Practice 14. imathesis.com By Carlos Sotuyo

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1 Practice 4 imathesis.com By Carlos Sotuyo Suggested solutions for Miscellaneous exercises 0, problems 5-0, pages 53 to 55 from Pure Mathematics, by Hugh Neil and Douglas Quailing, Cambridge University Press, Redefine x as φ; the x interval 0 x 80 becomes 0 x 360 that is 0 φ 360. Then, 3cosφ = implies cosφ = 3 ; therefore, cos ( 3 ) = φ = 48.0 (using the cos key in scientific a calculator). Now, since cos( θ) = cos(θ), then 48.0 is another root. Applying the periodic property of cos(θ ± 360) = cos(θ) we have = (Adding another 360 will take the angle outside the given interval ( 0 x 360). So, we have φ = 48.0 and φ = 3.80 ; the answer in terms of x is given by x = φ; therefore, the two solutions in the interval are: x = φ or x = 4. and x = (a) Period of the tangent and cotangent functions is 80 ; sine sin(bx ), and cosine, cosbx, have period T = 360 b ; hence, when b =, sinx and cosx both have period equal to 80 (b) sin3x = 0.5; interval, 0 < x < 80; redefine 3x = φ that is, φ interval is three times x interval; that is, 0 < φ < 540. We know that the angle whose sine is 0.5 is 30 ; or, using the calculator, sin (0.5) = 30 ; another solution is based on the property sin(80 θ) = sin(θ) which implies that = 50 is another solution. Considering that φ interval upper limit is 540, and applying the periodic property, we have two additional solutions: = 390 and = 50; then, solving for x = 3 φ we have: x = 0 ; x = 50 ; x 3 = 30 and x 4 = Given the interval: 0 θ 360 and the equation cos(θ + 30) = values of θ in the interval: make φ = θ + 30; our interval becomes 30 φ 390. The equation cos(φ) = simplifies to cos(φ) = 0.5; the angle whose cosine is 0.5 is 60 : cos (0.5) = 60. Another solution for the cosine function is given by cos( θ) = cos(θ); however, in our case, 60 in not in the interval under consideration. Let s apply the periodic property, therefore, = 300. ( = 40 is outside the interval). Then, solving for θ = φ 30 we have two solutions for the equation: x = 30 and x = (a) We know that the cofunctions of complementary angles are equal to each other: sin(90 θ) = cos(θ) and cos(90 θ) = sin(θ); that is cos(90 x) = sin(x) which implies that sinx + cos(90 x) = sinx + sinx = sin(x). (b) Hence, sin(x) =. Now, we define φ = x, and the interval 0 x 360 becomes 0 φ 70. Then, sin(φ) = and sin ( 0.5) = 30 ; which is not in the interval; applying the periodic property of the sine function we get: = 330 and, considering that sin(80 θ) = sin(θ)another solution is 80 ( 30) = 0 ; in addition to these two solutions (300 and 0 ) in the interval 0 to360 we have to consider adding 360 periodic property since our modified interval spreads from 0 to70 ; that is: = 690 and = 570. Now, solving for x we have: x = φ which yields x = 05, x = 65, x 3 = 85 and x 4 = 345.

2 9. We have o find the least positive value that is, zero is not included of the angle A for which: (a) sina = 0. and cosa < 0; sine is positive; cosine is negative: II quadrant. Since sina = 0.; sin =.5 which is located in the I quadrant; in the second quadrant we have: sin(80 θ) = sin(θ); that is, 80.5 = (b) tana < 0, sina < 0 : IV quadrant. tana = 0.5; therefore, tan ( 0.5) = 6.6; this result is not the least positive value, it is indeed, negative; but 6.6 is coterminal with = (c)cosa = sina both negative. III quadrant. Now, in the I quadrant we know that cosa = sina = for A = 45 ; in this case both, sine and cosine are negative, therefore cosa = sina = and since sin ( ) = 45 applying sin(80 θ) = sin(θ) we get 80 ( 45) = 5. (d) sina = 0.75 A > 360 by calculator sin ( 0.75) = 3.5 The smallest angle whose sine is negative lies in the III quadrant, that is sin(80 θ) = sin(θ), then 80 ( 3.5) = 93.5; but it is given that A > 360 thus, applying the periodic property, the answer is: = (a) Proving that: sin θ sinθ cos θ sinθ sinθ. Combining the two terms in the left hand side: tanθ tanθ ; from the Pythagorean identity we have: cos θ = sin θ; then, sinθ tanθ ; rewriting cos θ as realizing that since tanθ = sinθ then that is: sinθ tanθ sinθ tanθ which leads to: tanθ tanθ. (b) sinθ + sinθ Let s multiply the left hand side by sinθ ( + sinθ) ( + sinθ) + sinθ sin θ ( + sinθ) + sinθ cos θ ( + sinθ) + sinθ ( + sinθ) + sinθ (c) tanθ + tanθ sinθ + tan θ tanθ sinθ sinθ = tanθ we get: ( + sinθ) ( + sinθ) which produces a difference of two squares sin θ in the numerator: considering that, by the Pythagorean identity, cos θ = sin θ, we have: which simplifies to: combining the two terms on the left hand side: we know that sec θ tanθ substituting sinθ cos θ sinθ sinθ sinθ sinθ cos θ = sec θ which simplifies to: + tan θ = sec θ; then, and sinθ = tanθ, the left hand side becomes: (d) sin θ sinθ + sinθ the left hand side can be rewritten as:

3 sin θ sin θ sinθ; therefore, considering that sin θ = cos θ: + sinθ cos θ sin θ sinθ; the numerator on the left hand side, a difference of two squares, becomes: + sinθ ( sinθ)( + sinθ) sinθ; which simplifies to: + sinθ sinθ sinθ. The maximum and minimum values of y and the least positive values of x at which these occur: Note: the cosine and the sine functions (y =, y = sinθ) reach a minimum value y = and a maximum value y = ; in summary, the range of both functions is y. (a) y = + cosx reaches a maximum y = whenever cosx = ; therefore, x value is given by: cos () = 0 but we cannot consider zero as a valid solution since we are looking for the least positive values of x; then = 360, thus x = 360 and x = 80. y = + cosx reaches a minimum value of y = 0 whenever cosx = ; and it occurs at: cos ( ) = 80 therefore, x = 80 or x = 90. (b) y = 4sin(x + 30); maximum, y = 9 occurs at sin(x + 30) = and since sin ( ) = 70 the x values is given by: x + 30 = 70 or x = 40. The minimum value of y = occurs at sin(x+30) = ; that is, sin () = 90 which implies that x+30 = 90 or x = 60. (c) y = 9 0sin(3x 45) reaches a maximum value of y = 49 when sin(3x 45) = ; the x value is given by: sin ( ) = 70 that is 3x 45 = or x = = Minimum value of y = 9 occurs when sin(3x 45) = ; in this case the x value is given by: sin () = 90 ; which implies that x = 45. (d) y = 8 3cos x since cos x > 0 both, ( ) = () = ; the maximum value of y = 8 is reached when cosx = 0. And, cos (0) = 90 ; hence, x = 90. The minimum value of y = 5 occurs when cos x = ; which implies that cosx = or cosx = ±; since cos () = 360 (least positive value) and cos ( ) = 80 therefore, the least positive value of x is 80. (e) y = this time, the maximum value of y = 6 is reached when the denominator takes its smallest vale at cosx = ; 3 + cosx that is at cos ( ) = x = 80. The minimum value of y = 3 occurs when cosx = therefore cos () = x = 0. Considering the periodic property of the cosine function the the least positive values of x is = 360. (f) y = 60 + sin (x 5) The function reaches its maximum value, y = 60 when sin (x 5) = 0; that is, sin (0) = x = 0. Solving for x we have: x 5 = 0; hence x = 5 = 7.5. The minimum value y = 30 occurs when sin(x 5) = ±; therefore, there are two possibilities: sin ( ) = x = 90. and sin () = x = 90. We takes the least positive vale, ; then, x 5 = 90; hence, x = = (a) sinθ = tanθ The equality, sinθ = sinθ is true when = or sinθ = ; and it occurs when cos () = x = 0 or sin (0) = x = 0 and x = 360. (b) cos θ = sinθ considering the identity sin θ + cos θ = ; or cos θ = sin θ, and substituting, we obtain: ( sin θ) = sinθ equivalent to: 3

4 sin θ = sinθ which reduces to: sin θ = sinθ equating it to zero in order to factorize the equation: 0 = sinθ sin θ 0 = sinθ( sinθ) Therefore, either sinθ = 0 or sinθ = 0; in the first instance we have as solution sin (0) = 0 also sinθ = 0 in the given interval at x = 80 and x = 360. The factor sinθ is equal to zero when sinθ = ; that is, sin ( ) = 30 and, since sin(80 θ) = sinθ another solution is = 50. (c) tan θ tanθ = This equation can be treated as a quadratic equation where the variable, say, x, is equal to tanθ. By making the x substitution and setting it equal to zero we have, x x = 0 then, using the quadratic formula: x, = b ± b 4ac where a =, b = and c = the a solutions are: x = + and x =. Since we define x = tanθ, thus tan ( ) =.5 This angle is coterminal with which is a solution in the given interval ( = 337.5). Another solution takes into consideration the periodic property of the tangent function: tan(80 + θ) = tanθ, therefore 80 + (.5) = The root tanθ = +, leads to tan (+ ) = 67.5 and, again by the periodic property of the tangent function: = (d) sinθ 3cosθ = 0 since we are dealing with θ the given interval, 0 θ 360 becomes 0 θ 70. Then, rearranging the equation, we have: sinθ = 3cosθ sinθ cosθ = 3 tanθ = 3 tan ( 3) = 60 Two other solutions for θ result from the periodic property of tangent, = 40 and from adding another 360 we are working in the interval 0 θ 70 : = 40 and, = 600. That is, we have four solutions for θ, namely: 60, 40, 40, and 600 ; therefore, the corresponding values of θ = (θ) are: 30, 0, 0, and (a) t(x) = tan3x tanθ has a period of 80 ; in this case, θ = 3x ; therefore, the period, T = 80 3 = 60. (b) Given tan3x = therefore, a solution is tan ( ) = 6.6 another solution is given by tan(80 + θ) = tanθ that is: = 06.6 and, the third solution results from adding 360 : = ; now, solving for x (we have found solutions for 3x) we have: x = 3 (6.6) = 8.9 ; x = 3 (06.6) = 68.9 ; and x = 3 (386.6) = 8.9. (c) tan3x = ; tan ( ) = 6.6 which is not a solution in the interval under consideration; but, since tan(80 + θ) = tanθ we have = 53.4 as a solution for 3x; the x values is x = 3 (53.4) = 5.. (d) tan3x = ; therefore, tan () = 3x = The x value is given by x = 3 (63.4) =.. 4. (a) We are looking for a function, sine or cosine function, of the type: y = A + Bsinkt. Since the period is 4 hours k = 360 = 5; the maximum occurs when sin5t = ; the minimum, when sin5t = ; then we set up a simple system of 4 two equations: 3.6 = A B and 6 = A + B 4

5 If we add these equations together, the terms containing B add up to zero; we obtain: 9.6 = A Hence, A = 4.8 and, B = =. Our equation is: y = sinkt Note: similarly, we can model the phenomena using the cosine function. Or, instead of y = A + Bsinkt we may use y = A Bsinkt, considering that the maximum value is reached when sinkt = and the minimum value when sinkt =, etc. (b) Using the same approach as in (a): k = equation is, = 36; using he equations of the form y = A + Bsin36t our system of two 5000 = A B and 8000 = A + B = A That is, A = 500 and, B = = 6500 y = sin36t (c) k = 360 = ; y = A + Bsint The system of two equations: 360 Our equation becomes, = A B and = A + B 4 = A therefore A =, B = 0 y = + 0sint 5. The displacement, y centimeters, of the tip of one of the prongs from its rest position after t seconds is given by y = 0.sin(00000t) (a) The greatest displacement is 0. cm. It occurs when sine reaches its maximum value, one; that is, sin(00000t) = ; by taking the inverse sine of, we find at which time it occurs: sin () = 90 ; therefore, 00000t = 90, and t = 90 = seconds (b) The time for one complete oscillation is given by the period of the cosine function; thus, period, T, T = = seconds. (c) The number of completes oscillations per seconds is the frequency of the wave, that is, the reciprocal of the period (time for one complete oscillation): F requency = = , oscillations per second (d) See the sketch of the function s graph below. The total time during the first complete oscillation for which the tip is more than 0.06 cm from its rest position, is given by the time elapsed between t and t plus the time elapsed between t 3 and t 4. To find t and t we solve 0.06 = 0.sin(00000t) which reduces to, 0.6 = sin(00000t) sin (0.6) = t = = s Also, since sin(80 θ) = sinθ, the solution in the II quadrant is: = And t is given by: t = = s 5

6 The oscillation in the opposite direction the prong oscillates from rest in one direction; and, from rest in the opposite direction, -0.06, as indicated in our sketch. Therefore, 0.06 = 0.sin(00000t) 0.6 = sin(00000t) sin ( 0.6) = Again, since sin(80 θ) = sinθ, the solution in the III quadrant is: 80 ( 36.87) = And t 3 is given by: t 3 = = s Notice, by analyzing the graph, that the prong vibrates beyond until it reaches time t 4, in the IV quadrant. That angle is given by = Thus, t 4 = = s The total time tip is more than 0.06 cm from its rest position= (t t ) + (t 4 t 3 ) = ( ) + ( ) = s. 6. (a) d = cos500t The greatest depth occurs when cos500t =, y = () = 0 cm; the least, when cos500t =, y = ( ) = 90 cm. (b) The lowest depth is the highest position (90cm), it occurs cos500t = that is cos ( ) = 80; therefore, t = 80 = 0.36 secs. 500 (c) Time for a complete oscillation, that is, the period, is given by T = 360 = 0.7 secs. 500 (d) The depth of the ball is less than 99 cm during the time elapsed between t and t ; t is given by, 99 = 00+0cos500t which simplifies to = 0cos500t ; or, 0. = cos500t ; then, solving for t, cos ( 0.) = ; therefore, t = = 0.95 secs. The ball reaches d = 99 cm position again, at the instant when the angle is equal to Consider that = cos( θ); then, cos = cos( 95.74) t = oscillation for which the depth of the ball is less than 99 cm, is given by t t and = At that point, = secs. Given that the period see item (c) is 0.7 secs the proportion of time during a complete below depicts one complete oscillation, and the corresponding angles: 0.7 = = The graph 6

7 7. The given equation of a particle that displacement y meters is: y = a sin(kt + α) ; then, (a) k in terms of T (period, or time for a complete oscillation) is given by k = 360 T ; (b) The number, in terms of k of complete oscillations per second is the frequency, that is, the reciprocal of the period: = T 360 ; 8. The given function is: P = N C coswt ; (a) T, the period, is known to be 50 weeks, then w = 360 T = = 7.. (b) (i) At the start of the year, t = 0, the number of birds is given by: P = N C cos7.(0) = N C. (ii) The maximum number of birds: it occurs when cos7.t = ; that is, cos ( ) = 80; therefore, 7.t = 80, or t = 80 = 5. The maximum number of birds occurs at mid year, week (a) The fact that high tides occur every hours tells us that the period of the function is ; therefore, since the period T is given by T = , then k = k = 30. (b) Our function becomes h = 4.6cos30t. Let s analyze its graph (see below): the maximum height is reached at t = 0 and it is equal to 4.6 m. The dots indicate the time interval the road is covered; since the cosine function is symmetrical about the y-axis the road level is reached when the water is rising, then the water level reaches its maximum and starts falling. The 3 hours the road is closed means that the water reaches the road.5 hr before its maximum of 4.6 m, and stays above the road level during the second.5 hr. In the function, time is set to be time in hours from high tide; therefore, the road level is reached.5 hrs before the tide reaches its maximum, t =.5; substituting, h = 4.6cos30(.5) = 3.5 m (c) Because of the road repair, the road is now covered for hours 40 minutes; that is, the road level is reached hour and 0 min from the tide maximum height. One hr and 0 mins = 3 hr or 4 hr; thus, 3 ( ) 4 h = 4.6cos30 = 3.5 m 3 7

8 The road level has been raised by = 0.7m or 7cm. 0. The given equation is h = Acosαt + Bcosβt; where the term Acosα is due to the sub s effect, whose period is 360 days; therefore, α = = ; while the term Bcosβt is due to the moon, whose period is 30 days; then, β = =. The equation becomes h = Acost + Bcost. In determining the values of A and B, we take into consideration that h = 5 when t = 0; this fact gives us the following relationship between A and B: since cos(0) =, it reduces to: 5 = Acos(0) + Bcos(0) 5 = A + B We know that the magnitude of the attraction of the moon is assumed to be nine times the magnitude of the attraction of the sun; that information is translated into the equation 9A = B (A is a factor in the sun s term; B a factor in the moon s term). Now we have a simple system of two equations, namely: Substituting the second equation into the first, we get: And, since B = 9A, it follows that B = 9(0.5) = = A + B and 9A = B 5 = A + 9A or 5 = 0A that is A = 0.5 8

Practice 14. imathesis.com By Carlos Sotuyo

Practice 14. imathesis.com By Carlos Sotuyo Practice 4 imathesis.com By Carlos Sotuyo Suggested solutions for Miscellaneous exercises 0, problems 5-0, pages 53 to 55 from Pure Mathematics, by Hugh Neil and Douglas Quailing, Cambridge University