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Powerpoint slides copied from or based upon:

Connally,

Hughes-Hallett,

Gleason, Et Al.

Copyright 2007 John Wiley & Sons, Inc.

Functions Modeling Change

A Preparation for Calculus

Third Edition

CHAPTER 6

TRIGONOMETRIC FUNCTIONS

SECTION 6.6

OTHER TRIGONOMETRIC FUNCTIONS

The Tangent Function

Another useful trigonometric function is called

the tangent.

Suppose P = (x, y) is the point on the unit

circle corresponding to the angle θ. We define

the tangent of θ, or tan θ, by

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tany

x

The Tangent Function

The graphical interpretation of tan θ is as a

slope.

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tany

x

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0tan

0

y y ym

x x x

6Page 279 Example #1

Find the slope of the line passing through the

origin at an angle of

(a) 30°

(b) π/4

7Page 279 Example #1

Find the slope of the line passing through the

origin at an angle of

(a) 30°

(b) π/4

(a) In degree mode, tan(30) = .5773502692

8Page 279 Example #1

Find the slope of the line passing through the

origin at an angle of

(a) 30°

(b) π/4

(a) In degree mode, tan(30) = .5773502692

Thus, the slope of the line is = .5773502692

9Page 279 Example #1

Find the slope of the line passing through the

origin at an angle of

(a) 30°

(b) π/4

(b) In radian mode, tan (π/4) = 1

10Page 279 Example #1

Find the slope of the line passing through the

origin at an angle of

(a) 30°

(b) π/4

(b) In radian mode, tan (π/4) = 1

An angle of π/4 describes a ray halfway

between the x- and y-axes, or the line y = x,

which has slope of 1. (π/4 = 45°)

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Graph of the Tangent Function

Table 6.10 contains values of the tangent

function.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Graph of the Tangent Function

As Θ goes from 0 to π/2, the slope (tangent)

goes from 0 to +∞.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Graph of the Tangent Function

At precisely π/2 (90 ), this line is vertical, and

its slope is undefined.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Graph of the Tangent Function

Thus, tan(π/2) is undefined and the graph of

y = tan θ has a vertical asymptote at θ = π/2.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Graph of the Tangent Function

For θ between π/2 and π, the slopes are

negative.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Graph of the Tangent Function

The line is very steep near π/2, but becomes

less steep as θ approaches π, where it is

horizontal.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Graph of the Tangent Function

Thus, tan θ becomes less negative as θ

increases in the second quadrant and reaches

0 at θ = π.

θ 0 30 60 90 120 150 180

θ 0 π/6 π/3 π/2 2π/3 5π/6 π

tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0

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Let's graph the tangent function on our

calculator:

y1=tan(x)

Window Value

Xmin 0

Xmax π

Xscl π/2

Ymin -5

Ymax 5

Yscl 1

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y1=tan(x)

Trace...

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y1=tan(x)

x=0, y=0

x=1, y=1.5574077

x=1.5, y=14.10142

x=1.55, y=48.078482

x=1.57, y=1,255.76

x=1.5707963, y=37,320,535!!

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Let's change the window settings:

y1=tan(x)

Window Value

Xmin -2π

Xmax 2π

Xscl π/2

Ymin -5

Ymax 5

Yscl 1

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y1=tan(x)

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For values of θ between π and 2π, observe

that: tan(Θ+π) = tanΘ

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tan(Θ+π) = tanΘ because the angles θ and (θ + π)

determine the same line through the origin, and

hence the same slope.

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tan(Θ+π) = tanΘ

Thus, y = tan θ has period π.

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Notice the differences between the tangent function

and the sinusoidal functions. The tangent function

has a period of π, whereas the sine and cosine both

have periods of 2π.

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The tangent function has vertical asymptotes at odd

multiples of π/2 (1π/2, 3π/2, 5π/2, etc.); the sine and

cosine have no asymptotes.

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Relationships Between the Trigonometric

Functions

There is a relationship among the three

trigonometric functions we have defined.

If (x, y) are the coordinates of the point P

determined by the angle θ on the unit circle,

we defined: cosθ = x and sinθ = y

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Since tan θ = y/x, if x ≠ 0, we have:

sintan

cos

The Unit Circle

The unit circle is the circle of radius one that is

centered at the origin.

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X

y

The Unit Circle

Since the distance from the point P with coordinates

(x, y) to the origin is 1, we have:

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X

y

2 2 1x y

The Unit Circle

Therefore:

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X

y

2 2 1x y

The Unit Circle

Therefore:

Substituting x = cos θ and y = sin θ gives:

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2 2 1x y

2 2cos sin 1

The Reciprocals of the Trigonometric

Functions: Secant, Cosecant, Cotangent

The reciprocals of the trigonometric functions

are given special names. Where the

denominators are not equal to zero, we have

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1secant =sec

cos

1cosecant = csc

sin

coscotangent = cot

sin

Use a graph of g(θ) = cos θ to explain the

shape of the graph of f(θ) = sec θ.

38Page 283 Example #5

Let's use our calculator:

Window: Zoom 7 (ZTrig)

39Page 283 Example #5

11 cos and 2

cosy y

40Page 283 Ex #5

11 cos and 2

cosy y

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In the first quadrant, cos θ decreases from 1 to 0, so

the reciprocal of cos θ increases from 1 toward +∞.

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The values of cos θ are negative in the second

quadrant and decrease from 0 to −1 ...

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so the values of sec θ increase from −∞ to −1.

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y = cos θ is symmetric about the vertical line θ = π, so

the graph of f(θ) = sec θ is symmetric about the same

line.

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Thus, the graph of f(θ) = sec θ on the interval π≤θ≤2π

is the mirror image of the graph on 0≤θ≤π.

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Note that sec θ is undefined wherever cos θ = 0,

namely, at θ = π/2 and θ = 3π/2.

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The graph of f(θ) = sec θ has vertical asymptotes at

those values.

End of Section 6.6

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