5.1: An Introduction to the Trigonometric Functions
- Page ID
- 99743
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Because many applications involving circles also involve a rotation of the circle, it is natural to introduce a measure for the rotation, or angle, between two rays (line segments) emanating from the center of a circle. The angle measurement you are most likely familiar with is degrees, so we’ll begin there.
The measure 
of an angle is a measurement between two intersecting lines, line segments or rays, starting at the initial side and ending at the terminal side. It is a rotational measure, not a linear measure.
Measuring Angles
A degree is a measurement of angle. One full rotation around the circle is equal to 360 degrees, so one degree is 1/360 of a circle.
An angle measured in degrees should always include the unit “degrees” after the number, or include the degree symbol \(^{\circ}\). For example, 90 degrees = \(90^{\circ}\).
When measuring angles on a circle, unless otherwise directed, we measure angles in standard position: starting at the positive horizontal axis and with counter-clockwise rotation.
Give the degree measure of the angle shown on the circle.
Solution
The vertical and horizontal lines divide the circle into quarters. Since one full rotation is 360 degrees=\(360{}^\circ\), each quarter rotation is 360/4 = \(90^{\circ}\) or 90 degrees.
Show an angle of \(30^{\circ}\) on the circle.
Solution
An angle of \(30^{\circ}\) is 1/3 of \(90^{\circ}\), so by dividing a quarter rotation into thirds, we can sketch a line at \(30^{\circ}\).
Going Greek
When representing angles using variables, it is traditional to use Greek letters. Here is a list of commonly encountered Greek letters.
| \(\theta\) | \(varphi\) or \(\phi\) | \(\alpha\) | \(\beta\) | \(\gamma\) |
| theta | phi | alpha | beta | gamma |
Working with Angles in Degrees
Notice that since there are 360 degrees in one rotation, an angle greater than 360 degrees would indicate more than 1 full rotation. Shown on a circle, the resulting direction in which this angle’s terminal side points would be the same as for another angle between 0 and 360 degrees. These angles would be called coterminal.
After completing their full rotation based on the given angle, two angles are coterminal if they terminate in the same position, so their terminal sides coincide (point in the same direction).
Find an angle \(\theta\) that is coterminal with \(800^{\circ}\), where \(0^{\circ} \le \theta <360^{\circ}\)
Solution
Since adding or subtracting a full rotation, 360 degrees, would result in an angle with terminal side pointing in the same direction, we can find coterminal angles by adding or subtracting 360 degrees. An angle of 800 degrees is coterminal with an angle of 800 - 360 = 440 degrees. It would also be coterminal with an angle of 440 - 360 = 80 degrees.
The angle \(\theta =80^{\circ}\)is coterminal with \(800^{\circ}\).
By finding the coterminal angle between 0 and 360 degrees, it can be easier to see which direction the terminal side of an angle points in.
Find an angle \(\alpha\) that is coterminal with \(870^{\circ}\), where \(0^{\circ} \le \alpha <360^{\circ}\).
- Answer
-
\[\alpha =870-360-360=150^{\circ}\nonumber\]
On a number line a positive number is measured to the right and a negative number is measured in the opposite direction (to the left). Similarly a positive angle is measured counterclockwise and a negative angle is measured in the opposite direction (clockwise).
Show the angle \(-45^{\circ}\)on the circle and find a positive angle \(\alpha\) that is coterminal and \(0^{\circ} \le \alpha <360^{\circ}\).
Solution
Since 45 degrees is half of 90 degrees, we can start at the positive horizontal axis and measure clockwise half of a 90 degree angle.
Since we can find coterminal angles by adding or subtracting a full rotation of 360 degrees, we can find a positive coterminal angle here by adding 360 degrees:
\[-45^{\circ} +360^{\circ} =315^{\circ}\nonumber\]
Find an angle\(\beta\) coterminal with \(-300^{\circ}\) where \(0^{\circ} \le \beta <360^{\circ}\).
- Answer
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\[\beta =-300+360=60^{\circ}\nonumber\]
It can be helpful to have a familiarity with the frequently encountered angles in one rotation of a circle. It is common to encounter multiples of 30, 45, 60, and 90 degrees. These values are shown to the right. Memorizing these angles and understanding their properties will be very useful as we study the properties associated with angles
Angles in Radians
While measuring angles in degrees may be familiar, doing so often complicates matters since the units of measure can get in the way of calculations. For this reason, another measure of angles is commonly used. This measure is based on the distance around a circle.
Arclen
gth is the length of an arc, \(s\), along a circle of radius \(r\) subtended (drawn out) by an angle \(\theta\).
It is the portion of the circumference between the initial and terminal sides of the angle.
The length of the arc around an entire circle is called the circumference of a circle. The circumference of a circle is \(C=2\pi r\). The ratio of the circumference to the radius, produces the constant \(2\pi\). Regardless of the radius, this ratio is always the same, just as how the degree measure of an angle is independent of the radius.
To elaborate on this idea, consider two circles, one with radius 2 and one with radius 3. Recall the circumference (perimeter) of a circle is \(C=2\pi r\), where \(r\) is the radius of the circle. The smaller circle then has circumference \(2\pi (2)=4\pi\) and the larger has circumference\(2\pi (3)=6\pi\).
Drawing a 45 degree angle on the two circles, we might be interested in the length of the arc of the circle that the angle indicates. In both cases, the 45 degree angle draws out an arc that is \({1/8}^{th}\) of the full circumference, so for the smaller circle, the arclength = \(\dfrac{1}{8} (4\pi )=\dfrac{1}{2} \pi\), and for the larger circle, the length of the arc or arclength = \(\dfrac{1}{8} (6\pi )=\dfrac{3}{4} \pi\).
Notice what happens if we find the ratio of the arclength divided by the radius of the circle:
Smaller circle: \(\dfrac{\dfrac{1}{2} \pi }{2} =\dfrac{1}{4} \pi\)
Larger circle: \(\dfrac{\dfrac{3}{4} \pi }{3} =\dfrac{1}{4} \pi\)
The ratio is the same regardless of the radius of the circle – it only depends on the angle. This property allows us to define a measure of the angle based on arclength.
The radian measure of an angle is the ratio of the length of the circular arc subtended by the angle to the radius of the circle.
In other words, if s is the length of an arc of a circle, and \(r\) is the radius of the circle, then
radian measure\(=\dfrac{s}{r}\)
If the circle has radius 1, then the radian measure corresponds to the length of the arc.
Because radian measure is the ratio of two lengths, it is a unitless measure. It is not necessary to write the label “radians” after a radian measure, and if you see an angle that is not labeled with “degrees” or the degree symbol, you should assume that it is a radian measure.
Considering the most basic case, the unit circle (a circle with radius 1), we know that 1 rotation equals 360 degrees, \(360^{\circ}\). We can also track one rotation around a circle by finding the circumference, \(C=2\pi r\), and for the unit circle \(C=2\pi\). These two different ways to rotate around a circle give us a way to convert from degrees to radians.
1 rotation = \(360{}^\circ\)=\(2\pi\) radians
1/2 rotation = \(180{}^\circ\) = \(\pi\) radians
1/4 rotation = \(90{}^\circ\)= \(\dfrac{\pi }{2}\) radians
Find the radian measure of one third of a full rotation.
Solution
For any circle, the arclength along such a rotation would be one third of the circumference, \(C=\dfrac{1}{3} (2\pi r)=\dfrac{2\pi r}{3}\). The radian measure would be the arclength divided by the radius:
\[\text{Radian measure} = \dfrac{2\pi r}{3r} =\dfrac{2\pi }{3}\]
1 degree = \(\dfrac{\pi }{180}\) radians
or: to convert from degrees to radians, multiply by \(\dfrac{\pi \text{radians}}{180^{\circ}}\)
1 radian = \(\dfrac{180}{\pi }\) degrees
or: to convert from radians to degrees, multiply by \(\dfrac{180^{\circ}}{\pi \text{radians}}\)
Converting Between Radians and Degrees
Convert \(\dfrac{\pi }{6}\) radians to degrees.
Solution
Since we are given a problem in radians and we want degrees, we multiply by \(\dfrac{180^{\circ}}{\pi }\).
Remember radians are a unitless measure, so we don’t need to write “radians.”
\(\dfrac{\pi }{6}\) radians = \(\dfrac{\pi }{6} \cdot \dfrac{180^{\circ}}{\pi } =30\) degrees.
Convert 15 degrees to radians.
Solution
In this example, we start with degrees and want radians so we use the other conversion\(\dfrac{\pi }{180^{\circ}}\)so that the degree units cancel and we are left with the unitless measure of radians.
15 degrees = \(15^{\circ} \cdot \dfrac{\pi }{180^{\circ}} =\dfrac{\pi }{12}\)
Convert \(\dfrac{7\pi }{10}\) radians to degrees.
- Answer
-
\[\dfrac{7\pi }{10} \cdot \dfrac{180^{\circ} }{\pi } =126^{\circ}\nonumber\]
Just as we listed all the common angles in degrees on a circle, we should also list the corresponding radian values for the common measures of a circle corresponding to degree multiples of 30, 45, 60, and 90 degrees. As with the degree measurements, it would be advisable to commit these to memory.
We can work with the radian measures of an angle the same way we work with degrees.
Find an angle \(\beta\) that is coterminal with \(\dfrac{19\pi }{4}\), where \(0\le \beta <2\pi\).
Solution
When working in degrees, we found coterminal angles by adding or subtracting 360 degrees, a full rotation. Likewise, in radians, we can find coterminal angles by adding or subtracting full rotations of \(2\pi\) radians.
\[\dfrac{19\pi }{4} -2\pi =\dfrac{19\pi }{4} -\dfrac{8\pi }{4} =\dfrac{11\pi }{4}\nonumber\]
The angle\(\dfrac{11\pi }{4}\) is coterminal, but not less than \(2\pi\), so we subtract another rotation.
\[\dfrac{11\pi }{4} -2\pi =\dfrac{11\pi }{4} -\dfrac{8\pi }{4} =\dfrac{3\pi }{4}\nonumber\]
The angle \(\dfrac{3\pi }{4}\) is coterminal with \(\dfrac{19\pi }{4}\).
Find an angle \(\phi\)that is coterminal with \(-\dfrac{17\pi }{6}\) where \(0\le \phi <2\pi\).
- Answer
-
\[-\dfrac{17\pi }{6} +2\pi +2\pi =-\dfrac{17\pi }{6} +\dfrac{12\pi }{6} +\dfrac{12\pi }{6} =\dfrac{7\pi }{6}\nonumber\]
Focus: An Introduction to the Trigonometric Functions
While it is convenient to describe the location of a point on a circle using an angle or a distance along the circle, relating this information to the x and y coordinates and the circle equation we explored in Section 5.1 is an important application of trigonometry.
A distress signal is sent from a sailboat during a storm, but the transmission is unclear and the rescue boat sitting at the marina cannot determine the sailboat’s location. Using high powered radar, they determine the distress signal is coming from a distance of 20 miles at an angle of 225 degrees from the marina. How many miles east/west and north/south of the rescue boat is the stranded sailboat?
In a general sense, to investigate this, we begin by drawing a circle centered at the origin with radius \(r\)
, and marking the point on the circle indicated by some angle \(\theta\). This point has coordinates (\(x\), \(y\)).
If we drop a line segment vertically down from this point to the x axis, we would form a right triangle inside of the circle.
No matter which quadrant our angle \(\theta\) puts us in we can draw a triangle by dropping a perpendicular line segment to the \(x\) axis, keeping in mind that the values of \(x\) and \(y\) may be positive or negative, depending on the quadrant.
Additionally, if the angle \(\theta\) puts us on an axis, we simply measure the radius as the \(x\) or \(y\) with the other value being 0, again ensuring we have appropriate signs on the coordinates based on the quadrant.
Triangles obtained from different radii will all be similar triangles, meaning corresponding sides scale proportionally. While the lengths of the sides may change, as we saw in the last section, the ratios of the side lengths will always remain constant for any given angle.
\(\dfrac{y_{1} }{r_{1} } =\dfrac{y_{2} }{r_{2} }\) 
\(\dfrac{x_{1} }{r_{1} } =\dfrac{x_{2} }{r_{2} }\)
To be able to refer to these ratios more easily, we will give them names. Since the ratios depend on the angle, we will write them as functions of the angle \(\theta\).
For the point (\(x\), \(y\)) on a circle of radius \(r\) at an angle of \(\theta\), we can define two important functions as the ratios of the sides of the corresponding triangle:
The sine function: \(\sin (\theta )=\dfrac{y}{r}\)
The cosine function: \(\cos (\theta )=\dfrac{x}{r}\)
In the previous section, we defined the sine and cosine functions as ratios of the sides of a right triangle in a circle. Since the triangle has 3 sides there are 6 possible combinations of ratios. While the sine and cosine are the two prominent ratios that can be formed, there are four others, and together they define the 6 trigonometric functions.
For the point (\(x\), \(y\)) on a circle of radius \(r\) at an angle of \(\theta\), we can define four addit
ional important functions as the ratios of the sides of the corresponding triangle:
- The tangent function: \[\tan (\theta )=\dfrac{y}{x}\]
- The secant function: \[\sec (\theta )=\dfrac{r}{x}\]
- The cosecant function: \[\csc (\theta )=\dfrac{r}{y}\]
- The cotangent function: \[\cot (\theta )=\dfrac{x}{y}\]
Geometrically, notice that the definition of tangent corresponds with the slope of the line segment between the origin (0, 0) and the point (\(x\), \(y\)). This relationship can be very helpful in thinking about tangent values.
In this chapter, we will explore these functions using both circles and right triangles. In the next chapter, we will take a closer look at the behavior and characteristics of the sine and cosine functions.
The point (3, 4) is on the circle of radius 5 at some angle \(\theta\). Find \(\cos (\theta )\)and \(\sin (\theta )\).
Solution
Knowing the radius of the circle and coordinates of the point, we can evaluate the cosine and sine functions as the ratio of the sides.
\[\cos (\theta )=\dfrac{x}{r} =\dfrac{3}{5} \sin (\theta )=\dfrac{y}{r} =\dfrac{4}{5}\nonumber\]
There are a few cosine and sine values which we can determine fairly easily because the corresponding point on the circle falls on the \(x\) or \(y\) axis.
Find \(\cos (90{}^\circ )\) and \(\sin (90{}^\circ )\).
Solution
On any circle, the terminal side of a 90 degree angle points straight up, so the coordinates of the corresponding point on the circle would be (0, r). Using our definitions of cosine and sine,
\[\cos (90{}^\circ )=\dfrac{x}{r} =\dfrac{0}{r} =0\nonumber\]
\[\sin (90{}^\circ )=\dfrac{y}{r} =\dfrac{r}{r} =1\nonumber\]
Find cosine and sine of the angle \(\pi\).
- Answer
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\[\cos (\pi )=-1 \sin (\pi )=0\nonumber\]
Notice that the definitions above can also be stated as:
On a circle of radius \(r\) at an angle of \(\theta\), we can find the coordinates of the point (\(x\), \(y\)) Circles:Points on a Circle at that angle using
\[x=r\cos (\theta )\]
\[y=r\sin (\theta )\]
On a unit circle, a circle with radius 1, \(x=\cos (\theta )\) and \(y=\sin (\theta )\).
Utilizing the basic equation for a circle centered at the origin, \(x^{2} +y^{2} =r^{2}\), combined with the relationships above, we can establish a new identity.
\[x^{2} +y^{2} =r^{2}\nonumber\] substituting the relations above,
\[(r\cos (\theta ))^{2} +(r\sin (\theta ))^{2} =r^{2}\nonumber\] simplifying,
\[r^{2} (\cos (\theta ))^{2} +r^{2} (\sin (\theta ))^{2} =r^{2}\nonumber\] dividing by \(r^{2}\)
\[(\cos (\theta ))^{2} +(\sin (\theta ))^{2} =1\nonumber\] or using shorthand notation
\[\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\nonumber\]
Here \(\cos ^{2} (\theta )\) is a commonly used shorthand notation for \((\cos (\theta ))^{2}\). Be aware that many calculators and computers do not understand the shorthand notation.
In Section 5.1 we related the Pythagorean Theorem \(a^{2} +b^{2} =c^{2}\) to the basic equation of a circle \(x^{2} +y^{2} =r^{2}\), which we have now used to arrive at the Pythagorean Identity.
The Pythagorean Identity. For any angle \(\theta\), \[\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\nonumber\]
One use of this identity is that it helps us to find a cosine value of an angle if we know the sine value of that angle or vice versa. However, since the equation will yield two possible values, we will need to utilize additional knowledge of the angle to help us find the desired value.
If \(\sin (\theta )=\dfrac{3}{7}\) and \(\theta\) is in the second quadrant, find \(\cos (\theta )\).
Solution
Substituting the known value for sine into the Pythagorean identity,
\[\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\nonumber\]
\[\cos ^{2} (\theta )+\dfrac{9}{49} =1\nonumber\]
\[\cos ^{2} (\theta )=\dfrac{40}{49}\nonumber\]
\[\cos (\theta )=\pm \sqrt{\dfrac{40}{49} } =\pm \dfrac{\sqrt{40} }{7} =\pm \dfrac{2\sqrt{10} }{7}\nonumber\]
Since the angle is in the second quadrant, we know the \(x\) value of the point would be negative, so the cosine value should also be negative. Using this additional information, we can conclude that \[\cos (\theta )=-\dfrac{2\sqrt{10} }{7}\nonumber\]
Values for Sine and Cosine
At this point, you may have noticed that we haven’t found any cosine or sine values from angles not on an axis. To do this, we will need to utilize our knowledge of triangles.
First, consider a point on a circle at an angle of 45 de
grees, or \(\dfrac{\pi }{4}\). At this angle, the x and y coordinates of the corresponding point on the circle will be equal because 45 degrees divides the first quadrant in half. Since the \(x\) and \(y\) values will be the same, the sine and cosine values will also be equal. Utilizing the Pythagorean Identity,
\[\cos ^{2} \left(\dfrac{\pi }{4} \right)+\sin ^{2} \left(\dfrac{\pi }{4} \right)=1\nonumber\]
since the sine and cosine are equal, we can substitute sine with cosine
\[\cos ^{2} \left(\dfrac{\pi }{4} \right)+\cos ^{2} \left(\dfrac{\pi }{4} \right)=1\nonumber\] add like terms
\[2\cos ^{2} \left(\dfrac{\pi }{4} \right)=1\nonumber\] divide
\[\cos ^{2} \left(\dfrac{\pi }{4} \right)=\dfrac{1}{2}\nonumber\] since the \(x\) value is positive, we’ll keep the positive root
\[\cos \left(\dfrac{\pi }{4} \right)=\sqrt{\dfrac{1}{2} }\nonumber\] often this value is written with a rationalized denominator
Remember, to rationalize the denominator we multiply by a term equivalent to 1 to get rid of the radical in the denominator.
\[\cos \left(\dfrac{\pi }{4} \right)=\sqrt{\dfrac{1}{2} } \sqrt{\dfrac{2}{2} } =\sqrt{\dfrac{2}{4} } =\dfrac{\sqrt{2} }{2}\nonumber\]
Since the sine and cosine are equal, \(\sin \left(\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2} }{2}\) as well. The (\(x\), \(y\)) coordinates for a point on a circle of radius 1 at an angle of 45 degrees are \(\left(\dfrac{\sqrt{2} }{2} ,\dfrac{\sqrt{2} }{2} \right)\).
Find the coordinates of the point on a circle of radius 6 at an angle of \(\dfrac{\pi }{4}\).
Solution
Using our new knowledge that \(\sin \left(\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2} }{2}\) and \(\cos \left(\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2} }{2}\), along with our relationships that stated \(x=r\cos (\theta )\) and \(y=r\sin (\theta )\), we can find the coordinates of the point desired:
\[x=6\cos \left(\dfrac{\pi }{4} \right)=6\left(\dfrac{\sqrt{2} }{2} \right)=3\sqrt{2}\nonumber \] \[y=6\sin \left(\dfrac{\pi }{4} \right)=6\left(\dfrac{\sqrt{2} }{2} \right)=3\sqrt{2}\nonumber\]
Find the coordinates of the point on a circle of radius 3 at an angle of \(90{}^\circ\).
- Answer
-
\[\begin{array}{l} {x=3\cos \left(\dfrac{\pi }{2} \right)=3\cdot 0=0} \\ {y=3\sin \left(\dfrac{\pi }{2} \right)=3\cdot 1=3} \end{array}\nonumber\]
Next, we will find the cosine and sine at an angle of 30 degrees, or \(\frac{\pi }{6}\)
. To do this, we will first draw a triangle inside a circle with one side at an angle of 30 degrees, and another at an angle of -30 degrees. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be 60 degrees.
Since all the angles are equal, the sides will all be equal as well. The
vertical line has length \(2y\), and since the sides are all equal we can conclude that \(2y = r\), or \(y=\dfrac{r}{2}\). Using this, we can find the sine value:
\[\text{sin}(\dfrac{\pi}{6}) = \dfrac{y}{r} = \dfrac{r/2}{r} = \dfrac{r}{2} \cdot \dfrac{1}{r} = \dfrac{1}{2}\nonumber\]
Using the Pythagorean Identity, we can find the cosine value:
\[\cos ^{2} \left(\dfrac{\pi }{6} \right)+\sin ^{2} \left(\dfrac{\pi }{6} \right)=1\nonumber\]
\[\cos ^{2} \left(\dfrac{\pi }{6} \right)+\left(\dfrac{1}{2} \right)^{2} =1\nonumber\]
\[\cos ^{2} \left(\dfrac{\pi }{6} \right)=\dfrac{3}{4}\nonumber\] since the \(x\) value is positive, we’ll keep the positive root
\[\cos \left(\dfrac{\pi }{6} \right)=\sqrt{\dfrac{3}{4} } =\dfrac{\sqrt{3} }{2}\nonumber\]
The (\(x\), \(y\)) coordinates for the point on a circle of radius 1 at an angle of 30 degrees are \(\left(\dfrac{\sqrt{3} }{2} ,\dfrac{1}{2} \right)\).
By drawing a the triangle inside the unit circle with a 30 degree angle and reflecting it over the line \(y = x\), we can find the cosine and sine for 60 degrees, or \(\dfrac{\pi }{3}\), without any additional work.
By this symmetry, we can see the coordinates of the point on the unit circle at an angle of 60 degrees will be \(\left(\dfrac{1}{2} ,\dfrac{\sqrt{3} }{2} \right)\), giving
\(\cos \left(\dfrac{\pi }{3} \right)=\dfrac{1}{2}\) and \(\sin \left(\dfrac{\pi }{3} \right)=\dfrac{\sqrt{3} }{2}\)
We have now found the cosine and sine values for all the commonly encountered angles in the first quadrant of the unit circle.
| Angle | \(0\) | \(\dfrac{\pi }{6}\), or 30\(\mathrm{{}^\circ}\) | \(\dfrac{\pi }{4}\), or 45\(\mathrm{{}^\circ}\) | \(\dfrac{\pi }{3}\), or 60\(\mathrm{{}^\circ}\) | \(\dfrac{\pi }{2}\), or 90\(\mathrm{{}^\circ}\) |
| Cosine | 1 | \(\dfrac{\sqrt{3} }{2}\) | \(\dfrac{\sqrt{2} }{2}\) | \(\dfrac{1}{2}\) | 0 |
| Sine | 0 | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{2} }{2}\) | \(\dfrac{\sqrt{3} }{2}\) | 1 |
For any given angle in the first quadrant, there will be an angle in another quadrant with the same sine value, and yet another angle in yet another quadrant with the same cosine value. Since the sine value is the \(y\) coordinate on the unit circle, the other angle with the same sine will share the same \(y\) value, but have the opposite \(x\) value. Likewise, the angle with the same cosine will share the same \(x\) value, but have the opposite \(y\) value.
As shown here, angle \(\alpha\) has the same sine value as angle \(\theta\); the cosine values would be opposites. The angle \(\beta\) has the same cosine value as the angle \(\theta\); the sine values would be opposites.
It is important to notice the relationship between the angles. If, from the angle, you measured the smallest angle to the horizontal axis, all would have the same measure in absolute value. We say that all these angles have a reference angle of \(\theta\).
An angle’s reference angle is the size of the smallest angle to the horizontal axis.
A reference angle is always an angle between 0 and 90 degrees, or 0 and \(\dfrac{\pi }{2}\) radians.
Angles share the same cosine and sine values as their reference angles, except for signs (positive or negative) which can be determined from the quadrant of the angle.
Find the reference angle of 150 degrees. Use it to find \(\cos (150{}^\circ )\) and \(\sin (150{}^\circ )\).
Solution
150 degrees is located in the second quadrant. It is 30 degrees short of the horizontal axis at 180 degrees, so the reference angle is 30 degrees.
This tells us that 150 degrees has the same sine and cosine values as 30 degrees, except for sign. We know that \(\sin (30{}^\circ )=\dfrac{1}{2}\) and \(\cos (30{}^\circ )=\dfrac{\sqrt{3} }{2}\). Since 150 degrees is in the second quadrant, the \(x\) coordinate of the point on the circle would be negative, so the cosine value will be negative. The \(y\) coordinate is positive, so the sine value will be positive.
\[\sin (150{}^\circ )=\dfrac{1}{2}\text{ and }\cos (150{}^\circ )=-\dfrac{\sqrt{3} }{2}\nonumber\]
The (\(x\), \(y\)) coordinates for the point on a unit circle at an angle of \(150{}^\circ\) are \(\left(\dfrac{-\sqrt{3} }{2} ,\dfrac{1}{2} \right)\).
Using symmetry and reference angles, we can fill in cosine and sine values at the rest of the special angles on the unit circle. Take time to learn the (\(x\), \(y\)) coordinates of all the major angles in the first quadrant!
Find the coordinates of the point on a circle of radius 12 at an angle of \(\dfrac{7\pi }{6}\).
Solution
Note that this angle is in the third quadrant, where both x and y are negative. Keeping this in mind can help you check your signs of the sine and cosine function.
\[x=12\cos \left(\dfrac{7\pi }{6} \right)=12\left(\dfrac{-\sqrt{3} }{2} \right)=-6\sqrt{3}\nonumber \] \[y=12\sin \left(\dfrac{7\pi }{6} \right)=12\left(\dfrac{-1}{2} \right)=-6\nonumber\]
The coordinates of the point are \((-6\sqrt{3} ,-6)\).
Find the coordinates of the point on a circle of radius 5 at an angle of \(\dfrac{5\pi }{3}\).
- Answer
-
\[\left(5\cos \left(\dfrac{5\pi }{3} \right),5\sin \left(\dfrac{5\pi }{3} \right)\right)=\left(\dfrac{5}{2} ,\dfrac{-5\sqrt{3} }{2} \right)\nonumber\]
We now have the tools to return to the sailboat question posed at the beginning of this section.
Solution
A distress signal is sent from a sailboat during a storm, but the transmission is unclear and the rescue 
boat sitting at the marina cannot determine the sailboat’s location. Using high powered radar, they determine the distress signal is coming from a point 20 miles away at an angle of 225 degrees from the marina. How many miles east/west and north/south of the rescue boat is the stranded sailboat?
We can now answer the question by finding the coordinates of the point on a circle with a radius of 20 miles at an angle of 225 degrees.
\[x=20\cos \left(225{}^\circ \right)=20\left(\dfrac{-\sqrt{2} }{2} \right)\approx -14.142\text{ miles}\nonumber\]
\[y=20\sin \left(225{}^\circ \right)=20\left(\dfrac{-\sqrt{2} }{2} \right)\approx -14.142\text{ miles}\nonumber\]
The sailboat is located 14.142 miles west and 14.142 miles south of the marina.
The special values of sine and cosine in the first quadrant are very useful to know, since knowing them allows you to quickly evaluate the sine and cosine of very common angles without needing to look at a reference or use your calculator. However, scenarios do come up where we need to know the sine and cosine of other angles.
To find the cosine and sine of any other angle, we turn to a computer or calculator. Be aware: most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When you evaluate “cos(30)” on your calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode. Most computer software with cosine and sine functions only operates in radian mode.
Evaluate the cosine of 20 degrees using a calculator or computer.
Solution
On a calculator that can be put in degree mode, you can evaluate this directly to be approximately 0.939693.
On a computer or calculator without degree mode, you would first need to convert the angle to radians, or equivalently evaluate the expression \[\cos \left(20 \cdot \dfrac{\pi }{180} \right)\nonumber\]
Important Topics of This Section
- Degree measure of angle
- Radian measure of angle
- Conversion between degrees and radians
- Common angles in degrees and radians
- Coterminal angles