Algebra and Trigonometry

# Chapter 8

### 8.1Graphs of the Sine and Cosine Functions

1.

$6π 6π$

2.

$1 2 1 2$compressed

3.

$π 2 ; π 2 ;$right

4.

2 units up

5.

midline:$y=0; y=0;$amplitude:$| A |= 1 2 ; | A |= 1 2 ;$period:$P= 2π | B | =6π; P= 2π | B | =6π;$phase shift:$C B =π C B =π$

6.

$f( x )=sin(x)+2 f( x )=sin(x)+2$

7.

two possibilities:$y=4sin( π 5 x− π 5 )+4 y=4sin( π 5 x− π 5 )+4$or$y=−4sin( π 5 x+ 4π 5 )+4 y=−4sin( π 5 x+ 4π 5 )+4$

8. midline:$y=0; y=0;$amplitude:$| A |=0.8; | A |=0.8;$period:$P= 2π | B | =π; P= 2π | B | =π;$phase shift:$C B =0 C B =0$ or none

9. midline:$y=0; y=0;$amplitude:$| A |=2; | A |=2;$period:$P= 2π | B | =6; P= 2π | B | =6;$phase shift:$C B =− 1 2 C B =− 1 2$

10.

7 11.

$y=3cos( x )−4 y=3cos( x )−4$ ### 8.2Graphs of the Other Trigonometric Functions

1. 2.

It would be reflected across the line$y=−1, y=−1,$becoming an increasing function.

3.

$g(x)=4tan(2x) g(x)=4tan(2x)$

4.

This is a vertical reflection of the preceding graph because$A A$is negative. 5. 6. 7. ### 8.3Inverse Trigonometric Functions

1.

$arccos(0.8776)≈0.5 arccos(0.8776)≈0.5$

2.

a.$− π 2 ; − π 2 ;$b.$− π 4 ; − π 4 ;$c.$π; π;$ d.$π 3 π 3$

3.

1.9823 or 113.578°

4.

$sin −1 (0.6)=36.87°=0.6435 sin −1 (0.6)=36.87°=0.6435$radians

5.

$π 8 ; 2π 9 π 8 ; 2π 9$

6.

$3π 4 3π 4$

7.

$12 13 12 13$

8.

$4 2 9 4 2 9$

9.

$4x 16 x 2 +1 4x 16 x 2 +1$

### 8.1 Section Exercises

1.

The sine and cosine functions have the property that$f( x+P )=f( x ) f( x+P )=f( x )$for a certain$P. P.$This means that the function values repeat for every$P P$units on the x-axis.

3.

The absolute value of the constant$A A$(amplitude) increases the total range and the constant$D D$(vertical shift) shifts the graph vertically.

5.

At the point where the terminal side of$t t$intersects the unit circle, you can determine that the$sin t sin t$equals the y-coordinate of the point.

7. amplitude:$2 3 ; 2 3 ;$period:$2π; 2π;$midline:$y=0; y=0;$maximum:$y= 2 3 y= 2 3$occurs at$x=0; x=0;$minimum:$y=− 2 3 y=− 2 3$occurs at$x=π; x=π;$for one period, the graph starts at 0 and ends at$2π 2π$

9. amplitude: 4; period:$2π; 2π;$midline:$y=0; y=0;$maximum$y=4 y=4$occurs at$x= π 2 ; x= π 2 ;$minimum:$y=−4 y=−4$occurs at$x= 3π 2 ; x= 3π 2 ;$one full period occurs from$x=0 x=0$to$x=2π x=2π$

11. amplitude: 1; period:$π; π;$midline:$y=0; y=0;$maximum:$y=1 y=1$occurs at$x=π; x=π;$minimum:$y=−1 y=−1$occurs at$x= π 2 ; x= π 2 ;$one full period is graphed from$x=0 x=0$to$x=π x=π$

13. amplitude: 4; period: 2; midline:$y=0; y=0;$maximum:$y=4 y=4$occurs at$x=0; x=0;$minimum:$y=−4 y=−4$occurs at$x=1 x=1$

15. amplitude: 3; period:$π 4 ; π 4 ;$midline:$y=5; y=5;$maximum:$y=8 y=8$occurs at$x=0.12; x=0.12;$minimum:$y=2 y=2$occurs at$x=0.516; x=0.516;$horizontal shift:$−4; −4;$vertical translation 5; one period occurs from$x=0 x=0$to$x= π 4 x= π 4$

17. amplitude: 5; period:$2π 5 ; 2π 5 ;$midline:$y=−2; y=−2;$maximum:$y=3 y=3$occurs at$x=0.08; x=0.08;$minimum:$y=−7 y=−7$occurs at$x=0.71; x=0.71;$phase shift:$−4; −4;$vertical translation:$−2; −2;$one full period can be graphed on$x=0 x=0$to$x= 2π 5 x= 2π 5$

19. amplitude: 1 ; period:$2π; 2π;$midline:$y=1; y=1;$maximum:$y=2 y=2$occurs at$x=2.09; x=2.09;$maximum:$y=2 y=2$occurs at$t=2.09; t=2.09;$minimum:$y=0 y=0$occurs at$t=5.24; t=5.24;$phase shift:$− π 3 ; − π 3 ;$vertical translation: 1; one full period is from$t=0 t=0$to$t=2π t=2π$

21. amplitude: 1; period:$4π; 4π;$midline:$y=0; y=0;$maximum:$y=1 y=1$occurs at$t=11.52; t=11.52;$minimum:$y=−1 y=−1$occurs at$t=5.24; t=5.24;$phase shift:$− 10π 3 ; − 10π 3 ;$vertical shift: 0

23.

amplitude: 2; midline:$y=−3; y=−3;$period: 4; equation:$f(x)=2sin( π 2 x )−3 f(x)=2sin( π 2 x )−3$

25.

amplitude: 2; period: 5; midline:$y=3; y=3;$equation:$f(x)=−2cos( 2π 5 x )+3 f(x)=−2cos( 2π 5 x )+3$

27.

amplitude: 4; period: 2; midline:$y=0; y=0;$equation:$f(x)=−4cos( π( x− π 2 ) ) f(x)=−4cos( π( x− π 2 ) )$

29.

amplitude: 2; period: 2; midline$y=1; y=1;$equation:$f( x )=2cos( πx )+1 f( x )=2cos( πx )+1$

31.

$0,π 0,π$

33.

$sin(π2)=1 sin(π2)=1$

35.

$π2 π2$

37.

$f(x)=sinx f(x)=sinx$ is symmetric

39.

$π3,5π3 π3,5π3$

41.

Maximum: $1 1$ at $x= 0 x=0$ ; minimum: $-1 -1$ at $x= π x=π$

43.

A linear function is added to a periodic sine function. The graph does not have an amplitude because as the linear function increases without bound the combined function $h(x)=x+sinx h(x)=x+sinx$ will increase without bound as well. The graph is bounded between the graphs of $y=x+1 y=x+1$ and $y=x-1 y=x-1$ because sine oscillates between −1 and 1. 45.

There is no amplitude because the function is not bounded. 47.

The graph is symmetric with respect to the y-axis and there is no amplitude because the function’s bounds decrease as $|x| |x|$ grows. There appears to be a horizontal asymptote at $y=0 y=0$ . ### 8.2 Section Exercises

1.

Since$y=csc x y=csc x$is the reciprocal function of$y=sin x, y=sin x,$you can plot the reciprocal of the coordinates on the graph of$y=sin x y=sin x$to obtain the y-coordinates of$y=csc x. y=csc x.$The x-intercepts of the graph$y=sin x y=sin x$are the vertical asymptotes for the graph of$y=csc x. y=csc x.$

3.

Answers will vary. Using the unit circle, one can show that$tan( x+π )=tan x. tan( x+π )=tan x.$

5.

The period is the same:$2π. 2π.$

7.

IV

9.

III

11.

period: 8; horizontal shift: 1 unit to left

13.

1.5

15.

5

17.

$−cotxcosx−sinx −cotxcosx−sinx$

19. stretching factor: 2; period:asymptotes:

21. stretching factor: 6; period: 6; asymptotes:

23. stretching factor: 1; period:asymptotes:

25. Stretching factor: 1; period:asymptotes:

27. stretching factor: 2; period:asymptotes:

29. stretching factor: 4; period:asymptotes:

31. stretching factor: 7; period:asymptotes:

33. stretching factor: 2; period:asymptotes:

35. stretching factor:period:asymptotes:

37.

$y=tan( 3( x− π 4 ) )+2 y=tan( 3( x− π 4 ) )+2$ 39.

$f( x )=csc( 2x ) f( x )=csc( 2x )$

41.

$f( x )=csc( 4x ) f( x )=csc( 4x )$

43.

$f( x )=2cscx f( x )=2cscx$

45.

$f(x)= 1 2 tan(100πx) f(x)= 1 2 tan(100πx)$

47. 49. 51. 53. 55.
1. $( − π 2 , π 2 ); ( − π 2 , π 2 );$
2. 3. $x=− π 2 x=− π 2$and$x= π 2 ; x= π 2 ;$the distance grows without bound as$| x | | x |$approaches$π 2 π 2$—i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;
4. 3; when$x=− π 3 , x=− π 3 ,$the boat is 3 km away;
5. 1.73; when$x= π 6 , x= π 6 ,$the boat is about 1.73 km away;
6. 1.5 km; when$x=0 x=0$
57.
1. $h( x )=2tan( π 120 x ); h( x )=2tan( π 120 x );$
2. 3. $h( 0 )=0: h( 0 )=0:$after 0 seconds, the rocket is 0 mi above the ground;$h( 30 )=2: h( 30 )=2:$after 30 seconds, the rockets is 2 mi high;
4. As$x x$approaches 60 seconds, the values of$h( x ) h( x )$grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it.

### 8.3 Section Exercises

1.

The function$y=sinx y=sinx$is one-to-one on$[ − π 2 , π 2 ]; [ − π 2 , π 2 ];$thus, this interval is the range of the inverse function of$y=sinx, y=sinx,$$f(x)= sin −1 x. f(x)= sin −1 x.$The function$y=cosx y=cosx$is one-to-one on $[ 0,π ]; [ 0,π ];$thus, this interval is the range of the inverse function of$y=cosx,f(x)= cos −1 x. y=cosx,f(x)= cos −1 x.$

3.

$π 6 π 6$is the radian measure of an angle between$− π 2 − π 2$and$π 2 π 2$whose sine is 0.5.

5.

In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval$[ − π 2 , π 2 ] [ − π 2 , π 2 ]$so that it is one-to-one and possesses an inverse.

7.

True . The angle,$θ 1 θ 1$that equals$arccos(−x) arccos(−x)$,$x>0 x>0$, will be a second quadrant angle with reference angle,$θ 2 θ 2$, where$θ 2 θ 2$equals$arccosx arccosx$,$x>0 x>0$. Since$θ 2 θ 2$is the reference angle for$θ 1 θ 1$,$θ 2 =π− θ 1 θ 2 =π− θ 1$and$arccos(−x) arccos(−x)$=$π−arccosx π−arccosx$-

9.

$− π 6 − π 6$

11.

$3π 4 3π 4$

13.

$− π 3 − π 3$

15.

$π 3 π 3$

17.

1.98

19.

0.93

21.

1.41

23.

25.

0

27.

0.71

29.

-0.71

31.

$− π 4 − π 4$

33.

0.8

35.

$5 13 5 13$

37.

$x−1 − x 2 +2x x−1 − x 2 +2x$

39.

$x 2 −1 x x 2 −1 x$

41.

$x+0.5 − x 2 −x+ 3 4 x+0.5 − x 2 −x+ 3 4$

43.

$2x+1 x+1 2x+1 x+1$

45.

$2x+1 x 2x+1 x$

47.

$t t$

49. domain$[ −1,1 ]; [ −1,1 ];$range$[ 0,π ] [ 0,π ]$

51.

approximately$x=0.00 x=0.00$

53.

55.

57.

59.

61.

No. The angle the ladder makes with the horizontal is 60 degrees.

### Review Exercises

1.

amplitude: 3; period:$2π; 2π;$midline:$y=3; y=3;$no asymptotes 3.

amplitude: 3; period:$2π; 2π;$midline:$y=0; y=0;$no asymptotes 5.

amplitude: 3; period:$2π; 2π;$midline:$y=−4; y=−4;$no asymptotes 7.

amplitude: 6; period:$2π 3 ; 2π 3 ;$midline:$y=−1; y=−1;$no asymptotes 9.

stretching factor: none; period:midline:asymptotes:whereis an integer 11.

stretching factor: 3; period:midline:asymptotes:$x= π 8 + π 4 k, x= π 8 + π 4 k,$whereis an integer 13.

amplitude: none; period:$2π; 2π;$no phase shift; asymptotes:whereis an odd integer 15.

amplitude: none; period:no phase shift; asymptotes:whereis an integer 17.

amplitude: none; period:no phase shift; asymptotes:whereis an integer 19.

largest: 20,000; smallest: 4,000

21.

amplitude: 8,000; period: 10; phase shift: 0

23.

In 2007, the predicted population is 4,413. In 2010, the population will be 11,924.

25.

5 in.

27.

10 seconds

29.

$π 6 π 6$

31.

$π 4 π 4$

33.

$π 3 π 3$

35.

No solution

37.

$12 5 12 5$

39.

The graphs are not symmetrical with respect to the line$y=x. y=x.$They are symmetrical with respect to the$y y$-axis. 41.

The graphs appear to be identical. ### Practice Test

1.

amplitude: 0.5; period:$2π; 2π;$midline$y=0 y=0$ 3.

amplitude: 5; period:$2π; 2π;$midline:$y=0 y=0$ 5.

amplitude: 1; period:$2π; 2π;$midline:$y=1 y=1$ 7.

amplitude: 3; period:$6π; 6π;$midline:$y=0 y=0$ 9.

amplitude: none; period:midline:asymptotes:whereis an integer 11.

amplitude: none; period:midline: asymptotes: whereis an integer 13.

amplitude: none; period:$2π; 2π;$midline:$y=−3 y=−3$ 15.

amplitude: 2; period: 2; midline:$y=0; y=0;$$f( x )=2sin( π( x−1 ) ) f( x )=2sin( π( x−1 ) )$

17.

amplitude: 1; period: 12; phase shift:$−6; −6;$midline$y=−3 y=−3$

19.

$D( t )=68−12sin( π 12 x ) D( t )=68−12sin( π 12 x )$

21.

period:$π 6 ; π 6 ;$horizontal shift:$−7 −7$

23.

$f( x )=sec( πx ); f( x )=sec( πx );$period: 2; phase shift: 0

25.

$4 4$

27.

The views are different because the period of the wave is$1 25 . 1 25 .$Over a bigger domain, there will be more cycles of the graph. 29.

$3 5 3 5$

31.

On the approximate intervals$( 0.5,1 ),( 1.6,2.1 ),( 2.6,3.1 ),( 3.7,4.2 ),( 4.7,5.2 ),(5.6,6.28) ( 0.5,1 ),( 1.6,2.1 ),( 2.6,3.1 ),( 3.7,4.2 ),( 4.7,5.2 ),(5.6,6.28)$

33.

$f( x )=2cos( 12( x+ π 4 ) )+3 f( x )=2cos( 12( x+ π 4 ) )+3$ 35.

This graph is periodic with a period of$2π. 2π.$ 37.

$π 3 π 3$

39.

$π 2 π 2$

41.

$1− ( 1−2x ) 2 1− ( 1−2x ) 2$

43.

$1 1+ x 4 1 1+ x 4$

45.

$x+1 x x+1 x$

47.

False

49.