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, 00. This is actually a selection taken from practice 4, question 5 to 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 cos θ. Combining the two terms in the left hand side: ; from the Pythagorean identity we have: cos θ = sin θ; then, ; rewriting cos θ as realizing that since = then that is: which leads to:. (b) + Let s multiply the left hand side by ( + ) ( + ) + sin θ ( + ) + cos θ ( + ) + ( + ) + (c) + + tan θ = we get: ( + ) ( + ) 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 substituting cos θ cos θ = sec θ which simplifies to: + tan θ = sec θ; then, and =, the left hand side becomes: (d) sin θ + the left hand side can be rewritten as:

3 sin θ sin θ ; therefore, considering that sin θ = cos θ: + cos θ sin θ ; the numerator on the left hand side, a difference of two squares, becomes: + ( )( + ) ; which simplifies to: +. 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 = ) 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) = The equality, = is true when = or = ; and it occurs when cos () = x = 0 or sin (0) = x = 0 and x = 360. (b) cos θ = considering the identity sin θ + cos θ = ; or cos θ = sin θ, and substituting, we obtain: ( sin θ) = equivalent to: 3

4 sin θ = which reduces to: sin θ = equating it to zero in order to factorize the equation: 0 = sin θ 0 = ( ) Therefore, either = 0 or = 0; in the first instance we have as solution sin (0) = 0 also = 0 in the given interval at x = 80 and x = 360. The factor is equal to zero when = ; that is, sin ( ) = 30 and, since sin(80 θ) = another solution is = 50. (c) tan θ = This equation can be treated as a quadratic equation where the variable, say, x, is equal to. 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 =, 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) =, therefore 80 + (.5) = The root = +, leads to tan (+ ) = 67.5 and, again by the periodic property of the tangent function: = (d) 3cosθ = 0 since we are dealing with θ the given interval, 0 θ 360 becomes 0 θ 70. Then, rearranging the equation, we have: = 3cosθ cosθ = 3 = 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 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) = 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) = 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. 5