Chapter 7. 1 a The length is a function of time, so we are looking for the value of the function when t = 2:

Transcription

1 Practice questions Solution Paper type a The length is a function of time, so we are looking for the value of the function when t = : L( ) = 0 + cos ( ) = 0 + cos ( ) = 0 + = cm We are looking for the minimum value of a function of the form y = acos [ ( x + c) ]+ d. Since the smallest value of cosine is, the minimum value of L will e: L min = 0 + ( ) = 8 cm Note: We have found the minimum value of the function using the fact that the minimum value of cosine is. We could have found the result differently: The minimum value of a function of the form y = acos [ ( x + c) ]+ d is equal to the vertical shift d minus the amplitude a ; so we have: min = d a = 0 = 8. c We have to find the least value of t such that L = 8. From part we can see that the lowest value will e otained when the cosine equals ; hence, t = t =. Note: We could find t y solving the equation: 0 + cos ( t) = 8 cos ( t) = 8 0 cos ( t ) =. t = t = sec d We have to determine the period of a function of the form y = acos [ ( x + c) ]+ d. So, the period is = = sec. Solution Paper type We will oserve the function L = 0 + cos ( t). Firstly, we have to select radian measure and then input the function. a We have to find L(): c For the remaining parts of the question, we will use the graph of the function. So, we have to set a suitale window. The minimum value is the vertical shift minus the amplitude, and the maximum value is the vertical shift plus the amplitude. Therefore, we can use y-values from (less than) 0 = 8 to (greater than) 0 + =.

2 Another way would e to use the ZoomFit feature. The window will not e suitale for all the calculations, ut we can read out the range and then just extend it a it in oth directions. We can solve oth parts and c y finding the minimum value with the smallest x. d We can find the period of a trigonometric function y finding two successive minimum points, or two successive maximum points, and then sutracting their x-coordinates. We already know the smallest positive minimum has x = 0., so we have to find the next positive minimum. Hence, we can conclude that the period is. 0. =. sin x cos x + = 0 ( cos x) cos x + = 0 Use the Pythagorean identity sin x = cos x. cos x cos x + = 0 t t + = 0 t =., t = Sustitute cos x = t. cos x =. has no solution cos x = x = 0, Solutions are: 0,. The perimeter of the shaded sector contains two radii and an arc. So, the length of the arc can e calculated: = + s s =. We can use the arc length formula to find the angle θ in radians: s = rθ = θ θ = For angles θ and θ : θ+ θ = θ =.

3 a i The amplitude of function f is, so the minimum value of the function is. ii Period of g = =. It is quickest to find the numer of solutions from the graph. We can see that there are four points of intersection; hence, there are four solutions to the equation. For d = p + q m t cos : a To find p we have to determine the mid-line; therefore, we have to find the average of the function s maximum and minimum value: + p = = To determine q, we have to determine the amplitude. The amplitude is the difference etween the function s maximum value and the mid-line: q = = 9. So, q is 9 or 9. From the given data, we can estalish that the graph starts from the maximum value, so q is positive; hence, q = 9. c From the data, we can see that the distance etween two successive maximum points (or minimum points) is 0. seconds. So, the period is 0.. Using the formula for period, we have: period = = m. Hence, m = 0.. m Note: It is useful to highlight the asic data using a rough sketch Solutions are: 0,.0,.0. Note: The question does not tell us to use a specific method to solve the equation, so we can choose any method. The most suitale method is graphical, ecause we can see the numer of solutions and find them all.

4 7 a Use the sustitution cos x = t : t + t + = 0 t =, t = cos x = has no solutions cos x = x =, Solutions are:,. Use doule angle identity for sine: - y x sinx cos x cos x = 0 Factorize: cos x = 0 x =, cosx( sin x )= 0 sin x = 0 sin x = x =, Solutions are:,,,. 8 Given < x <, it follows that sin x > 0 and cos x < 0. a Using the Pythagorean identity for sine, we have: sin cos 8 x = x = =. 9 9 Using the fact that sin x > 0, we have: sin x = =. 9 Using the doule angle identity for cosine, we have: cos cos 8 7 x = x = = = c Using cos x < 0, we have: cos x = = = 9. Hence, sin sin cos x = x x =. = 9 9 We have to interpret the data as points on a graph: High (maximum tide) at :00 with depth of.8 m ( 8,. ) (,.8) y x Low (minimum) tide at 0:0 with depth. m ( 0.,. ) a The function can e written in the form: d = Asin ( B( x + C) )+ D (or similar with the cosine function). Firstly, we will determine D, so we have to find the average of the function s maximum and minimum value: D = =. Next we will determine A. The amplitude of the function is the difference etween the function s maximum value and the mid-line: A = 8.. =.. So, A is. or.. From the data, we can see that the earlier minimum is reached at ( 0.,. ), so the graph after the mid-line first reaches the maximum value, then the minimum value; therefore, A is positive, A =.. (-0.,.) 0 (0.,.) (,.8) (0.,.) 0

5 We can also see from the data that the distance etween successive maximum and minimum points is. hours. So, half of theperiod =.. Using the formula for period, we have: period = = B =. B B To determine the horizontal translation (phase shift), we have to find the points of the graph on the mid-line. Ascissae of those points are on a distance of a of the period from the ascissa of the maximum point. Hence, ±, so 9 and. Thus, C = 9 and the function is: (.,.) (,.8) (7.7,.) 0 (0.,.) 9 d =. sin x +.. We have to find the value of the function at noon, that is, hours after midnight; therefore, d( ). sin.. 9 = + metres. c We have to find the time interval in which the value of d is greater than.. We have to convert from decimal numers to minutes: Therefore, from aout :7 pm to 7: pm the oat can dock safely. We can also see that around midnight the oat can dock safely again.

6 0 Use the sustitution tan x = t : t + t = 0 t =, t = tan x = x = tan x = cannot e solved exactly Solutions are:, 89.. a We will use the arc length formula: s = rθ = 0 = cm The angle in the shaded region is ; therefore, using the area of a sector formula: A = r = 0 θ 9 cm The function f ( x) = cos x reaches a maximum value of and a minimum value of. That means that f ( x ). The function will not reach values greater than or less than. Hence, for k > and k <, the equation will have no solutions. From the graph, we can see that the point (0, ) is on the mid-line, so k =. To determine a, we need to find the amplitude. The amplitude is the difference etween the function s maximum value and the mid-line: a = = ; so, a is or. We can see that the graph, starting from initial position at x = 0, first reaches the minimum value, and then the maximum value, so a is negative; hence, a =. We can write tan sin α cos α a using secant only: tan α = = = = sec α cos α cos α cos α So, the equation is: sec α sec α 0 = 0 sec α sec α = 0 Using the sustitution sec α = t, we have: t t = 0 t =, t =. Since α is in the second quadrant, secant should e negative and the solution is sec α =. a Using the compound angle formula for sine, we have: sin( α + β) = sin αcos β + cos α sin β From the upper triangle, we can see that: sin α = 7, cos α = 8 7 From the lower triangle, we can see that: sin β = 0, cos β = So, sin( α + β) = + = = cos( α + β) = cos αcos β sin α sin β = 8 8 = sin ( α + β) c tan( α + β) = cos ( α + β) = = = 8 Note: We can find tanα and tan β from the drawing, and then use the compound formula for tangent. From the diagram, we can see that: sin p = = and cos p = =. + + Hence: sinp = sin p sin p = =. For sinp, we can use the compound angle formula: sinp = sin p + p sinp cos p cosp sin p ( ) = +

7 So, we have to find cosp = cos p sin p = =. Finally, we have: sinp = + = =. 7 If B is otuse, then the sine is positive and cosine negative. a sin B = = + cos B = = + c sin sin cos 0 B = B B = = 9 d 9 cosb = cos B sin B = = 9 Note: In the solution, we have used the property of angles and sides in a right triangle. We can find the solution y using the Pythagorean identity for sine and cosine: sin x = sinx = cos x cos x 9 sin x + cos x = cosx cos x cos + = x = cos x = = 9 Finally, sin x = =. Now we can continue using the doule angle formulae. tan θ 8 Using the doule angle formula for tangent, we have: tan θ = = 8 tanθ = tan θ tan θ Using the sustitution tan θ = t, we will have a quadratic equation: t + 8t = 0 t =, t =. Hence, the possile values of tan θ are:,. 9 We will use the compound angle formula for sine: sinx cos α cosx sin α = k ( sinx cos α + cosx sin α) Then we will divide oth sides y cosx cos α : sinx cos α cosx sin α sinx cos α cosx sin α = k + cosx cos α cosx cos α cos x cos α cos cos α x tanx tan α = k ( tanx + tan α) tanx k tan x = k tanα + tan α tanx ( k)= tan α ( k + ) tan α( k + ) tan x = k ( k + ) Note: If we rearrange the equation differently, we otain the result tan x = tan α, which is equivalent to the k aove result. 0 We can see that either tanθ = or tanθ =. So, tan θ = θ = + k θ = + k θ =, + = tan θ = θ = + k θ = + k θ =, + = Hence, the solutions are: ±, ±. 8 8 a We have to set the window on the GDC to e the same as on the grid. 7

8 The domain of the functions is x, so we just have to draw this part of the graph. c No solutions Cosine function has a range from. But, here, its domain is restricted. Since cosine is an even function, it is enough to oserve its ehaviour on the interval from 0 to. Hence, the function is decreasing on this interval; its largest value is cos0=, and the smallest value cos. Hence, the range is the set [ cos, ]. Let CAD ˆ = θ and AC = x. From the drawing, we can see that: tan θ = x = and tan θ = = x tan θ x x tan θ. Hence: = tanθ tan θ ( tan θ) Using the doule angle formula for tangent, we have: = = = tan θ tan θ = tan θ tan θ tan θ Since θ is an angle in a right triangle, its tangent has to e positive. So, tan θ = θ.. Note: We can solve the equation = graphically. tanθ tan θ We are looking for the angle when those two sides are the same. We have to solve the equation sec x 0 = +. Hence, x x 8 sec cos = = = 8 x 8 cos = We need the first positive solution, so: x 8 8 = arccos x = arccos Hence, the solution of the equation is: 8 arccos, and the width of the water surface in the channel is: arccos = arccos cm. 8