Chapter 5: Trigonometric Functions of Angles Homework Solutions

Transcription

1 Chapter : Trigonometric Functions of Angles Homework Solutions Section.1 1. D = ( ( 1)) + ( ( )) = + 8 = 100 = 10. D + ( ( )) + ( ( )) = + = 1. (x + ) + (y ) =. (x ) + (y + 7) = r To find the radius, we need to know the distance between the center of the circle and the point on the circle. 10. r = ( 1) + ( 7 1) = = So, our equation is (x ) + (y + 7) = The equation of the circle is (x ) + (y ) =. To find the y-intercepts, we set x = 0 in our equation. (0 ) + (y ) = + (y ) = (y ) = y = y = ± So, the intercepts are (0, + ) and (0, ).

2 1. The equation of the circle is (x ) + (y ) = 1. To find the x-intercepts, we set y = 0 in our equation. (x ) + (0 ) = 1 (x ) + = 1 (x ) = 7 x = ± 7 So, the intercepts are ( + 7, 0) and ( 7, 0). x = ± 7 1. The circle has equation x + (y ) =. We want to know where y = x + intersects with this circle. x + ((x + ) ) = x + (x) = x = x = ± We only want the point in the first quadrant, so x = x = gives a positive y value. y = + > 0 is out. We need to make sure Since ( the point is indeed in the first quadrant, the intersection point of the line and circle is, ) The equation of the circle we need is (x + 1) + y =. We want to know where this circle intersects with the line y = x + in the first quadrant. (x + 1) + (x + ) = x + x x + x + = 0 x + x 1 = 0 x = ± + 8 x = ± 8 The only possible x value would be x = + 8 as the other root would take us out of the first quadrant. We now need to make sure that this point lies in the first quadrant, but since this x > 0 and we are only adding ( to obtain y, we will be in the first quadrant. Therefore, ) the point of intersection we seek is + 8,

3 18. Assume the transmitter is at the origin City City 1 To solve this problem, we need to find the length of the trip (distance between cities) and the length of the trip in the radius of the transmitter. First, the distance between cities is D = ( 0) + (0 ( ) = The equation of the circle of transmission is x + y = x + y = miles Next, we need the equation of the line containing these two points. m = 0 0 ( ) = y = x + b Note: the y-intercept is the location of city 1. y = x To find the intersection, we substitute in this equation for y. x + ( x ) = 1 x x + 7 x + 1 = 1 80 x + 7 x = 0 x 1.1,.1

4 We need the ordered pairs associated with these points so that we can find the distance. It is easier to use the linear equation. : x = 1.1 gives y 1. : x =.1 gives y 10.8 Now, the distance between these two points is D = ( 1.1 (.1)) + ( 1. ( 10.8)). miles So, the part of the trip in range of the transmitter is Therefore, approximately.% of the trip will be in range. Section.. π π π rad 180 = π rad. π rad 180 π rad = = = (0 ) = (0 ) = π 18π = 8π

5 1. 17π ( ) π = π 1. π + π = π 1. 7π + 1π = π Section.. We use the Pythagorean Identity to solve this. Since the y coordinate is the same as sin θ, then sin θ =. Then, ( ) + cos θ = 1 + cos θ = 1 cos θ = 1 cos θ = ± Since we want the point in quadrant II, we need x =.. We use the Pythagorean Identity to solve this. Since the x coordinate is the same as cos θ, then cos θ = 1. Then, ( ) 1 sin θ + = 1 sin θ + 1 = 1 sin θ = sin θ = ± Since we want the point in quadrant IV, we need x =.. Using the Pythagorean Identity, we have sin θ + cos θ = 1 ( ) 1 sin θ + θ = 1 7 sin θ + 1 = 1 sin θ = 8 sin θ = ± 7 Since we want the angle in quadrant IV, we need sin θ = 7.

6 . Using the Pythagorean Identity, we have sin θ + cos θ = 1 ( ) sin θ + θ = 1 sin θ + 81 = 1 sin θ = sin θ = ± Since we want the angle in quadrant I, we need sin θ = 7. Using the Pythagorean Identity, we have 77. sin θ + cos θ = 1 ( ) θ + cos θ = cos θ = 1 cos θ = cos θ = ± 8 Since we want the angle in quadrant II, we need cos θ = Using the Pythagorean Identity, we have sin θ + cos θ = 1 ( 1 θ + cos ) θ = cos θ = 1 cos θ = cos θ = ±. Since we want the angle in quadrant III, we need cos θ = 1. (a) lies in quadrant III and the appropriate reference angle is. (b) 00 lies in quadrant IV and the appropriate reference angle is 0.

7 (c) 1 lies in quadrant II and the appropriate reference angle is. (d) 10 lies in quadrant III and the appropriate reference angle is π lies in quadrant III and the appropriate reference angle is π. π lies in quadrant II and the appropriate reference angle is π. π lies in quadrant III and the appropriate reference angle is π. 7π lies in quadrant IV and the appropriate reference angle is π (a) π lies in the quadrant III. sin ( π ) = cos ( π ) = (b) π is coterminal with 11π and so it has the same trig values. The angle lies in the quadrant IV. sin ( ) π = 1 (c) π cos ( π ) = lies on the negative y-axis. sin ( π ) = 1 cos ( π ) = 0 (d) π lies on the negative x-axis and has the same sine and cosine as π. sin (π) = 0 cos (π) = 1 (a) π lies in the quadrant III. The best choice for reference angle is π. sin ( π ) = cos ( π ) = 1 (b) 17π is coterminal with π and so it has the same trig values. The angle lies in the quadrant I. sin ( ) 17π = ) = cos ( 17π (c) π lies in quadrant IV. The best choice for reference angle is π. sin ( π ) = 1 ) = cos ( π (d) 10π lies on the positive x-axis and has the same sine and cosine as 0.

8 sin (π) = 0 cos (10π) = 1 1. For each, we are looking for an angle with the same sine value. The two quadrants with a positive sine are I and II. For quadrants III and IV, the sine is negative. So, to find another angle, in each case we need to reflect across the y-axis. (a) π gives the same sine as π. (b) 80 gives the same sine as 100. (c) 10 gives the same sine as 0. (d) π gives the same sine as π. (e) 0 gives the same sine as. 1. For each, we are looking for an angle with the same sine value. The two quadrants with a positive sine are I and II. For quadrants III and IV, the sine is negative. So, to find another angle, in each case we need to reflect across the y-axis. (a) π gives the same sine as π. (b) 1 gives the same sine as 1. (c) 10 gives the same sine as 0. (d) 7π 11π gives the same sine as. (e) 0 gives the same sine as For each, we are looking for an angle with the same cosine value. The two quadrants with a positive cosine are I and IV. For quadrants II and III, the cosine is negative. So, to find another angle, in each case we need to reflect across the x-axis. (a) π gives the same cosine as π. (b) 80 gives the same cosine as 80. (c) 10 gives the same cosine as 0. (d) π gives the same cosine as π. (e) 0 gives the same cosine as. 18. For each, we are looking for an angle with the same cosine value. The two quadrants with a positive cosine are I and IV. For quadrants II and III, the cosine is negative. So, to find another angle, in each case we need to reflect across the x-axis (a) π gives the same cosine as 7π. (b) 1 gives the same cosine as. (c) 10 gives the same cosine as 00. (d) 7π gives the same cosine as π. (e) 0 gives the same cosine as This was the only calculator one assigned to see what you would do with it. If we use the calculator, we find

9 cos(0 ).7 sin(0 ). It is not a unit circle, but that just means we have to factor the radius into the calculation. x = rcos(0 ) = 1(.7) = 11. y = rsin(0 ) = 1(.) =. Therefore, the coordinate we seek is ( 11.,.). Section. 1. θ = π. θ = π sec(θ) = 1 cos(θ) = = csc(θ) = 1 sin(θ) = = tan(θ) = sin(θ) cos(θ) = 1 cot(θ) = cos(θ) sin(θ) = sec(θ) = 1 cos(θ) = csc(θ) = 1 sin(θ) = tan(θ) = sin(θ) cos(θ) = 1 cot(θ) = cos(θ) sin(θ) = (a) sec(1 1 ) = cos(1 ) = 1 cos( ) = 1 (b) csc(10 1 ) = sin(10 ) = 1 sin(0 ) = 1 1 cos(0 ) = 1 (c) tan(0 ) = sin(0 ) = sin( ) = (d) cot( ) = cos( ) sin( ) = cos( ) = = = = 1 ( sin ) θ + cos θ = 1 + cos θ = cos θ = 1 cos θ =

10 cos(θ) = 7 sec(θ) = 7 csc(θ) = 7 tan(θ) = cot(θ) = 1. Since tan(θ) = = 1, we have sin(θ) = 17 cos(θ) = 1 17 sec(θ) = 17 csc(θ) = cot(θ) = 1 17 hyp = + 1 = 17 hyp = 17 Section = hyp 1 = hyp hyp = 1 sin(a) = 10 1 cos(a) = 8 1 tan(a) = sec(a) = 1 8 csc(a) = 1 10 cot(a) = = hyp hyp = 11 hyp = sin(a) = 10 = cos(a) = = tan(a) = 10 =. sec(a) = csc(a) = cot(a) = 10 = B = 0 by the Triangle Sum theorem

11 tan(0 ) = 7 b b = 7 tan(0 ) = 7 1 = 7 c = 7 + (7 ) = + 17 = 1 c = 1. A = 0 by the Triangle Sum theorem sin(0 ) = 10 c c = 10 sin(0 ) = 10 = 0 ( ) c = c = = c = 700. Time for the calculator.... A = 8 by the Triangle Sum theorem sin( ) = 10 c c = 10 sin( a = 11.7 a = B = by the Triangle Sum theorem sin( ) = 7 c c = 7 sin( ) 1.0 tan( ) = 7 b b = 7 tan( ).7 B = by the Triangle Sum theorem sin( ) = a 10 a = 10sin( ).0 cos( ) = b 10 b = 10cos( ). B = 80 by the Triangle Sum theorem sin(10 ) = a 1 a = 1sin(10 ).08 cos(10 ) = b 1 b = 1cos(10 ) Radio tower x 1 x Window 00 ft

12 tan( ) = x 1 00 x 1 = 00tan( ) = 0. feet tan( ) = x 00 x = 00tan( ) = 1.7 feet x 1 + x = = 0.1 feet ft x 1 x x ft 1 window Let the distance from the window to the monument is x feet. We don t have enough information to find the answer directly, so we will use two trig functions to find what we need. First, notice that tan(1 ) = x 1 x x 1 = xtan(1 ) tan( ) = x x x = xtan( ) Now, we know x 1 + x = 00 since those tw distances equal that of the monument. So, we have 00 = x 1 + x = xtan(1 ) + xtan( ) 00 = x (tan(1 ) + tan( )) 00 x = (tan(1 ) + tan( )) x 0. ft ft x 1 x x ft 18 window

13 Let the distance from the window to the monument is x feet. We don t have enough information to find the answer directly, so we will use two trig functions to find what we need. First, notice that tan(18 ) = x 1 x x 1 = xtan(18 ) tan( ) = x x x = xtan( ) Now, we know x 1 + x = 00 since those tw distances equal that of the monument. So, we have 00 = x 1 + x = xtan(18 ) + xtan( ) 00 = x (tan(18 ) + tan( )) 00 x = (tan(18 ) + tan( )) x 17. ft 17. Antenna h Top of Building h ft The way to solve this is to find the height of the building with and without the antenna. We have the base (adjacent side) associated with both angles, and the difference between heights will be that of the antenna. tan(0 ) = h 1 00 h 1 = 00tan(0 ) tan( ) = h 1 + h 00 h 1 + h = 00tan( ) h 1 + h h 1 = 00(tan( ) tan(0 )) h 8.0 ft 1.

14 8 a b x Based on what we are given, we have to do this similar earlier where we find separate pieces and add them to get what we need. x = a + b = tan( ) = 8 a 8 a = tan( ) tan( ) = 8 a 8 b = tan( ) 8 tan( ) + 8 tan( ) y x The total distance x + y can be found by using the tangent of the angle measuring. We can find y by using the tangent of 70. The value we want, then is the difference between these values. tan( ) = 11 x + y x + y = 11 tan( ) x + y. tan(70 ) = 11 x + y y = 11 tan(70 ) y.1 x = x + y y..1 = 00.8