Unit 6 trigonometric identities and equations homework 2 answer key

Try It

6.1 Graphs of the Sine and Cosine Functions

5.

midline: y=0;y=0 ; amplitude: |A|=12;|A|=12; period: P =2π|B|=6π;P=2π|B|=6π; phase shift: CB=πCB=π

6.

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

7.

two possibilities: y=4sin(π5x− π5)+4y=4sin(π5x−π5)+4 or y=−4sin (π5x+4π5)+4y=−4sin(π5x+4π5)+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: CB=0CB=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: CB=−1 2CB=−12

11.

y=3cos(x)−4 y=3cos(x)−4

6.2 Graphs of the Other Trigonometric Functions

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.

6.3 Inverse Trigonometric Functions

1.

arccos(0.8776)≈0.5 arccos(0.8776)≈0.5

2.

1. ⓐ−π2;− π2;
2. ⓑ−π4;−π4;
3. ⓒπ;π;
4. ⓓπ3π3

4.

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

9.

4x16x2+14x 16x2+1

6.1 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 PP units on the x-axis.

3.

The absolute value of the constant AA (amplitude) increases the total range and the constant DD (vertical shift) shifts the graph vertically.

5.

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

7.

amplitude: 23;23; period: 2π;2π; midline: y=0;y=0; maximum: y=23y=23 occurs at x=0;x=0 ; minimum: y=−23y=−23 occurs at x=π;x=π; for one period, the graph starts at 0 and ends at 2π2π

9.

amplitude: 4; period: 2π;2π; midline: y=0;y=0; maximum y=4y=4 occurs at x=π2;x=π2; minimum: y=−4y= −4 occurs at x=3π2;x=3π2; one full period occurs from x=0x=0 to x=2πx=2π

11.

amplitude: 1; period: π;π; midline: y=0;y=0; maximum: y=1y=1 occurs at x=π;x=π; minimum: y=−1y=−1 occurs at x=π2;x=π2; one full period is graphed from x=0x= 0 to x=πx=π

13.

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

15.

amplitude: 3; period: π4;π4; midline: y=5;y=5; maximum: y=8y=8 occurs at x=0.12;x=0.12; minimum: y=2y=2 occurs at x=0.516;x=0.516; horizontal shift: −4;−4; vertical translation 5; one period occurs from x=0x=0 to x=π4x=π4

17.

amplitude: 5; period: 2π5; 2π5; midline: y=−2;y=−2; maximum: y=3y=3 occurs at x=0.08;x=0.08; minimum: y=−7y=−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=0x=0 to x=2π5x=2π5

19.

amplitude: 1 ; period: 2π;2π ; midline: y=1;y=1; maximum: y=2y=2 occurs at x=2.09;x=2.09; maximum: y=2y=2 occurs at t=2.09;t=2.09; minimum: y=0y=0 occurs at t=5.24;t=5.24; phase shift: −π3;− π3; vertical translation: 1; one full period is from t=0t=0 to t=2πt=2π

21.

amplitude: 1; period: 4π;4π; midline: y=0;y=0; maximum: y=1y=1 occurs at t=11.52;t=11.52; minimum: y=−1y=−1 occurs at t=5.24;t=5.24; phase shift: −10π3;−10π3; vertical shift: 0

23.

amplitude: 2; midline: y=−3; y=−3; period: 4; equation: f(x)=2sin(π2x)−3f(x )=2sin(π2x)−3

25.

amplitude: 2; period: 5; midline: y=3;y=3; equation: f(x)=−2cos(2π5x)+3 f(x)=−2cos(2π5x)+3

27.

amplitude: 4; period: 2; midline: y=0;y=0; equation: f(x)=−4cos(π(x−π2))f(x)=−4cos(π(x−π2))

29.

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

37.

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

41.

Maximum: 11 at x=0x=0 ; minimum: -1-1 at x=πx =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+sinxh(x)=x+sinx will increase without bound as well. The graph is bounded between the graphs of y=x+1y=x+1 and y=x-1y=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=0y=0 .

6.2 Section Exercises

1.

Since y=cscxy=cscx is the reciprocal function of y=sinx,y=sinx, you can plot the reciprocal of the coordinates on the graph of y=sinxy=sinx to obtain the y-coordinates of y=cscx.y=cscx. The x-intercepts of the graph y=sinx y=sinx are the vertical asymptotes for the graph of y=cscx.y=cscx.

3.

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

5.

The period is the same: 2π. 2π.

11.

period: 8; horizontal shift: 1 unit to left

17.

−cotxcosx−sinx−cotxcosx−sinx

19.

stretching factor: 2; period: π4;π4 ; asymptotes: x=14(π2+πk)+8, where k is an integerx=1 4(π2+πk)+8, where k is an integer

21.

stretching factor: 6; period: 6; asymptotes: x=3k, where k is an integer x=3k, where k is an integer

23.

stretching factor: 1; period: π;π; asymptotes: x=πk, where k is an integerx=πk, where k is an integer

25.

Stretching factor: 1; period: π;π; asymptotes: x=π4+πk, where k is an integerx=π4+πk, where k is an integer

27.

stretching factor: 2; period: 2π;2π; asymptotes: x=πk, where k is an integerx=πk,  where k is an integer

29.

stretching factor: 4; period: 2π3;2π3; asymptotes: x=π6k, where k is an odd integer x=π6k, where k is an odd integer

31.

stretching factor: 7; period: 2π5; 2π5; asymptotes: x=π10k, where k is an odd integerx=π10 k, where k is an odd integer

33.

stretching factor: 2; period: 2π;2π; asymptotes: x=−π4+πk, where k is an integerx=−π4+πk, where k is an integer

35.

stretching factor: 75;7 5; period: 2π;2π; asymptotes: x=π4+πk, where  k is an integerx=π4+πk, where k is an integer

37.

y=tan(3(x−π4 ))+2y=tan(3(x−π4))+2

39.

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

41.

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

43.

f(x)=2cscxf(x)=2cscx

45.

f(x)=12tan (100πx)f(x)=12tan(100πx)

55.

1. ⓐ(−π2,π 2);(−π2,π2);
2. ⓑ
3. ⓒx=−π2x=−π2 and x=π2;x=π2; the distance grows without bound as |x||x| approaches π2π2 —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=−π3,x=−π3, the boat is 3 km away;
5. ⓔ 1.73; when x=π6, x=π6, the boat is about 1.73 km away;
6. ⓕ 1.5 km; when x=0x=0

57.

1. ⓐh(x)=2 tan(π120x);h(x)=2tan(π120x);
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 xx 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.

6.3 Section Exercises

1.

The function y=sinxy=sin x is one-to-one on [−π2,π2];[−π2,π2]; thus, this interval is the range of the inverse function of y=sinx,y=sinx, f(x)= sin−1x.f(x)=sin−1x. 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−1x.y=cosx,f(x)=cos−1x.

3.

π6π6 is the radian measure of an angle between −π2−π2 and π2π2 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 [−π2,π2] [−π2,π2] so that it is one-to-one and possesses an inverse.

7.

True . The angle, θ1θ1 that equals arccos(−x)arccos(−x), x>0 x>0, will be a second quadrant angle with reference angle, θ2θ2, where θ 2θ2 equals arccosxarccosx, x>0x>0. Since θ2θ2 is the reference angle for θ1θ1, θ2=π−θ1θ2=π−θ1 and arccos(−x)arccos(−x) = π−arccosxπ −arccosx-

37.

x−1−x2+2xx−1−x2+2x

41.

x+0.5− x2−x+34x+0.5−x2−x+34

49.

domain [−1,1]; [−1,1]; range [0,π][0,π]

51.

approximately x=0.00x=0.00

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: 2π3;2π3; midline: y=−1;y=−1; no asymptotes

9.

stretching factor: none; period: π;π; midline: y=−4;y=−4; asymptotes: x=π2+πk,x=π2+πk, where kk is an integer

11.

stretching factor: 3; period: π4; π4; midline: y=−2;y=−2; asymptotes: x=π8+ π4k,x=π8+π4k, where kk is an integer

13.

amplitude: none; period: 2π; 2π; no phase shift; asymptotes: x=π2k,x=π2k, where kk is an odd integer

15.

amplitude: none; period: 2π5;2π5; no phase shift; asymptotes: x=π5k,x=π5k, where kk is an integer

17.

amplitude: none; period: 4π; 4π; no phase shift; asymptotes: x=2πk,x=2πk, where kk is 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.

39.

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

41.

The graphs appear to be identical.

Practice Test

1.

amplitude: 0.5; period: 2π;2 π; midline y=0y=0

3.

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

5.

amplitude: 1; period: 2π;2π; midline: y=1y=1

7.

amplitude: 3; period: 6π;6π; midline: y=0y=0

9.

amplitude: none; period: π;π; midline: y=0,y=0, asymptotes: x=2π3+πk, x=2π3+πk, where kk is an integer

11.

amplitude: none; period: 2π3;2 π3; midline: y=0,y=0, asymptotes: x=π3k, x=π3k, where kk is an integer

13.

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

19.

D(t)=68−12sin (π12x)D(t)=68−12sin(π12x)

21.

period: π6;π6; horizontal shift: −7−7

23.

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

27.

The views are different because the period of the wave is 125. 125. Over a bigger domain, there will be more cycles of the graph.

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+π4))+3f(x)=2cos(12(x+π4))+ 3

35.

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

41.

1−(1−2 x)21−(1−2x)2

49.

approximately 0.07 radians