Chapter 3. Radian Measure and Circular Functions. Section 3.1: Radian Measure. π 1.57, 1 is the only integer value in the

Transcription

1 Chapter Radian Measure and Circular Functions Section.: Radian Measure. Since θ is in quadrant I, 0 < θ <. Since.57, is the only integer value in the interval. Thus, the radian measure of θ is radian.. Sinceθ is in quadrant II, < θ <. Since.57 and., and are the only integers in the interval. Since θ is closer to, the radian measure of θ is radians. 5. Since θ is an angle in quadrant III drawn in a clockwise direction, < θ <. Also. and.57, so and are the only integers in the interval. θ is closer to, so the radian measure of θ is radians = 0 radian = radians = 90 radian = radians = 50 radian = radians 5 00 = 00 radian = radians 5 50 = 50 radian = radians = 800 radian = 0 radians 9. Multiply the degree measure by radian 80 and reduce. Your answer will be in radians. Leave the answer as a multiple of, unless otherwise directed... Answers will vary = = = = 5 = = 0 = = = = 0 0 = = = = = 5 = = 9 radian.8 radian..5 =.5 radian.7 radians ( 9 ) = radian. radians 7..9 =.9 radian. radians 5

2 5 Chapter : Radian Measure and Circular Functions ( 5 ) = radian.987 radians = 7.95 radian.89 radian sin = sin = sin50 = sin 0 = cos = cos = cos 50 = cos 50 0 = cos= ( ) radians = = + ( ) 5.7 radians = = 99 + (.955 0) 99.7 radian = = 9 + ( ) = radian = = 87 (.08 0) Without the degree symbol on the 0, it is assumed that 0 is measured in radians. Thus, the approximate value of sin 0 is.9880, not sin = sin = sin 0 = tan = tan = tan 5 = sec = sec = sec0 = sin = sin = sin 90 = 5 5 tan = tan = tan 00 = tan 0 = sin = sin = sin ( 80 ) = sin ( ) = sin 0 = sin 0 = sin = sin = sin ( 0 ) = sin ( ) = sin50 = sin 0 = 8. tan = tan = tan ( 80 ) = tan ( ) = tan 0 = tan 0 = 8. Begin the calculation with the blank next to 0º, and then proceed counterclockwise from there. 0 = 0 radian = radians radians = = 5 0 = 0 radian = radians radians = = 0 radians = = = 50 radian = radians = 80 radian = radians 7 0 = 0 radian = radians

3 Section.: Applications of Radian Measure = 5 radian = radians radians = = radians = = = 5 radian = radians 0 = 0 radian = radians 85. (a) In hours, the hour hand will rotate twice around the clock. One complete rotation is radians, the two rotations will measure = radians. (b) In hours, the hour heand will rotate = of the way around the clock, which is = radians. 87. In each rotation around Earth, the space vehicle would rotate radians. (a) In.5 orbits, the space vehicle travels.5 = 5 radians. (b) In of an orbit, the space vehicle travels 8 = radians. Section.: Applications of Radian Measure Connections (page ) Answers will vary. The longitude at Greenwich is 0º. Exercises. r =, θ = = =. 5 r =, θ = 5 = = 0 5. s =, θ = s r = = = = θ 7. r =, s = s θ = = = r 9. s = 0, r = 0 s 0 θ = = = r 0. r =. cm, θ = radians =. = cm 5. r =.8 ft, θ = radians 5 =.8 =.5 ft. ft (rounded to three significant digits) 5. r =.8 m, θ = 0 Converting θ to radians, we have θ = 0 = 0 radian = radians. Thus, the arc is.8 =.8 = 5.05 m. (rounded to three significant digits) 7. r = 5. in., θ = 0 Converting θ to radians, we have 7 θ = 0 = 0 radian = radians. Thus, the arc is = 5. = 55. in. (rounded to three significant digits) 9. The formula for arc length is. Substituting r for r we obtain s = ( r) θ = ( rθ). The length of the arc is doubled.

4 5 Chapter : Radian Measure and Circular Functions For Exercises, note that since 00 has two significant digits and the angles are given to the nearest degree, we can have only two significant digits in the answers.. 9 N, 0 N θ = 0 9 = = radian = radian 80 = km. N, S S = N θ = ( ) = 5 = 5 radian 5 = radian 80 5 = km 5. r = 00 km, s = 00 km = 00θ θ = = 00 Converting radian to degrees, we have θ =. The north-south distance between the two cities is. Let x = the latitude of Madison. x = x = N The latitude of Madison is N. 7. The arc length on the smaller gear is 5 =.7 00 = = cm An arc with this length on the larger gear corresponds to an angle measure θ where = 7.θ = θ θ = 5. The larger gear will rotate through approximately 5º. 9. A rotation of θ = 0.0 radian = radians on the smaller wheel moves through an arc length of 5. = 5. = cm. (holding on to more digits for the intermediate steps) Since both wheels move together, the larger wheel moves cm, which rotates it through an angle θ, where 5. = 8.θ 5. θ = radian.8 5. = The larger wheel rotates through The arc length s represents the distance traveled by a point on the rim of a wheel. Since the two wheels rotate together, s will be the same for both wheels. For the smaller wheel, θ = 80 = 80 = radians and 9 =.7 = 5. cm. 9 For the larger wheel, 5 θ = 50 = 50 radian = radian. 8 Thus, we can solve 5 5. = r 8 8 r = 5. = The radius of the larger wheel is 8.7 cm. (rounded to significant digits). (a) The number of inches lifted is the arc length in a circle with r = 9.7 in. and θ = = ( 7+ 0 ) 50 s = 9.7 ( 7+ 0 ). The weight will rise. in. (rounded to three significant digits)

5 Section.: Applications of Radian Measure 57 (b) When the weight is raised in., we have = 9.7θ θ = radian = = ( ) The pulley must be rotated through A rotation of θ = 80 radian = radians. The chain 80 moves a distance equal to half the arc length of the larger gear. So, for the large gear and pedal,.7. Thus, the chain moves.7 in. The small gear rotates through an angle as follows. s.7 θ = θ =. r.8 θ for the wheel and θ for the small gear are the same, or.. So, for the wheel, we have r =.(.). The bicycle will move in. (rounded to three significant digits) 7. Let t = the length of the train. t is approximately the arc length subtended by 0. First convert θ = 0 to radians. 0 θ = 0 = ( + 0 ) = 0 = ( ) radian = radian = radian 5 The length of the train is t = rθ t =.5.0 km long. 5 (rounded to two significant digits) 9. r =, s = = θ θ = = A= r θ A = ( ) = ( ) =. r =, s = = θ θ = = A= r θ A = ( ) = 7. A = sq units, r = A= r θ = ( ) θ = ( ) θ = 8θ θ = = radian 8 80 radian = = 0 The measure of the central angle is 0º. 5. A = sq units, r = A= r θ = ( ) θ = ( ) θ = θ θ = =.5 radians In Exercises 7 5, we will be rounding to the nearest tenth r = 9. m, θ = radians A= r θ 5 5 A = ( 9.) = ( 85.).0 The area of the sector is. m. ( 0 m rounded to three significant digits) 9. r = 0.0 ft, θ = radians A= r θ A = ( 0.0) = ( 900) = The area of the sector is 70.9 ft. ( 707 ft rounded to three significant digits) 5. r =.7 cm, θ = 8 The formula A = r θ requires that θ be measured in radians. Converting 8 to radians, we have 9 θ = 8 radian = radians. Since A = (.7) = (.9).009, 0 0 the area of the sector is.0 cm. ( cm rounded to three significant digits)

6 58 Chapter : Radian Measure and Circular Functions 5. r = 0.0 mi, θ = The formula A = r θ requires that θ be measured in radians. Converting 5 to radians, we have θ = 5 radian = radians. ( ) A = 0.0 = ( 00) = The area of the sector is mi. ( 880 mi rounded to three significant digits) A= in., r =.0 in. () 9 A= r θ = θ = θ θ = =. radians 9 9 (rounded to two significant digits) 57. (a) The central angle in degrees measures 0 =. Converting to radians, we 7 have the following. = ( ) radian 0 = radian = radian 7 (b) Since C = r, and r = 7 ft, we have ( ) C = 7 = The circumference is 78 ft. (c) Since r = 7 ft and θ =, we have 7 5 = 7 = Thus, the length of the arc is 7.7 ft. Note: If this measurement is approximated to be 0, then the 9 approximated value would be 7.8 ft, rounded to three significant digits. (d) The area of a sector with r = 7 ft and θ = is A= r θ A = ( 7 ) = ( 577) = ft 59. (a) (b) (c) The triangle formed by the central angle and the chord is isosceles. Therefore, the bisector of the central angle is also the perpendicular bisector of the chord sin = r = 0 ft r sin 50 r = ; θ = sin Converting θ to radians, we have 7 radians = radians. Solving 0 for the arc length, we have s = = 0 ft sin 0 sin The area of the portion of the circle can be found by subtracting the area of the triangle from the area of the sector. From the figure in part (a), we have tan = so h =. h tan Asector = r θ 50 7 Asector = 75 ft sin 0 and Atriangle = bh 50 Atriangle = ( 00) 5 ft tan The area bounded by the arc and the chord is 75 5 = ft.. Use the Pythagorean theorem to find the hypotenuse of the triangle, which is also the radius of the sector of the circle. r = r = r = 500 r = 50 The total area of the lot is the sum of the areas of the triangle and the sector.

7 Section.: The Unit Circle and Circular Functions 59 Converting θ = 0 to radians, we have 0 radian = radians. Atriangle = bh = ( 0 )( 0 ) = 00 yd Asector = r θ = ( 50 ) 50 = ( 500 ) = yd Total area 50 Atriangle + Asector = or 900 yd, rounded to two significant digits.. Converting ( ) θ = 7 = 7+ = 7. to 0 radians, we have radian = = radian Solving for the radius with the arc length formula, we have 9 = r 5 5, 00 r = 9 = 97.0 Thus, the radius is approximately 950 mi. (rounded to three significant digits) Using this approximate radius, we can find the circumference of the Earth. Since C = r, we have C ( 950),800. Thus, the approximate circumference is,800 mi. (rounded to three significant digits) 5. The base area is Asector = r θ. Thus, the r θ h volume is V = r θ h or V =, where θ is in radians. 7. L is the arc length, so L = rθ. Thus, θ 9. d = r h d = r rcos θ d = r cos L r =. θ 7. If we let r =, r then Asector = ( r ) θ = ( r) θ Thus, the area, = ( r ) θ = r θ r θ, is quadrupled. Section.: The Unit Circle and Circular Functions Connections (page ) Answers will vary. Exercises θ = radians intersects the unit circle at the point ( 0, ).. An angle of (a) sinθ = y = (b) cosθ = x = 0 y (c) tan θ = = ;undefined x 0. An angle of θ = radians intersects the unit circle at the point (, 0 ). (a) sinθ = y = 0 (b) cosθ = x = y 0 (c) tanθ = = = 0 x 5. An angle of θ = radians intersects the unit circle at the point (, 0 ). (a) sinθ = y = 0 (b) cosθ = x = y 0 (c) tanθ = = = 0 x

8 0 Chapter : Radian Measure and Circular Functions For Exercises 7, use the following copy of Figure on page 9 of the text. 7. Since 7 is in quadrant III, the reference angle is 7 7 = =. In quadrant III, the sine is negative. Thus, 7 sin = sin =. Converting 7 to 7 7 degrees, we have = ( ) = 0. The reference angle is 0 = 0. Thus, 7 sin = sin 0 = sin 0 =.. Converting to degrees, we have = ( ) = 0. The reference angle is 0 0 = 0. Thus, csc = csc0 = csc0 =. is coterminal with + = + =. Since is in quadrant II, the reference angle is = =. In quadrant II, the cosine is negative. Thus, cos = cos = cos =. Converting to degrees, we have = ( ) = 0. The reference angle is 0 = 0. Thus, cos = cos = cos0 = cos 0 = 9. Since is in quadrant II, the reference angle is = =. In quadrant II, the tangent is negative. Thus, tan = tan =. Converting to degrees, we have = ( ) = 5. The reference angle is 5 = 5. Thus, tan = tan5 = tan 5 =. 5. Since 7 is in quadrant IV, the reference angle is = =. In quadrant IV, the cosine is positive. Thus, 7 cos = cos =. Converting 7 to 7 7 degrees, we have = ( ) = 5. The reference angle is 0 5 = 5. Thus, 7 cos = cos5 = cos 5 =.. Since is in quadrant IV, the reference angle is = =. In quadrant IV, the cosecant is negative. Thus, csc = csc =.

9 Section.: The Unit Circle and Circular Functions is coterminal with + = + =. Since is in quadrant II, the reference angle is = =. In quadrant II, the sine is positive. Thus, sin = sin = sin =. Converting to degrees, we have = ( ) = 0. The reference angle is 0 = 0. Thus, sin = sin = sin0 = sin 0 = is coterminal with = =. Since is in quadrant IV, the reference angle is = =. In quadrant IV, the secant is positive. Thus, sec = sec = sec =. Converting to degrees, we have = ( ) = 0. The reference angle is 0 0 = 0. Thus, sec = sec = sec0 = sec0 =. For Exercises, 9 5, and 70, your calculator must be set in radian mode. Keystroke sequences may vary based on the type and/or model of calculator being used. As in Example, we will set the calculator to show four decimal digits.. sin cos (.59 ) tan csc( 9.9).8. Since 5 is in quadrant II, the reference angle 5 5 is = =. In quadrant II, the tangent is negative. Thus, 5 tan = tan =. Converting 5 to 5 5 degrees, we have = ( ) = 50. The reference angle is 50 = 0. 5 Thus, tan = tan50 = tan 0 =.. sec.80.00

10 Chapter : Radian Measure and Circular Functions. cot cos sin.9 9. sin.8.. cos θ =.5 x =.5 θ. radians or θ.0 radians. sin θ =.7 y =.7 θ.8 radians or θ. radians 5. cos.57 and., so < <. Thus, an angle of radians is in quadrant II. (The figure for Exercises 5 8 also shows that radians is in quadrant II.) Because values of the cosine function are negative in quadrant II, cos is negative. 7. sin 5.7 and.8, so < 5<. Thus, an angle of 5 radians is in quadrant IV. (The figure for Exercises 5 8 also shows that 5 radians is in quadrant IV.) Because values of the sine function are negative in quadrant IV, sin 5 is negative. 9. tan.9.8 and 5 + = + = 7.85, so 5 <.9 <. Notice that is coterminal with 0 and 5 is coterminal with. Thus, an angle of.9 radians is in quadrant I. Because values of the tangent function are positive in quadrant I, tan.9 is positive. 5. secθ = = = = = x cscθ = = = = = y 5 sin θ = y = ; cosθ = x = y tanθ = = = = x x 5 5 cotθ = = = = y sec θ = = = ; cscθ = = = x 5 5 y 55. tan s =. s sin s =.998 s. 59. sec s =.080 s.887., ; sins = Recall that sin = and in quadrant II, sin s is positive. Therefore, 5 5 sin = sin =, so s =. 5. sin θ = y = ; cosθ = x = x θ y y tanθ = = = ; cot = = = x

11 Section.: The Unit Circle and Circular Functions. 5., ; tans = Recall that tan = and in quadrant III, tan s is positive. Therefore, tan + = tan =, so s =., ;tan s = Recall that tan = and in quadrant IV, tan s is negative. Therefore, 7 7 tan = tan =, so s =. 7. s = 5 Since cosine is negative and sine is positive, an angle of 5 radians lies in quadrant II. 75. ( xy, ) (.5599,.805) = cos s =.5599 and sin s =.805 s s = the length of an arc on the unit circle =.5 x = cos s x = cos.5 y = sin s y = sin.5 ( xy=, ) (.80,.5985) 9. s = 7. x = cos s x = cos 7. ( ) ( ) y = sin s y = sin 7. ( xy=, ) (.85,.8987) 7. s = 5 Since cosine and sine are both positive, an angle of 5 radians lies in quadrant I. 77. (a) New Orleans has latitude L = 0º or.5 radians. Since the day and time have not changed, D =.5 and ω =.785 sinθ = cos Dcos Lcosω + sin Dsin L sinθ = cos (.5) cos (.5) cos (.785) + sin.5 sin.5 ( ) ( ) ( ) ( ) ( ) sin (.5) sin (.5) θ = sin cos.5 cos.5 cos radians or. (b) Answers will vary. x 79. t = 0 0cos (a) January: x = 0; 0 t = 0 0cos = 0 0cos 0 = 0 0() = 0 0 = 0 (b) April: x = ; t = 0 0cos = 0 0cos = 0 0(0) = 0 (c) May: x = ; t = 0 0cos = 0 0cos = 0 0 = 75

12 Chapter : Radian Measure and Circular Functions (d) June x = 5; 5 5 t = 0 0cos = 0 0cos = 0 0 = (e) August x = 7; 7 7 t = 0 0cos = 0 0cos = 0 0 = (f) October x = 9; 9 t = 0 0cos = 0 0cos = = 0 ( ) (d) June 5 is day ( = ) T ( ) = 7 sin ( 0) = 7 sin F 5 (e) September is day ( = ) T ( ) = 7 sin ( 0) = 7sin F 5 (f) October is day 0 ( = 0) T ( 0) = 7 sin ( 0 0) = 7sin + 5 F 5 Chapter : Quiz (Sections..) = 5 radian = radians 5 5 = = r = 00, s = = 00θ θ = = Since 7 is in quadrant IV, the reference angle is = =. In quadrant IV, the cosine is positive. Thus, 7 cos = cos =. Converting 7 to degrees, we have 7 7 = ( ) = 5. The reference angle is 5 = 5. Thus, 7 cos = cos5 = cos 5 =. 9. is coterminal with =. So tan = tan = 0 Converting to degrees, we have 80 = 50. The reference angle is 50º 0º = 80º. Thus tan = tan 50 = tan= 0. Section.: Linear and Angular Speed. The circumference of the unit circle is. ω = radian per sec, θ = radians θ ω = = t = sec t t. r = 0 cm, 5. (a) ω = radian per sec, t = sec ω = θ = θ θ = radians t (b) s = 0 = 0 cm rθ (c) v = v = = = cm per t sec ω = radians per sec, t = sec θ θ ω = = θ = radians t

13 Section.: Linear and Angular Speed 5 7. θ = radians, t = 8 sec θ ω = θ = = = radian per t 8 8 sec 5 9. θ = radian, ω = radian per min 9 7 θ ω = = = t 7 t 7 9t 5 5t = 5 t = = min 5 5. θ =.87 radians, t =.9sec θ ω = t.87 ω =.80 radian per sec.9. r = m, ω = radians per sec v = rω v= = 8 m per sec 5. v = 9 m per sec, r = 5 m 9 v = rω 9 = 5 ω ω = radians per sec 5 7. v = 07.9 m per sec, r = 58.7 m v = rω 07.9 = 58.7ω 07.9 ω =.8 radians per sec r = cm, ω = radians per sec, t = 9 sec s = rωt s = ( 9) = 8 cm. s = cm, r = cm, ω = radians per sec s = rωt = t = t t = = sec. s = km, r = km, t = sec s = rω t = ω = 8ω ω = = radian per sec 8 5. The hour hand of a clock moves through an angle of radians (one complete revolution) in hours, so θ ω = = = radian per hr. t 7. The minute hand makes one revolution per hour. Each revolution is radians, so we have ω = ( ) = radians per hr. There are 0 minutes in hour, so ω = = 0 0 radians per min. 9. The minute hand of a clock moves through an angle of radians in 0 min, and at the tip of the minute hand, r = 7 cm, so we have rθ 7 ( ) 7 v = v = = cm per min t 0 0. The flywheel making rotations per min turns through an angle ( ) = 8 radians in minute with r = m. So, rθ 8 ( ) v = v = = 8 m per min t. At 500 rotations per min, the propeller turns through an angle of θ = 500() = 000 radians in min with r = =.5 m, we have rθ.5( 000 ) v = v = = 500 m per min. t 5. At 5 revolutions per minute, the bicycle tire is moving 5 () = 0 radians per min. This is the angular velocity ω. The linear velocity of the bicycle is v = rω = ( 0) = 5590 in. per min. Convert this to miles per hour: 5590 in. 0 min ft mi v = min hr in. 580 ft. mph 7. (a) θ ( ) (b) = = radian 5 5 ω = radian per day 5 = radian per hr 5 = radian per hr 80

14 Chapter : Radian Measure and Circular Functions (c) v = rω v = ( 9,000,000) 7,000 mph (a) Since s = 5 cm of belt go around in t = 8 sec, the linear velocity is s 5 8 v = v = =. cm per sec t 8 9 (b) Since the 5 cm belt goes around in 8 sec, we have 5 v = rω = (.9) ω 8 8 = (.9) ω ω =. radian per sec.9. ω = ( 5)( ) = 0 radians per min 0 = radians per sec 0 7 = radians per sec 5 7 v = rω 59. = r 5 5 r = cm 7 Expressing the velocity v = 0 mph in ft per sec, we have 0 v = 0 mph = mi per sec 00 ( 0) 580 = ft per sec = ft per sec 00 s 00 v = = t = 00 t t t = = 9 sec. In one minute, the propeller makes 5000 revolutions. Each revolution is radians, so we have 5000() = 0,000 radians per min. There are 0 sec in a minute, so 0, ω = = 5. radians per sec 0 Chapter : Review Exercises. An angle of º is of the way around a 0 circle. radian is the measurement of an angle with its vertex at the center of a circle that intercepts an arc on the circle equal in length to the radius of the circle. Thus, radian is of the way around the circle. Since <, the angular measurement of 0 radian is larger than º.. To find a coterminal angle, add or subtract multiples of. Three of the many possible answers are +,+, and = 5 radian = radians = 75 radian = radians = 800 radian = radians = = = = 80 = = Since 5 = rotation, we have 0 θ = ( ) =. Thus, s = = in. θ = =, we have 9. Since ( ) s r s ( ). = θ = = in. r = 5. cm, θ = s = 5. =. 5.8 cm

15 Chapter : Review Exercises 7. r = 8.97 cm, θ = 9.0 First convert θ = 9.0 to radians: θ = 9.0 = 9.0 = s = cm 5. r = 8.0 m, θ = 0 First convert θ = 0 to radians: 0 θ = 0 = ( + 0) 5 = = 08 A= r θ A = ( 8.0) 7 m The cities are at 8ºN and ºS. ºS = ºN, so θ = 8 ( ) = 0 = 0 = 9 radians = km (rounded to 9 two significant digits) 9. r =, s = = θ θ = = radian A= r θ A = ( ) = ( ) = =.5 sq units. (a) The hour hand of a clock moves through an angle of radians in hours, so θ ω = = = radian per hour t In two hours, the angle would be = radians. (b) The distance s the tip of the hour hand travels during the time period from o clock to o clock is the arc length for θ = and r = in. Thus, = = in.. tan Converting to degrees, we have = ( ) = 0 tan = tan 0 = 5 sin is coterminal with + =. Since 7 is in quadrant III, the reference angle is 7 =. In quadrant III, the sine is negative. Thus, 5 7 sin = sin = sin = Converting 5 to degrees, we have 7 = 0. The reference angle is 0º 80º = 0º. Thus, 5 7 sin = sin = sin 0 = sin 0 = csc is coterminal with + = + =. Since is in quadrant I, we have csc = csc =. Converting to degrees, we have = ( ) = 0. Thus, csc = csc = csc 0 =.

16 8 Chapter : Radian Measure and Circular Functions 9. Since 0< <, sin and cos are both positive. Thus, tan is positive. Also, since < <, sin is positive and cos is negative. Thus, tan is negative. Therefore, tan > tan.. Snce < <, sin is positive and cos is negative. Thus, sin > cos.. sin cos(.) sec cos s =.950 s sin s =.9 s cot s =.50 s ,, cos s = Because cos s =, the reference angle for s must be since cos =. For s to be in the interval 0,, s must be the reference angle. Therefore, s =.,, secs = Because sec s =, the reference angle for s must be since sec =. For s to be in the interval,, we must add the reference angle to. Therefore, 7 s = + = θ =, ω = radians per sec 9 5 θ ω = = = t 9 t 9 t 5 5 9t = 5 t = = sec 9. t = 8 sec, θ = radians 5 θ 5 ω = ω = = = radians per t sec. r =. cm, ω =.8 radians per sec, t = 5.8 sec = rωt = (.)(.8)( 5.8) 85. cm

17 Chapter Test Since t = 0 sec and θ =, radians we 5 θ 5 have ω = ω = = = radian t 0 0 per sec. 7. (a) Because alternate interior angles of parallel lines with transversal have the same measure, the measure of angle ABC is equal to the measure of angle BCD (the angle of elevation). Moreover, triangle BAC is a right triangle and we can write h the relation sin θ =. d Chapter Test. 0 = 0 radian = radians. 5 = 5 radian = radian. 5 = 5 radian =.09 radian. = = = = 0. = r = 50 cm, s = 00 cm Solving for d we have, h sinθ = dsinθ = h d d = h d = hcscθ sinθ (b) Since d = hcsc θ, we substitute h for d and solve. h = hcscθ = cscθ sinθ = θ = d is double h when the sun is radians (0 ) above the horizon. (c) csc = and csc =.5 When the sun is lower in the sky θ =, sunlight is filtered by more atmosphere. There is less ultraviolet light reaching the earth s surface, and therefore, there is less likelihood of becoming sunburned. In this case, sunlight passes through 5% more atmosphere. 8. (a) (b) = 50θ θ = = 50 A= r θ A = ( 50) = (, 500) = 5,000 cm r = in., s = in. = θ θ = radians For Exercises 9, refer to Figure on page 9 of the text.

18 70 Chapter : Radian Measure and Circular Functions sin Since is in quadrant II, the reference angle is = =. In quadrant II, the sine is positive. Thus, sin = sin =. Converting to degrees, we have = ( ) = 5. The reference angle is 5 = 5. Thus, sin = sin5 = sin 5 =. 7 cos 7 is coterminal with = + =. Since 5 is in quadrant II, the reference angle is 5 5 = =. In quadrant II, the cosine is negative. Thus, 7 5 cos = cos = cos =. Converting 5 to degrees, we have 5 5 = ( ) = 50. The reference angle is 50 = Thus, cos = cos = cos50 = cos 0 = tan = tan 70 is undefined. 8 sec 8 8 is coterminal with =. Since is in quadrant II, the reference angles is =. In quadrant II, the secant is negative. 8 Thus, sec = sec = sec =. Converting to degrees, we have = = 0. The reference angle is 80º 0º = 0º. Thus, 8 sec = sec = sec0 = sec 0 =.. tan = tan= 0. cos = cos 70 = 0 5. s = 7 Since 7 is in quadrant III, the reference angle is 7 =. In quadrant III, the sine and cosine are negative. 7 sin = sin = 7 cos = cos = 7 tan = sin =. For any point (x, y) on the unit circle, y x sin s = = y and cos s =, both of which are defined for all values of x and y. Thus, the domains of the sine and cosine functions are y both (, ). tan s = and sec x =, so x x these functions are not defined for x = 0. This occurs at 5 5 s =,,,,,,, Therefore, the domains of the tangent and secant functions are both s s ( n+ ), where n is any integer. x cot s = and csc s = are not defined for y y y = 0. This occurs at s =,,,,0,,,, Therefore, the domains of the cotangent and cosecant functions are both s s n, where n is any integer { }

19 Chapter Test 7 7. (a) sin s =.858 s.97 (b) Since cos = and 0, s =. 8. (a) The speed of ray OP is ω = radian θ per sec. Since ω =, then in 8 sec, t 8 ω = θ = θ θ = = t 8 radians (b) From part (a), P generates an angle of radians in 8 sec. The distance traveled by P along the circle is s = 0 = 0 cm 0. (a) Suppose the person takes a seat at point A. When the person travels radians, the person is 50 ft above the ground. When the person travels more radians, we can let x be the additional vertical distance traveled: x sin = x = 50 sin = 50 = 5 50 Thus, the person traveled an additional 5 ft above the ground, for a total of 75 ft above the ground. (b) The Ferris wheel goes radians per 0 sec or = radians per second s 0 (c) v = = 5 cm per sec. t 8 9. r = 8,00,000 mi In. years, Jupiter travels radians. θ ω = ω = t. v = rω v = 8, 00, 000.,0,78. miles per year Now convert this to miles per second:,0,78. mi yr v = yr 5 days day hr min hr 0 min 0 sec 8.78 mi per sec