The definition of the trigonometric functions introduced in this section focuses on trigonometric functions of any angle (in degrees) - not just acute, obtuse, or positive angles; however, if given an angle (rather than a coordinate) and asked to evaluate a trigonometric function, the angle will strictly be a special angle. Therefore, using technology (e.g., calculators) to evaluate any trigonometric function in this section is strictly avoided so as to get the student familiar with these functions at a fundamental level. Simply stated, students should not require a calculator to evaluate any of the trigonometric functions within this entire chapter!
To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics.
Prerequisite Skills and Support Topics (click to expand)
Simplifying Expressions
Rationalizing Denominators and Numerators: This skill is required and is discussed in a note within the section regarding conventions for simplifying trigonometric values.
An Overview of Functions
The Definition of a Function: This section defines the trigonometric ratios as functions of an angle, so a prior understanding of what a function is (with independent and dependent variables) is necessary.
Evaluating Functions: This skill is explicitly tested in the "Skills Refresher" exercises for polynomial, radical, rational, and exponential functions, and the concept is extended to trigonometric functions throughout the section.
Defining Trigonometric Functions in Terms of Coordinates
Armed with our knowledge from the previous chapter, we now define the trigonometric functions. We will be viewing the trigonometric functions from three different perspectives in this course - coordinate trigonometry, right triangle trigonometry, and unit circle trigonometry.1 For the most part, each perspective is mathematically equivalent to the others; however, each has an inherent weakness as a standalone perspective of the trigonometric functions. While you can find a favorite perspective and stick with that one for most of the course, understanding all three is the only way to succeed in Trigonometry.
Let \( \theta \) be an angle in standard position with the point \( P\left( x,y \right) \) on the terminal side of \( \theta \), where \( P \) is not the origin. Define \( r = \sqrt{x^2 + y^2} \). The trigonometric functions of the angle \( \theta \) are defined as follows:\[ \begin{array}{|cccc|cccc|}
\hline
\textbf{Function} & \textbf{Function} & & & \textbf{Function} & \textbf{Function} & & \\
\textbf{Name} & \textbf{Notation} & & \textbf{Definition} & \textbf{Name} & \textbf{Notation} & & \textbf{Definition} \\
\hline
\text{The sine of }\theta & \sin\left( \theta \right) & = & \dfrac{y}{r} & \text{The cosecant of }\theta & \csc\left( \theta \right) & = & \dfrac{r}{y} \\[6pt]
\text{The cosine of }\theta & \cos\left( \theta \right) & = & \dfrac{x}{r} & \text{The secant of }\theta & \sec\left( \theta \right) & = & \dfrac{r}{x} \\[6pt]
\text{The tangent of }\theta & \tan\left( \theta \right) & = & \dfrac{y}{x} & \text{The cotangent of }\theta & \cot\left( \theta \right) & = & \dfrac{x}{y} \\[6pt]
\hline \end{array} \nonumber \]
A few notes are in order. First, the trigonometric functions are simply the ratios of \( x \), \( y \), and \( r \), where \( P\left( x,y \right) \) is a point on the terminal side of \( \theta \) and \( r \) is the distance from the origin to the point \( P \) (see Figure \( \PageIndex{ 1 } \) below).
Figure \( \PageIndex{ 1 } \)
Second, because it's the distance from the origin to \(P\), \( r \) is always positive. However, \(x\) and \(y\) can be positive, negative, or zero (but not both zero), depending on the angle \(\theta\). For example, in the second quadrant, \(x\) is negative, but \(y\) is positive, so the cosine and the tangent of angles between \(90^{\circ}\) and \(180^{\circ}\) are negative, but their sines are positive.
Third, referencing Figure \( \PageIndex{ 1 } \), since the tangent function is defined to be \( \frac{y}{x} \) and the terminal side of \( \theta \) is always a ray emanating from the origin, we get that\[ \dfrac{y}{x} = \dfrac{\text{rise}}{\text{run}} = \text{slope of the terminal side of }\theta. \nonumber \]The interpretation of the value of the tangent function at \( \theta \) being the slope of the ray that is the terminal side of \( \theta \) grants us a potent conceptual understanding of the tangent function. Specifically, since a vertical line has an undefined slope, it makes sense that the tangent function is undefined when \( x = 0 \). Any ray emanating from the origin and going through a point \( P\left( 0,y \right) \), where \( y \neq 0 \), will be vertical. While the secant function is also undefined when \( x = 0 \) (and the cosecant and cotangent functions are undefined when \( y=0 \)), we will hold off on gaining a conceptual understanding of why until we familiarize ourselves with more of the basics of the trigonometric functions.
Note: Angles are in Standard Position
Unless otherwise stated, all angles from this point forward will be considered angles in standard position.
Example \(\PageIndex{1}\)
Find the values of the six trigonometric functions if \( P\left( -3, -8 \right) \) is on the terminal side of \( \theta \).
In Algebra, we learned to rationalize denominators so we could "simplify" expressions like \( -\frac{8}{\sqrt{73}} \); however, in Trigonometry, such manipulations often are pointless. The rule of thumb I will adhere to in this textbook is to rationalize denominators if the denominator involves a variable (e.g., \( \frac{3}{\sqrt{x} - 2} \)). In all other cases (specifically, when the numerator and denominator do not contain variables), I will only rationalize denominators if there is an "obvious" benefit.
In the case of \( -\frac{8}{\sqrt{73}} \), if we tried to rationalize the denominator, we would get\[ -\dfrac{8}{\sqrt{73}} \cdot \dfrac{\sqrt{73}}{\sqrt{73}} = - \dfrac{8 \sqrt{73}}{73}, \nonumber \]which is no better than the original fraction.
Checkpoint \(\PageIndex{1}\)
Find the equation of the terminal side of the angle in the previous example. (Hint: The terminal side lies on a line that goes through the origin and the point \( \left( -3,-8 \right) \)).
Show that the point \( P^{\prime}\left( -6,-16 \right) \) also lies on the terminal side of the angle.
Compute the trigonometric ratios for \( \theta \) using the point \( P^{\prime} \) instead of \( P \).
Answers
\(y=\dfrac{8}{3} x\)
\((-6,-16)\) satisfies \(y=\frac{8}{3} x\). That is, the equation \(-16=\frac{8}{3}(-6)\) is true.
Find the values of \(\cos \left(\theta\right)\) and \(\tan \left(\theta\right)\) if \(\theta\) is an obtuse angle with \(\sin \left(\theta\right)=\frac{1}{3}\).
Solution
Because \(\theta\) is obtuse, the terminal side of the angle lies in the second quadrant, as shown in Figure \( \PageIndex{ 3 } \).
Figure \( \PageIndex{ 3 } \)
Because \(\sin \left(\theta\right)=\frac{1}{3}\), we know that \(\frac{y}{r}=\frac{1}{3}\), so we can choose a point \(P\) with \(y=1\) and \(r=3\). To find \(\cos \left(\theta\right)\) and \(\tan \left(\theta\right)\), we need to know the value of \(x\). From the Pythagorean Theorem,\[\begin{array}{rrrclcl}
& & x^2+1^2 & = & 3^2 & \quad & \left( \text{substitute known values} \right) \\
\scriptscriptstyle\mathrm{Arithmetic} & \implies & x^2 + 1 & = & 9 & \quad & \left( \text{Laws of Exponents} \right) \\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & x^2 & = & 8 & \quad & \left( \text{subtract }1\text{ from both sides} \right) \\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & x & = & \pm \sqrt{8} & \quad & \left( \text{Extraction of Roots} \right) \\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & x & = & \pm 2 \sqrt{2} & \quad & \left( \text{simplifying radicals} \right) \\
\end{array} \nonumber \]Remembering that \(x\) is negative in the second quadrant, we get that \( x = -2\sqrt{2} \). Thus,\[\cos \left(\theta\right)=\dfrac{x}{r}=\dfrac{-2\sqrt{2}}{3} \quad \text { and } \quad \tan \left(\theta\right)=\dfrac{y}{x}=\dfrac{-1}{2\sqrt{2}}\nonumber \]
Example \( \PageIndex{ 2 } \) illustrates a common theme in Trigonometry - we will often use the quadrant of the terminal side of an angle to inform our decision on the sign of \( x \)- and \( y \)-values.
If given a negative trigonometric ratio, as in Checkpoint \( \PageIndex{ 2 } \), it is important to remember that the value of \( r \) is always non-negative.2
Checkpoint \( \PageIndex{ 2 } \)
Sketch an obtuse angle \(\theta\) whose cosine is \(\frac{-8}{17}\).
Our use of function notation (e.g., the parentheses in \( \sin\left( \theta \right) = \frac{y}{r} \)) is purposeful. This is because, a given angle \( \theta \) (which is considered the independent variable) returns a single value for the associated trigonometric ratio (which we consider to be the dependent variable). Thus, by definition, the trigonometric functions are... functions (of angles).
Caution: Notational Abuse
We will sometimes write \(\cos \theta\) when we really mean \(\cos\left(\theta\right)\). This is a common, but accepted, abuse of notation reserved specifically for the trigonometric functions (and, at times, the logarithmic functions).
A common mistake is to think of \(\cos \theta\) or \(\cos(\theta)\) as a product, cos times \(\theta\), but this makes no sense, because ”cos” by itself has no meaning. Remember that \(\cos \theta\) represents a single number, namely the output of the cosine function.
Trigonometric Values at Special Angles
As was mentioned earlier in this textbook, special triangles play a crucial role in Trigonometry. In fact, the angles \(30^{\circ}\), \(45^{\circ}\), and \(60^{ \circ }\) are just a few of the so-called special angles because we can express the exact values of their trigonometric functions in terms of radicals. In the following example, we explore the values of the trigonometric functions at one of the special angles.
Example \(\PageIndex{3}\)
Find the values of all six trigonometric functions of \( 30^{ \circ } \).
Solution
Like any good Trigonometry student, we start with a sketch.
Figure \( \PageIndex{ 5 } \)
Unfortunately, we were not given a value for a point on the terminal side of \( \theta = 30^{ \circ } \); however, recall the following figure from our exploration of the special triangles last chapter.
Figure \( \PageIndex{ 6 } \)
Rotating this special triangle counterclockwise by \( 90^{ \circ } \), we can visualize fitting it to our first sketch, almost like a jigsaw puzzle.
Figure \( \PageIndex{ 7 } \)
If we let \( a = 1 \) (a completely arbitrary choice), we get\[ x = \sqrt{3}, \quad y = 1, \text{ and} \quad r = 2. \nonumber \]Therefore,\[ \begin{array}{rccclcrcccl}
\sin\left( 30^{ \circ } \right) & = & \dfrac{y}{r} & = & \dfrac{1}{2} & \qquad & \csc\left( 30^{ \circ } \right) & = & \dfrac{r}{y} & = & 2 \\
\cos\left( 30^{ \circ } \right) & = & \dfrac{x}{r} & = & \dfrac{\sqrt{3}}{2} & \qquad & \sec\left( 30^{ \circ } \right) & = & \dfrac{r}{x} & = &\dfrac{2}{\sqrt{3}} \\
\tan\left( 30^{ \circ } \right) & = & \dfrac{y}{x} & = & \dfrac{1}{\sqrt{3}} & \qquad & \cot\left( 30^{ \circ } \right) & = & \dfrac{x}{y} & = & \sqrt{3} \\
\end{array} \nonumber \]
At this point, you should be asking yourself why we arbitrarily chose \( a = 1 \) in Example \( \PageIndex{ 3 } \) to "lock in" the value of the point \( P\left( \sqrt{3},1 \right) \). Had we chosen, for example, \( a = 5 \), we would have gotten\[ x = 5\sqrt{3}, \quad y = 5, \text{ and} \quad r = 10. \nonumber \]Therefore,\[ \begin{array}{rccclcrcccl}
\sin\left( 30^{ \circ } \right) & = & \dfrac{y}{r} & = & \dfrac{5}{10} = \dfrac{1}{2} & \qquad & \csc\left( 30^{ \circ } \right) & = & \dfrac{r}{y} & = & \dfrac{10}{5} = 2 \\
\cos\left( 30^{ \circ } \right) & = & \dfrac{x}{r} & = & \dfrac{5\sqrt{3}}{10} = \dfrac{\sqrt{3}}{2} & \qquad & \sec\left( 30^{ \circ } \right) & = & \dfrac{r}{x} & = & \dfrac{10}{5\sqrt{3}} = \dfrac{2}{\sqrt{3}} \\
\tan\left( 30^{ \circ } \right) & = & \dfrac{y}{x} & = & \dfrac{5}{5\sqrt{3}} = \dfrac{1}{\sqrt{3}} & \qquad & \cot\left( 30^{ \circ } \right) & = & \dfrac{x}{y} & = & \dfrac{5\sqrt{3}}{5} = \sqrt{3} \\
\end{array} \nonumber \]Notice that the values of the trigonometric functions didn't change? In fact, if we had chosen to leave \( a \) "as is," we would have arrived at\[ x = a\sqrt{3}, \quad y = a, \text{ and} \quad r = 2a. \nonumber \]Therefore,\[ \begin{array}{rccclcrcccl}
\sin\left( 30^{ \circ } \right) & = & \dfrac{y}{r} & = & \dfrac{a}{2a} = \dfrac{1}{2} & \qquad & \csc\left( 30^{ \circ } \right) & = & \dfrac{r}{y} & = & \dfrac{2a}{a} = 2 \\
\cos\left( 30^{ \circ } \right) & = & \dfrac{x}{r} & = & \dfrac{a\sqrt{3}}{2a} = \dfrac{\sqrt{3}}{2} & \qquad & \sec\left( 30^{ \circ } \right) & = & \dfrac{r}{x} & = & \dfrac{2a}{a\sqrt{3}} = \dfrac{2}{\sqrt{3}} \\
\tan\left( 30^{ \circ } \right) & = & \dfrac{y}{x} & = & \dfrac{a}{a\sqrt{3}} = \dfrac{1}{\sqrt{3}} & \qquad & \cot\left( 30^{ \circ } \right) & = & \dfrac{x}{y} & = & \dfrac{a\sqrt{3}}{a} = \sqrt{3} \\
\end{array} \nonumber \]The choice of \( a \) doesn't change the values of the trigonometric functions at all! This demonstration implies (but does not prove) that the values of the trigonometric functions do not depend on which point you choose on the terminal side of \( 30^{ \circ } \) - just as long as that point is not the origin.
Note
The result of this last statement is beneficial. If we have angles from our special triangles, we can just set \( a = 1 \) in those special triangles to get the \( x \)- and \( y \)-values of the related point.
Checkpoint \(\PageIndex{3}\)
Find the values of the six trigonometric functions at \( \theta = 45^{ \circ } \).
We do not need to limit our use of the special angles and special triangles to angles given in the first quadrant. The following example illustrates our tactics if given an angle terminating in a quadrant other than the first, where we can use our knowledge of the special triangles.
Example \( \PageIndex{ 4 } \)
Find exact values for all six trigonometric functions of \(240^{\circ}\).
Solution
An angle of \(240^{\circ}\) lies in the third quadrant, as seen in Figure \( \PageIndex{ 8 } \).
Figure \( \PageIndex{ 8 } \)
If we want to work with a smaller angle, we consider the angle between the terminal side of \( 240^{ \circ } \) and the negative \( x \)-axis. This can be found by taking the larger angle (\( 240^{ \circ } \)) and subtracting off the smaller angle (the \( 180^{ \circ } \) angle that gets to the negative \( x \)-axis).\[ 240^{ \circ } - 180^{ \circ } = 60^{ \circ } \nonumber \]This action is illustrated in Figure \( \PageIndex{ 9 } \).
Figure \( \PageIndex{ 9 } \)
That \( 60^{ \circ } \) angle between \( 240^{ \circ } \) and \( 180^{ \circ } \) should be familiar.
Figure \( \PageIndex{ 10 } \)
Superimposing this \( 30^{ \circ } \)-\( 60^{ \circ } \)-\( 90^{ \circ } \) special triangle into our graph, we get Figure \( \PageIndex{ 11 } \).
Figure \( \PageIndex{ 11 } \)
Using our new knowledge that we can let \( a \) be any convenient number, we choose \( a = 1 \). However, here is where we need to be careful! If \( a = 1 \), it is not true that \( x = a = 1 \). This is because the point \( P\left( x,y \right) \) on the terminal side of \( 240^{ \circ } \) must be in \( \mathrm{QIII} \). Therefore, \( x = -1 \). Likewise, \( y = -\sqrt{3} \). No matter what, however, \( r > 0 \). Thus, \( r = 2a = 2 \). We now compute the values of the trigonometric functions.\[ \begin{array}{rccclcrcccl}
\sin\left( 240^{ \circ } \right) & = & \dfrac{y}{r} & = & -\dfrac{\sqrt{3}}{2} & \qquad & \csc\left( 240^{ \circ } \right) & = & \dfrac{r}{y} & = & -\dfrac{2}{\sqrt{3}} \\
\cos\left( 240^{ \circ } \right) & = & \dfrac{x}{r} & = & -\dfrac{1}{2} & \qquad & \sec\left( 240^{ \circ } \right) & = & \dfrac{r}{x} & = & -2 \\
\tan\left( 240^{ \circ } \right) & = & \dfrac{y}{x} & = & \sqrt{3} & \qquad & \cot\left( 240^{ \circ } \right) & = & \dfrac{x}{y} & = & \dfrac{1}{\sqrt{3}} \\
\end{array} \nonumber \]
Checkpoint \( \PageIndex{ 4 } \)
Find exact values for the sine, cosine, and tangent of \(300^{\circ}\).
All special angles between \( 0^{ \circ } \) and \( 360^{ \circ } \) are shown in Figure \( \PageIndex{ 12 } \) below.
Figure \( \PageIndex{ 12 } \)
The astute student will notice that these can be succinctly written using the following definition.
Definition: Special Angles
The special angles are defined to be any angle that can be written as either \( \pm 30^{ \circ } k\) or \( \pm 45^{ \circ } k \), where \( k \in \mathbb{Z} \).
It would be best if you memorized these values or be able to calculate them quickly.
Trigonometric Values at Quadrantal Angles
Evaluating trigonometric functions at quadrantal angles is much easier than at non-quadrantal angles. However, we have to watch out for division by zero.
Example \(\PageIndex{5}\)
Find the values of the six trigonometric functions at \( -270^{ \circ } \).
Solution
We get Figure \( \PageIndex{ 13 } \) by creating a quick sketch for this situation and remembering that negative angles open clockwise.
Figure \( \PageIndex{ 13 } \)
We took the liberty to label a point on the terminal side of \( -270^{ \circ } \). In this case, \( x = 0 \), \( y = 1 \), and \( r = 1 \). Therefore,\[ \begin{array}{rccclcrcccl}
\sin\left( -270^{ \circ } \right) & = & \dfrac{y}{r} & = & \dfrac{1}{1} = 1 & \qquad & \csc\left( -270^{ \circ } \right) & = & \dfrac{r}{y} & = & \dfrac{1}{1} = 1 \\
\cos\left( -270^{ \circ } \right) & = & \dfrac{x}{r} & = & \dfrac{0}{1} = 0 & \qquad & \sec\left( -270^{ \circ } \right) & = & \dfrac{r}{x} & = &\dfrac{1}{0} = \text{ UNDEFINED} \\
\tan\left( -270^{ \circ } \right) & = & \dfrac{y}{x} & = & \dfrac{1}{0} = \text{ UNDEFINED} & \qquad & \cot\left( -270^{ \circ } \right) & = & \dfrac{x}{y} & = & \dfrac{0}{1} = 0 \\
\end{array} \nonumber \]
Example \( \PageIndex{ 5 } \) illustrates the fact that there are angles for which some of the trigonometric functions are not defined. In terms of functions, we would say that these functions have domain restrictions. We will discuss this in more detail later in this textbook. It suffices to know that the tangent, cotangent, secant, and cosecant functions each have angles for which the functions are undefined. You can see from the definitions of these trigonometric functions where these domain issues occur (either when \( x = 0 \) or when \( y = 0 \), depending on the trigonometric function).
Checkpoint \(\PageIndex{5}\)
Find the values of the six trigonometric functions at \( 540^{ \circ } \).
Very technically, "\( \cot\left( 540^{ \circ } \right) = \text{ UNDEFINED}\)" is not correct. It not only sounds odd (read the following sentence out loud, and you will understand what we mean, "This value equals undefined"), but it also is not mathematically accurate. Once you use the equals sign (\( = \)), you are assigning a value; however, "UNDEFINED" is not a value.
A mathematically accurate response would be, "\( \cot\left( 540^{ \circ } \right) \) is undefined."
Comparing Values of the Trigonometric Functions
Understanding how to compare the values of a trigonometric function for different angles can be beneficial in future math courses. For example, in Calculus, you will need to have an innate understanding that \( \sin \left(50^{ \circ }\right) < \sin\left( 80^{ \circ } \right) \). The following example illustrates the thought process.
Example \(\PageIndex{6}\)
Without reaching for technology, can you determine which is greater, \( \sin\left( 50^{ \circ } \right) \) or \( \sin\left( 80^{ \circ } \right) \)?
Solution
Quickly sketching two angles, \( \alpha = 50^{ \circ } \) and \( \beta = 80^{ \circ } \), we get the following graph.
Figure \( \PageIndex{ 14 } \)
By itself, this graph is not helpful; however, if we recall that the sine of an angle is defined to be \( \frac{y}{r} \), where \( \left( x,y \right) \) is any point on the terminal side of the angle and \( r \) is the distance from the origin to the point, then we could choose points on the terminal sides of \( \alpha = 50^{ \circ }\) and \( \beta = 80^{ \circ } \) such that \( r = 1 \). This gives us \( \sin\left( 50^{ \circ } \right) = \frac{y_1}{r} = \frac{y_1}{1} = y_1 \) and \( \sin\left( 80^{ \circ } \right) = \frac{y_2}{r} = \frac{y_2}{1} = y_2 \) (see Figure \( \PageIndex{ 15 } \)).
Figure \( \PageIndex{ 15 } \)
From Figure \( \PageIndex{ 15 } \), we can see that \( y_1 < y_2\). Therefore, \( \sin\left( 50^{ \circ } \right) < \sin\left( 80^{ \circ } \right)\).
Checkpoint \(\PageIndex{6}\)
Without reaching for technology, can you determine which is greater, \( \cos\left( 100^{ \circ } \right) \) or \( \cos\left( 170^{ \circ } \right) \)?
Answer
\( \cos\left( 100^{ \circ } \right) \) is greater than \( \cos\left( 170^{ \circ } \right) \)
Quadrants and the Signs of the Trigonometric Functions
Example \(\PageIndex{7}\)
Give the sign of the sine, cosine, and tangent of the angle.
\(200^{\circ}\)
\(300^{\circ}\)
Solutions
In standard position, the terminal side of an angle of \(200^{\circ}\) lies in \( \mathrm{QIII} \) (see Figure \( \PageIndex{ 16(a)} \) below). In the third quadrant, \(x < 0\) and \(y < 0\), but \(r > 0\). Thus, \(\sin \left( 200^{\circ}\right) \) is negative, \(\cos \left(200^{\circ}\right)\) is negative, and \(\tan \left(200^{\circ}\right)\) is positive.
Figure \( \PageIndex{ 16 } \)
\(300^{\circ} \in \mathrm{QIV}\), so \(x > 0\) and \(y < 0\), and \(r > 0\). Thus, \(\sin \left(300^{\circ}\right)\) is negative, \(\cos \left(300^{\circ}\right)\) is positive, and \(\tan \left(300^{\circ}\right)\) is negative.
Example \( \PageIndex{ 7 } \) demonstrates a common need in Trigonometry - to determine the sign of a trigonometric function. Luckily, our task is fairly easy.
Theorem: Signs of the Trigonometric Functions
The signs of the trigonometric functions depend on the trigonometric function in question and the quadrant where the terminal side of \( \theta \) lies.
All trigonometric functions are positive when the terminal side of \( \theta \) is in the first quadrant.
Sine and the cosecant functions are positive when the terminal side of \( \theta \) is in the second quadrant.
Tangent and the cotangent functions are positive when the terminal side of \( \theta \) is in the third quadrant.
Cosine and the secant functions are positive when the terminal side of \( \theta \) is in the fourth quadrant.
Proof
Since the value of \( r \) is always positive (unless the point is the origin), the signs of the trigonometric functions are completely dependent on the signs of the \( x \)- and \( y \)-coordinates of the point on the terminal side of \( \theta \).\[ \begin{array}{|c|c|c|}
\hline \text{The sign of...} & \text{... which is...} & \text{...depends on the value(s) of...} \\
\hline \sin\left( \theta \right) & \dfrac{y}{r} & y \\
\hline \cos\left( \theta \right) & \dfrac{x}{r} & x \\
\hline \tan\left( \theta \right) & \dfrac{y}{x} & x \text{ and } y \\
\hline \csc\left( \theta \right) & \dfrac{r}{y} & y \\
\hline \sec\left( \theta \right) & \dfrac{r}{x} & x \\
\hline \cot\left( \theta \right) & \dfrac{x}{y} & x \text{ and } y \\
\hline \end{array} \nonumber \]Therefore, the signs of the sine and cosecant functions completely depend on the signs of the \( y \)-value of the point on the terminal side of \( \theta \). Hence, the sine and cosecant are positive in \( \mathrm{QI} \) and \( \mathrm{QII} \). On the other hand, these functions are negative when the angle terminates in either \( \mathrm{QIII} \) or \( \mathrm{QIV} \).
Likewise, the signs of the cosine and secant functions completely depend on the signs of the \( x \)-value of the point on the terminal side of \( \theta \). Therefore, the cosine and secant functions are positive in \( \mathrm{QI} \) and \( \mathrm{QIV} \) - they are negative in \( \mathrm{QII} \) and \( \mathrm{QIII} \).
Finally, since the tangent and cotangent functions rely on the values of both \( x \) and \( y \), we must consider what would make these ratios positive and negative. If the values of \( x \) and \( y \) are both positive or both negative, their ratios will be positive. This occurs in the first and third quadrants. Thus, the tangent and cotangent functions are positive in \( \mathrm{QI} \) and \( \mathrm{QIII} \) - they are negative in \( \mathrm{QII} \) and \( \mathrm{QIV} \).
Figure \( \PageIndex{ 17 } \) summarizes these results.
Figure \( \PageIndex{ 17 } \)
The bold lettering in the previous theorem is purposeful. Figure \( \PageIndex{ 18 } \) gives the lovely mnemonic, "All Student Take Calculus," to remember that all trigonometric functions are positive in the first quadrant, the sine and cosecant functions are positive in the second quadrant, the tangent and cotangent functions are positive in the third quadrant, and the cosine and secant functions are positive in the fourth quadrant.
Figure \( \PageIndex{ 18 } \)
Example \(\PageIndex{8}\)
Given that \( \cos\left( \theta \right) = -\frac{3}{5} \) and \( \theta \) terminates in \( \mathrm{QII} \), find the values of the remaining trigonometric functions of \( \theta \).
Solution
Since we are given \( \cos\left( \theta \right) = -\frac{3}{5} \), we know the ratio of \( x \) to \( r \) is \( -\frac{3}{5} \). Letting \( x = -3 \) and \( r = 5 \), we find \( y \) by solving\[ \begin{array}{rrrclcl}
& & r & = & \sqrt{x^2 + y^2} & \quad & \left( \text{definition of }r \right)\\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 5 & = & \sqrt{(-3)^2 + y^2} & \quad & \left( \text{substitute values} \right)\\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 25 & = & 9 + y^2 & \quad & \left( \text{Laws of Exponents and squaring both sides} \right) \\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & 16 & = & y^2 & \quad & \left( \text{subtracting }9\text{ from both sides} \right)\\
\scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \pm 4 & = & y & \quad & \left( \text{Extraction of Roots} \right) \\
\end{array} \nonumber \]Using the fact that \( \theta \in \mathrm{QIII} \), we throw out the positive possibility and set \( y = -4 \). We now have everything we need!\[ \begin{array}{rclcrcl}
\sin\left( \theta \right) & = & - \dfrac{4}{5} & \qquad & \csc\left( \theta \right) & = & -\dfrac{5}{4} \\
\cos\left( \theta \right) & = & - \dfrac{3}{5} & \qquad & \sec\left( \theta \right) & = & -\dfrac{5}{3} \\
\tan\left( \theta \right) & = & \dfrac{3}{4} & \qquad & \cot\left( \theta \right) & = & \dfrac{4}{3} \\
\end{array} \nonumber \]
Checkpoint \(\PageIndex{8}\)
In which quadrant(s) is the sine negative but the cosine positive?
Answer
\( \mathrm{QIV} \)
Proving that Values of the Trigonometric Functions are Independent of \( P \)
Up to this point, we have avoided proving that the choice of \( P \) on the terminal side of \( \theta \) makes no difference (other than choosing the origin - which is not allowed). We now officially prove the result so we can use it from this point forward.This proves the following theorem.
Theorem: Trigonometric Ratios are Independent of \( P \)
Let \( P\left( x,y \right) \) and \( \bar{P}\left( \bar{x}, \bar{y} \right) \) be two distinct points on the terminal side of \( \theta \), neither of which are the origin, and let \( r = \sqrt{x^2 + y^2} \) and \( \bar{r} = \sqrt{\bar{x}^2 + \bar{y}^2} \). Then\[ \begin{array}{rccclcrcccl}
\sin\left( \theta \right) & = & \dfrac{y}{r} & = & \dfrac{\bar{y}}{\bar{r}} & \qquad & \csc\left( \theta \right) & = & \dfrac{r}{y} & = & \dfrac{\bar{r}}{\bar{y}} \\
\cos\left( \theta \right) & = & \dfrac{x}{r} & = & \dfrac{\bar{x}}{\bar{r}} & \qquad & \sec\left( \theta \right) & = & \dfrac{r}{x} & = &\dfrac{\bar{r}}{\bar{x}} \\
\tan\left( \theta \right) & = & \dfrac{y}{x} & = & \dfrac{\bar{y}}{\bar{x}} & \qquad & \cot\left( \theta \right) & = & \dfrac{x}{y} & = & \dfrac{\bar{x}}{\bar{y}} \\
\end{array} \nonumber \]
Proof
Let \( P\left( x,y \right) \) and \( \bar{P}\left( \bar{x}, \bar{y} \right) \) be two distinct points on the terminal side of \( \theta \), neither of which are the origin, and let \( r = \sqrt{x^2 + y^2} \) and \( \bar{r} = \sqrt{\bar{x}^2 + \bar{y}^2} \) (see the following figure).
1 Since we are not covering right triangle trigonometry and unit circle trigonometry in this chapter, I chose to only bold coordinate trigonometry.
2 For those of you who have seen advanced topics in Trigonometry before, you know that the claim, "\( r \) is always non-negative" is a bit of an exaggeration. When we introduce the polar coordinate system near the end of this course, we allow \( r \) to be negative.
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