi}{12}\\ x &= -\frac{11\pi}{12}, -\frac{1\pi}{4}, \frac{1\pi}{12}, \frac{3\pi}{4}\\ \end{aligned}\\ \]

\[ \text{Example 6.5: Solve } 2\sin(2(x-\frac{\pi}{6}))+1 = 0 \text{ (Find the general solutions)}\\ \text{Steps 1-5 remain the same as the previous question.}\\ \sin(\frac{7\pi}{6}) = \sin(\frac{11\pi}{6}) = -\frac{1}{2} \text{ (Last line of step 5)}\\ \text{ }\\ \text{ }\\ \text{6. Add or subtract } 2\pi\\ 2x-\frac{\pi}{3} =\frac{7\pi}{6} + 2\pi k , \frac{11\pi}{6} + 2\pi k ,\, k \in \mathbb{Z}\\ \text{ }\\ \text{ }\\ \text{7. Solve for } x\\ \begin{aligned} 2x &=\frac{7\pi}{6} + \frac{\pi}{3} + 2\pi \cdot k , \frac{11\pi}{6} + \frac{\pi}{3} + 2\pi \cdot k , k \in \mathbb{Z}\\ 2x &=\frac{9\pi}{6} + 2\pi \cdot k , \frac{13\pi}{6} + 2\pi \cdot k , \,k \in \mathbb{Z}\\ x &=\frac{9\pi}{12} + \pi \cdot k , \frac{13\pi}{12} + \pi \cdot k , \,k \in \mathbb{Z}\\ \end{aligned} \]

Note: the textbook combines these two solutions into one (for cosmetic reasons I guess). This is unnecessary and should not be done. The reason being that if you happen to make a mistake when combining the two solutions you will lose 1 mark for no reason.

\[ \text{Example 6.6: Solve } \sqrt{3}\cos(2x)=\sin(2x) \text{ for } x \in [0, 2\pi]\\ \text{Step 1 Divide by cosine to get tan}\\ \begin{aligned} \frac{\sqrt{3}\cos(2x)}{\sqrt{3}\cos(2x)}&=\frac{\sin(2x)}{\sqrt{3}\cos(2x)}\\ \frac{\sin(2x)}{\sqrt{3}\cos(2x)}&=1\\ \frac{1}{\sqrt{3}}\tan(2x)&=1\\ \end{aligned} \text{ }\\ \text{ }\\ \text{Using the same process outlined above, you will get:}\\ x = \frac{\pi}{6} , \frac{2\pi}{3}, \frac{7\pi}{6}, \frac{5\pi}{3}\\ \]

Determining Rules of Circular Functions

The general form of sine, cosine and tangent functions are shown below. However, the question may provide us with the form we should use, this means that the equation provided may or may not have as many unknowns as the ones below. \[y = A\sin(nx+b)+c\] \[y = A\cos(nx+b)+c\] \[y = A\tan(nx+b)+c\] To find the value of any other unknowns, simply substitute the x and y-coordinates of any additional points provided into the equation (having found some unknowns as detailed below) and solve them (simultaneously).

For sine and cosine:
The value of A is the amplitude - half the vertical distance from peak to trough.
The value of n can be calculated by examining the period of the function. \[\frac{2\pi}{n} = \text{period}\ \rightarrow n = \frac{2\pi}{\text{period}}\] The value of c will be the value of the y-intercept of the centreline (the horizontal line which lies in between the peaks and troughs).
To find the value of b use the additional point provided. When solving the equation simply find any two consecutive solutions.

For tangent
The value of n can be calculated by examining the period of the function. \[\frac{\pi}{n} = \text{period}\ \rightarrow n = \frac{\pi}{\text{period}}\] Asymptotes will be present when \(\cos(nx+b) = 0\). Substituting in the x-intercept value of the asymptotes may allow you to find the value for n or b