Convert angle measures between degrees and radians
Recognize the triangular and circular definitions of the basic trigonometric functions
Write the basic trigonometric identities
Radian Measure
To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle [latex]\theta[/latex], let [latex]s[/latex] be the length of the corresponding arc on the unit circle (Figure 1). We say the angle corresponding to the arc of length 1 has radian measure 1.
Figure 1. The radian measure of an angle [latex]\theta [/latex] is the arc length [latex]s[/latex] of the associated arc on the unit circle.
Since an angle of 360° corresponds to the circumference of a circle, or an arc of length [latex]2\pi [/latex], we conclude that an angle with a degree measure of 360° has a radian measure of [latex]2\pi [/latex]. Similarly, we see that 180° is equivalent to [latex]\pi [/latex] radians. The table below shows the relationship between common degree and radian values.
Table 1. Common Angles Expressed in Degrees and Radians
Degrees
Radians
Degrees
Radians
0
0
120
[latex]2\pi/3[/latex]
30
[latex]\pi/6[/latex]
135
[latex]3\pi/4[/latex]
45
[latex]\pi/4[/latex]
150
[latex]5\pi/6[/latex]
60
[latex]\pi/3[/latex]
180
[latex]\pi [/latex]
90
[latex]\pi/2[/latex]
Though the common radian and degree equivalents in the table above are worth memorizing, if you can’t remember them, don’t forget that we have a conversion formula!
Recall: radian and degree conversions
Converting between Radians and Degrees
Because degrees and radians both measure angles, we need to be able to convert between them. We can easily do so using a proportion.
This proportion shows that the measure of angle [latex]\theta [/latex] in degrees divided by 180 equals the measure of angle [latex]\theta [/latex] in radians divided by [latex]\pi . [/latex] Or, phrased another way, degrees is to 180 as radians is to [latex]\pi [/latex].
Express [latex]\dfrac{5\pi}{3}[/latex] rad using degrees.
Show Solution
Use the fact that 180° is equivalent to [latex]\pi [/latex] radians as a conversion factor: [latex]1=\dfrac{\pi \, \text{rad}}{180^{\circ}}=\dfrac{180^{\circ}}{\pi \, \text{rad}}[/latex].
Express 210° using radians. Express [latex]\dfrac{11\pi}{6}[/latex] rad using degrees.
Hint
[latex]\pi [/latex] radians is equal to [latex]180^{\circ}[/latex].
Show Solution
[latex]\dfrac{7\pi}{6}[/latex] rad; 330°
Try It
The Six Basic Trigonometric Functions
Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.
To define the trigonometric functions, first consider the unit circle centered at the origin and a point [latex]P=(x,y)[/latex] on the unit circle. Let [latex]\theta [/latex] be an angle with an initial side that lies along the positive [latex]x[/latex]-axis and with a terminal side that is the line segment [latex]OP[/latex]. An angle in this position is said to be in standard position (Figure 2). We can then define the values of the six trigonometric functions for [latex]\theta [/latex] in terms of the coordinates [latex]x[/latex] and [latex]y[/latex].
Figure 2. The angle [latex]\theta [/latex] is in standard position. The values of the trigonometric functions for [latex]\theta [/latex] are defined in terms of the coordinates [latex]x[/latex] and [latex]y[/latex].
Definition
Let [latex]P=(x,y)[/latex] be a point on the unit circle centered at the origin [latex]O[/latex]. Let [latex]\theta [/latex] be an angle with an initial side along the positive [latex]x[/latex]-axis and a terminal side given by the line segment [latex]OP[/latex]. The trigonometric functions are then defined as
If [latex]x=0[/latex], then [latex]\sec \theta [/latex] and [latex]\tan \theta [/latex] are undefined. If [latex]y=0[/latex], then [latex]\cot \theta [/latex] and [latex]\csc \theta [/latex] are undefined.
We can see that for a point [latex]P=(x,y)[/latex] on a circle of radius [latex]r[/latex] with a corresponding angle [latex]\theta[/latex], the coordinates [latex]x[/latex] and [latex]y[/latex] satisfy
The values of the other trigonometric functions can be expressed in terms of [latex]x, \, y[/latex], and [latex]r[/latex] (Figure 3).
Figure 3. For a point [latex]P=(x,y)[/latex] on a circle of radius [latex]r[/latex], the coordinates [latex]x[/latex] and [latex]y[/latex] satisfy [latex]x=r \cos \theta [/latex] and [latex]y=r \sin \theta[/latex].
The table below shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of [latex]\sin \theta [/latex] and [latex]\cos \theta[/latex].
Values of [latex]\sin \theta [/latex] and [latex]\cos \theta [/latex] at Major Angles [latex]\theta[/latex] in the First Quadrant
On the unit circle, the angle [latex]\theta =\large\frac{2\pi}{3}[/latex] corresponds to the point [latex]\Big(-\large\frac{1}{2}, \frac{\sqrt{3}}{2}\Big)[/latex]. Therefore, [latex]\sin \Big(\large\frac{2\pi}{3}\Big) \normalsize = y = \large\frac{\sqrt{3}}{2}[/latex].
Figure 4. Unit circle depiction of [latex]\theta[/latex]
An angle [latex]\theta =-\large\frac{5\pi}{6}[/latex] corresponds to a revolution in the negative direction, as shown. Therefore, [latex]\cos \Big(-\large\frac{5\pi}{6}\Big)[/latex] [latex]=x=-\large\frac{\sqrt{3}}{2}[/latex].
Figure 5. Unit circle depiction of [latex]\theta[/latex]
An angle [latex]\theta =\large\frac{15\pi}{4} \normalsize = 2\pi +\large\frac{7\pi}{4}[/latex]. Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of [latex]\large\frac{7\pi}{4}[/latex] corresponds to the point [latex]\Big(\large\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\Big)[/latex], we can conclude that [latex]\tan \Big(\large\frac{15\pi}{4}\Big)\normalsize =\large\frac{y}{x}[/latex] [latex]=-1[/latex].
Figure 6. Unit circle depiction of [latex]\theta[/latex]
Try It
Evaluate [latex]\cos \left(\dfrac{3\pi}{4}\right)[/latex] and [latex]\sin \left(\dfrac{−\pi}{6}\right)[/latex].
As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let [latex]\theta [/latex] be one of the acute angles. Let [latex]A[/latex] be the length of the adjacent leg, [latex]O[/latex] be the length of the opposite leg, and [latex]H[/latex] be the length of the hypotenuse. By inscribing the triangle into a circle of radius [latex]H[/latex], as shown in Figure 7, we see that [latex]A, \, H[/latex], and [latex]O[/latex] satisfy the following relationships with [latex]\theta[/latex]:
Figure 7. By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at [latex]\theta[/latex].
Example: Constructing a Wooden Ramp
A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be [latex]10^{\circ}[/latex], how long does the ramp need to be?
Show Solution
Let [latex]x[/latex] denote the length of the ramp. In the following image, we see that [latex]x[/latex] needs to satisfy the equation [latex]\sin(10^{\circ})=\dfrac{4}{x}[/latex]. Solving this equation for [latex]x[/latex], we see that [latex]x=\frac{4}{ \sin(10^{\circ})} \approx 23.035[/latex] ft.
Figure 8. Sketch of the ramp and staircase.
Watch the following video to see the worked solution to Example: Constructing a Wooden Ramp
Closed Captioning and Transcript Information for Video
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A house painter wants to lean a 20-ft ladder against a house. If the angle between the base of the ladder and the ground is to be [latex]60^{\circ}[/latex], how far from the house should she place the base of the ladder?
Hint
Draw a right triangle with hypotenuse 20 ft.
Show Solution
10 ft
Trigonometric Identities
A trigonometric identity is an equation involving trigonometric functions that is true for all angles [latex]\theta [/latex] for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.
Remember, solving a trigonometric equation is not very different from solving an algebraic equation.
Recall: Given a trigonometric equation, solve using algebra.
Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
Substitute the trigonometric expression with a single variable, such as [latex]x[/latex] or [latex]u[/latex].
Solve the equation the same way an algebraic equation would be solved.
Substitute the trigonometric expression back in for the variable in the resulting expressions.
Solve for the angle.
Try It
Additionally, trigonometric equations will have an infinite number of solutions. If we need to find all possible solutions, then we must add [latex]2\pi k[/latex], where [latex]k[/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\pi :[/latex]
For each of the following equations, use a trigonometric identity to find all solutions.
[latex]1+\cos(2\theta)=\cos \theta [/latex]
[latex]\sin(2\theta)=\tan \theta [/latex]
Show Solution
a. Using the double-angle formula for [latex]\cos(2\theta)[/latex], we see that [latex]\theta [/latex] is a solution of
[latex]1+\cos(2\theta)=\cos \theta [/latex]
if and only if
[latex]1+2\cos^2 \theta -1=\cos \theta[/latex],
which is true if and only if
[latex]2\cos^2 \theta -\cos \theta =0[/latex].
To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by [latex]\cos \theta[/latex]. The problem with dividing by [latex]\cos \theta[/latex] is that it is possible that [latex]\cos \theta[/latex] is zero. In fact, if we did divide both sides of the equation by [latex]\cos \theta[/latex], we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that [latex]\theta [/latex] is a solution of this equation if and only if
b. Using the double-angle formula for [latex]\sin(2\theta)[/latex] and the reciprocal identity for [latex]\tan(\theta)[/latex], the equation can be written as
To solve this equation, we multiply both sides by [latex]\cos \theta [/latex] to eliminate the denominator, and say that if [latex]\theta [/latex] satisfies this equation, then [latex]\theta [/latex] satisfies the equation
However, we need to be a little careful here. Even if [latex]\theta [/latex] satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by [latex]\cos \theta[/latex]. However, if [latex]\cos \theta =0[/latex], we cannot divide both sides of the equation by [latex]\cos \theta[/latex]. Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor [latex]\sin \theta [/latex] out of both terms on the left-hand side instead of dividing both sides of the equation by [latex]\sin \theta[/latex]. Factoring the left-hand side of the equation, we can rewrite this equation as
[latex]\sin \theta (2\cos^2 \theta -1)=0[/latex]
Therefore, the solutions are given by the angles [latex]\theta [/latex] such that [latex]\sin \theta =0[/latex] or [latex]\cos^2 \theta =1/2[/latex]. The solutions of the first equation are [latex]\theta =0, \pm \pi, \pm 2\pi, \cdots[/latex]. The solutions of the second equation are [latex]\theta =\pi /4, \, (\pi/4) \pm (\pi/2), \, (\pi/4) \pm \pi, \cdots[/latex]. After checking for extraneous solutions, the set of solutions to the equation is
Watch the following video to see the worked solution to Example: Solving Trigonometric Equations
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
[latex]\pi [/latex] radians is equal to [latex]180^{\circ}[/latex].