]> Trigonometry

Trigonometry

9.1 Radian and Degree Measure

Angle Measures

I am aware of three ways to measure angles: degrees, radians, and gradians. In all cases, an angle in standard position has its vertex at the origin, one ray (the initial side) along the positive x-axis, and the other ray (the terminal side) pointing in some direction.

Degrees divide a circle into 360 parts, each of which is called a degree. These are very familiar to most students. The symbol to denote a degree is a small circle in the superscript position: 15°.

Gradians divide a circle into 400 parts, so that a right angle is 100 gradians. I have only seen this used in military applications.

Radians don't divide the circle into parts; they measure the size of the central angle in a sector of a unit circle with a certain arc length (whew!). Thus, radians really measure the length of an arc.

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Figure 1 - Radian Measure

In mathematics, the radian is the preferred (and in many cases, only) option. In applied situations, degrees are preferred—although certain applications require that radians be used; which forces conversions at the beginning and end of the problem!

You should be able to convert between degrees and radians. Since radians deal with arc length, then a full circle angle (360°) is equal to a number of radians that is the same as the circumference (2πr). The angle part of that is just 2π…just remember that 180° = π (radians), and you'll always be able to set up a proportion to convert between degrees and radians.

Arc Length

The length of the arc subtended by an angle of measure θ in a circle of radius r is $AL = r θ$ (provided the angle is measured in radians). If it isn't, then you need to set up a proportion: $\frac{AL}{2 π r} = \frac{θ}{360 °}$ (where the angle is now in degrees). Of course, this is the same thing as just converting to radians, and then using the easier formula…

Examples

[1.] Convert 18° to radians.

$18 ° \times \frac{π}{180 °} = \frac{π}{10}$ .

[2.] Convert $\frac{4 π}{5}$ to degrees.

$\frac{4 π}{5} \times \frac{180 °}{π} = 4 \times 36 ° = 144 °$ .

[3.] Find the length of the arc subtended by an angle of 12° in a circle of radius 7.

$12 ° \times \frac{π}{180 °} = \frac{π}{15}$ ; $AL = r θ = 7 (\frac{π}{15}) = \frac{7 π}{15}$ .

9.2 The Unit Circle

Definition

The unit circle is a circle of radius one, centered at the origin. Many students have dark nightmares about the unit circle, mainly because they don't understand what it's for: a memorization tool. The special right triangles that you (should have) learned in Geometry do exactly the same thing! Forget all this "wrapping" stuff; it doesn't help.

Special Right Triangles

This knowledge is essential. Through the unit circle, or special right triangles, knowing the relationships between angles and lengths/coordinates is a must.

30-60-90

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45-45-90

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The Connection

As it turns out, every angle intersects the unit circle at a point, which creates a right triangle with hypotenuse of length 1. That is useful once you know how to deal with right triangles and points in space…so keep reading, and it will all become clear eventually. Of course, that's why I'm saving this section for later, as we actually take notes.

9.3 Right Triangle Trigonometry

The Definitions

We begin with a triangle labeled in the usual manner.

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Figure 2 - Standard Right Triangle

Then we have: $sin θ = \frac{opposite}{hypotenuse}$ , $cos θ = \frac{adjacent}{hypotenuse}$ , $tan θ = \frac{opposite}{adjacent}$ , $csc θ = \frac{hypotenuse}{opposite}$ , $sec θ = \frac{hypotenuse}{adjacent}$ , $cot θ = \frac{adjacent}{opposite}$ .

Use these definitions, along with your special right triangles (and/or unit circle) to find sines, cosines, and tangents of the special angles.

Some Identities

Reciprocal

$csc θ = \frac{1}{sin θ}$ , $sec θ = \frac{1}{cos θ}$ , $cot θ = \frac{1}{tan θ}$

Quotient

$tan θ = \frac{sin θ}{cos θ}$ , $cot θ = \frac{cos θ}{sin θ}$

Pythagorean

${sin}^{2} θ + {cos}^{2} θ = 1$ , $1 + {cot}^{2} θ = {csc}^{2} θ$ , ${tan}^{2} θ + 1 = {sec}^{2} θ$

A Note about Squaring

${sin}^{2} θ = {(sin θ)}^{2} \neq sin (θ^{2})$

Examples

[4.] Find the exact values of all six trig functions of the angle θ given in the figure.

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Notice that you must use the Pythagorean Theorem to find the length of the third side: $x^{2} + 8^{2} = 12^{2} ⇒ x^{2} + + 64 = 144 ⇒ x^{2} = 80 ⇒ x = \sqrt{80}$ .

So $sin θ = \frac{8}{12} = \frac{2}{3}$ , $cos θ = \frac{\sqrt{80}}{12}$ , $tan θ = \frac{8}{\sqrt{80}}$ , $csc θ = \frac{12}{8} = \frac{2}{3}$ , $sec θ = \frac{12}{\sqrt{80}}$ , $cot θ = \frac{\sqrt{80}}{8}$ .

[5.] Sketch a right triangle where $tan θ = \frac{3}{4}$ .

Well, since we know the lengths of two sides, we automatically know the third—not that we really need it, in order to draw the triangle.

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[6.] If $tan θ = \frac{3}{4}$ , then find the exact value of $sin θ$ .

Now we do need to know the length of the hypotenuse! $3^{3} + 4^{3} = x^{2} ⇒ 9 + 16 = x^{2} ⇒ x^{2} = 25 ⇒ x = 5$ . Don't forget that 3 is the length of the side opposite, and 4 is the length of the side adjacent. $sin θ = \frac{3}{5}$ .

9.4 Trigonometric Functions of Any Angle

The Definitions

Consider a point (x, y) out in the coordinate plane. Connect that point to the x-axis with a vertical line, and connect the point to the origin—you now have a triangle! Thus, any point in the coordinate plane defines a triangle (well, almost any point…), and you can then define the six trig functions in terms of that point.

Note that r is distance from the point to the origin, and θ is the counter-clockwise angle between the positive x-axis and the line connecting the point to the origin.

$sin θ = \frac{y}{r}$ , $cos θ = \frac{x}{r}$ , $tan θ = \frac{y}{x}$ , $csc θ = \frac{r}{y}$ , $sec θ = \frac{r}{x}$ , $cot θ = \frac{x}{y}$ .

Notice that this definition means that lots of angles (an infinite number) all have the same sine and cosine values—because the angle can wrap around the origin (angles of greater than 360° are allowed).

Reference Angles

Some people have a problem referring to the angle θ in the way that I did…in particular, this allows angles with measures greater than 180°. Thus, some people talk about reference angles—the smallest angle between the line connecting the point and the origin, and the x-axis (positive or negative).

The Unit Circle

As mentioned previously, every angle crosses the unit circle at a point (x, y). The radius (distance from that point to the center) is 1, so we can rewrite the earlier definitions…

$sin θ = y$ , $cos θ = x$ , $tan θ = \frac{y}{x}$ , $csc θ = \frac{1}{y}$ , $sec θ = \frac{1}{x}$ , $cot θ = \frac{x}{y}$ . Thus, we just need to know the coordinates of the point for each angle, and we can use those coordinates to find the values of all six trigonometric functions.

See? The unit circle is just a memorization tool!

Examples

[7.] Find the value of all six trigonometric functions for the angle determined by the point (2, 5).

First of all, find r: $2^{3} + 5^{3} = r^{2} ⇒ 4 + 25 = r^{2} ⇒ r^{2} = 29 ⇒ r = \sqrt{29}$ .

$sin θ = \frac{5}{\sqrt{29}}$ , $cos θ = \frac{2}{\sqrt{29}}$ , $tan θ = \frac{5}{2}$ , $csc θ = \frac{\sqrt{29}}{5}$ , $sec θ = \frac{\sqrt{29}}{2}$ , $cot θ = \frac{2}{5}$ .

[8.] Evaluate $cos 405 °$ without a calculator.

Notice that 405° = 360° + 45°. Thus, 405° and 45° point in the same direction—rays along those angles pass through the same points. Thus, $cos 405 ° = cos 45 ° = \frac{1}{\sqrt{2}}$ .

9.5 Graphs of Sine and Cosine Functions

First up: when we start graphing trigonometric functions, we MUST use radians.

Hint: if you are sketching these functions without a calculator, then use a few choice points—x-intercepts, maxima and minima. You should know those points from earlier memorization!

The Basic Graphs

The horizontal scale on these graphs is $\frac{π}{4}$ , and the vertical scale is 1.

y = sin x

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y = cos x

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Transformations

Naturally, you can transform these graphs. The templates are $y = a \cdot sin (b (x - c)) + d$ and $y = a \cdot cos (b (x - c)) + d$ . The parameters a, b, c and d all do exactly what they have always done—but in trigonometry, they sometimes get special names. For example, the amplitude of the function is |a|, and the phase shift is c.

Remember that the trigonometric functions are periodic—they repeat a pattern every so often. Sine and cosine normally have a period of 2π; when transformed, this changes (don't forget that the period is a horizontal property!).

Examples

[9.] Find the period and amplitude of y = -2cos (4x + 1).

The amplitude is |-2| = 2. The new period is the original period (2π) divided by b (4): $\frac{2 π}{4} = \frac{π}{2}$ .

[10.] Describe a sequence of transformations that changes the graph of y = sin x into y = 3sin (x + π) - 1.

Multiply the y-values by 3; shift π units to the left; shift down 1 unit.

9.6 Graphs of Other Trigonometric Functions

Sine and Cosine are continuous—no gaps or breaks. The other trigonometric functions are not so nice. Here are their graphs:

The horizontal scale on these graphs is $\frac{π}{4}$ , and the vertical scale is 1.

y = csc x

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y = sec x

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y = tan x

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y = cot x

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Of course, you can transform each of these…the templates all look the same. One difference is that |a| is not called the amplitude for these graphs.

If you need to sketch these graphs without a calculator, then you should begin with the asymptotes and x-intercepts/extrema. These should be fairly easy to find if you don't have them memorized…

If you need to graph these with a calculator, then you need to know the Quotient identities—since calculators only have Sine, Cosine and Tangent available.

Examples

[11.] Sketch the graph of $y = 2 \cdot cot x$ .

Enter $2 (1 / tan x)$ into your calculator…

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[12.] Solve (graphically) $sec x = -2$ for $x \in [-2 π, 2 π]$ .

Enter $1 / cos x$ and y = -2, then find the intersection.

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The solutions are -4.1888, -2.0944, 2.0944, and 4.1888.

9.7 Inverse Trigonometric Functions

The Need

We often need to "undo" a function—take what is usually the output, and make it the input, in order to find what is usually given (whew!). We use inverse functions for this. We now have six (well, three, at least) brand-new functions…what about inverse functions?

The Problem

As we learned (reviewed) in Chapter 2, not every function has an inverse function. In particular, a function must be one-to-one (pass the horizontal line test) in order to have an inverse function.

Unfortunately, none of the trigonometric functions is one-to-one!

The Solution

Something that we mentioned only in passing back in Chapter 2 was that you can force two things to be inverse functions if you restrict the domain of one or both functions. And that is precisely what we must do to the trigonometric functions!

Definitions and Limitations

The inverse function of $y = sin x$ is $y = arcsin x = {sin}^{-1} x$ . Since sine is a function, the "exponent" of -1 really isn't an exponent—it denotes the inverse. Some people get around this by calling the function arcsine. You may (at some point) be told that there is a difference between arcsine and Arcsine. There is—but the difference is not terribly important for most of us, and we won't linger on them. We will be using the lowercase version exclusively.

The restricted domain of sine is $[- \frac{π}{2}, \frac{π}{2}]$ , which still produces a range of [-1, 1]. That means that the domain of arcsine is [-1, 1], and its range is $[- \frac{π}{2}, \frac{π}{2}]$ .

Inverse Function	Domain	Range
arcsine	[-1, 1]	$[- \frac{π}{2}, \frac{π}{2}]$
arccosine	[-1, 1]	[0, π]
arctangent	$ℝ$	$[- \frac{π}{2}, \frac{π}{2}]$

Very few (if any) people refer to inverse functions for secant, cosecant and cotangent—so we won't, either.

Examples

[13.] arcos(½) = ?

Since $cos (\frac{π}{3}) = \frac{1}{2}$ , $arccos (\frac{1}{2}) = \frac{π}{3}$ .

[14.] arcsin(-½) = ?

Careful! You may be tempted to say that since $sin (\frac{11 π}{6}) = - \frac{1}{2}$ , the answer must be $\frac{11 π}{6}$ —but that can't be; the range of arcsine is $[- \frac{π}{2}, \frac{π}{2}]$ , and $\frac{11 π}{6}$ isn't in that interval! The answer will be an angle that is coterminal with $\frac{11 π}{6}$ , but in the interval $[- \frac{π}{2}, \frac{π}{2}]$ . That angle is $- \frac{π}{6}$ .

[15.] $arctan (tan \frac{π}{4})$ = ?

Since they are inverses, they just wipe each other out—the answer is $\frac{π}{4}$ .

[16.] $arccos (cos \frac{4 π}{3})$ = ?

Alas, $\frac{4 π}{3}$ isn't in the range of arccosine, so that can't be the answer! We need an angle that has the same cosine value as $\frac{4 π}{3}$ , but in the range of arccosine ( [0, π] ). That angle is $\frac{2 π}{3}$ .

[17.] $cos (arcsin \frac{5}{13})$ = ?

This requires a specific technique, since $\frac{5}{13}$ isn't a special fraction from the unit circle / special right triangles. remember that the arc functions take in real numbers, and spit out angles. Thus, what we ultimately want to do is take the cosine of some angle—if only we knew what the angle was! We can draw it…let $arcsin \frac{5}{13} = θ$ .

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Now—the question asks for $cos (arcsin \frac{5}{13})$ , which is the same as cos θ…can you find cos θ? I should hope so! $cos (arcsin \frac{5}{13}) = \frac{12}{13}$ .

Page last validated 2010-08-15

Trigonometry

9.1 Radian and Degree Measure

Angle Measures

Arc Length

Other Topics

Examples

9.2 The Unit Circle

Definition

Special Right Triangles

30-60-90

45-45-90

The Connection

9.3 Right Triangle Trigonometry

The Definitions

Some Identities

Reciprocal

Quotient

Pythagorean

A Note about Squaring

Examples

9.4 Trigonometric Functions of Any Angle

The Definitions

Reference Angles

The Unit Circle

Examples

9.5 Graphs of Sine and Cosine Functions

The Basic Graphs

Transformations

Examples

9.6 Graphs of Other Trigonometric Functions

Examples

9.7 Inverse Trigonometric Functions

The Need

The Problem

The Solution

Definitions and Limitations

Examples