Solutions. MATH 1060 Exam 3 Fall (10 points) For 2 sin(3x π) + 1 give the. amplitude. period. phase shift. vertical shift.

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Clifton Godfrey Goodman
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1 MATH 060 Exam Fall 008 Solutions. (0 points) For sin(x π) + give the amplitude period phase shift vertical shift amplitude= period= π phase shift= π vertical shift=. (5 points) Consider a sine curve with amplitude=, period=π, phase shift=, and vertical shift=-. Sketch the curve. y K.5 p Kp K 0.5 p p.5 p x p.5 p p K K K

2 y K.5 p Kp K 0.5 p p.5 p x p.5 p p K K K. (5 points) There are two values of x between 0 and π for which sin(x) =. One of these is x = arcsin( ). What is the other? π arcsin( ). (5 points) Consider a sine curve with amplitude=, period= π, phase shift=π, and vertical shift=. Give a formula for this sine curve. The formula is of the form a sin(bx c) + d where amplitude==a vertical shift = =d period = π = π b = b = phase shift = π = c b = c = c = π so sin(x π) + 5. (5 points) The following sine curve passes through the points (0, ) and (π, ). Find:

3 amplitude period vertical shift phase shift 6 5 K p K p Kp 0.5 p p.5 p p.5 p p K K x The maximum is 6 and the minimum is -. The amplitude is half the difference between these, so the amplitude is 6 ( ) = 8 = The equilibrium line is half way between the maximum and minimum, which is, so the vertical shift is. Pick a vertical starting line somewhere where the curve crosses the equilibrium line headed up. The y-axis would be a good choice, but any multiple of π will do. The phase shift is the x-coordinate of this line, so the phase shift could be any multiple of π. The period looks like π. To be sure that it is exactly π, notice that the curve crosses the equilibrium line going up at the points (0, ) and (π, ). The curve completes three complete cycles in this time, so the period is π = π. 6. (5 points) For the curve described by the formula sin(x π), find:

4 amplitude period vertical shift phase shift amplitude= period= π = π vertical shift=0 phase shift= π 7. (5 points) Simplify the expression ( sin x)( + sin x) ( sin x)( + sin x) = sin x = cos x 8. (0 points) cos u = and u is in the st quadrant. Find sin u sin(u + π) sin(u π) cos( u) cos(u + π) cos(u π) cos(u + 00π) sin(u + π)

5 sin u sin u + cos u = sin u = cos u sin u = ± cos u since u is in the first quadrant we want the positive root sin u = cos u = ( ) = 6 = 5 6 sin(u + π ) = cos u = sin(u π ) = sin(u π + π) = sin(u + π ) = cos(u) = cos( u) = cos u = since cosine is an even function. cos(u + π) = cos u = cos(u π) = cos u = cos(u + 00π) = cos u = since cosine is π-periodic. sin(u + π ) = sin(u + π + π) = sin(u + π ) = cos u = 9. (0 points) Find all solutions to: cos(x) = cos(x) = cos(x) = = There are two angles between π and π that have cosine=. arccos( ). They are arccos( ) and 5

6 So Either x = arccos( ) or x = arccos( ) x = arccos( ) x = arccos( ) Finally, the preiod is π, so we get all solutions by adding multiples of the period to the two particular solutions. So all solutions are of the form: x = arccos( ) + π n or x = arccos( ) + π n 0. (0 points) Find all solutions to cos (x) =. sin (x) cos (x) =. sin (x) =. + sin (x) =. + cos (x) using Pythagorean identity cos (x) =. cos(x) = ±. So again we get two linear equations in cos cos(x) =. or cos(x) =. Let s solve the first one: cos(x) =. cos(x) =. x = arccos(.) or x = arccos(.) x = arccos(.) x = arccos(.) Now add multiples of the period to these particular solutions to get the general solution for this linear equation. The period here is π. x = arccos(.) + π n or x = arccos(.) + π n 6

7 Now go back and solve the second linear equation cos(x) =. and find more solutions of the form: x = arccos(.) + π n or x = arccos(.) + π n. (0 points) Use the formula to find all x such that Using the formula, sin(u) + sin(v) = sin( u + v 0 = sin(5x) + sin(x) = sin( sin(5x) + sin(x) = 0 ) cos( u v ) 5x + x 5x x ) cos( ) = sin(x) cos(x) A product of numbers is equal to zero only if one of the factors is zero, so either sin(x) = 0 or cos(x) = 0. Now we have two different linear equations to solve. sin(x) = 0 when x = arcsin(0) = 0 or when x = π arcsin(0) = π. So we get particular solutions x = 0 or x = π. We get all solutions by adding multiples of the period, which is π = π, so the solutions for this linear equation are of the form: x = 0 + π n or x = π + π n The other possibility was cos(x) = 0. We get particular solutions x = arccos(0) = π and x = arccos(0) = π. We get all solutions by adding multoples of the period, π, to these. So all solutions are of the form x = arccos(0) + πn = π + πn or x = arccos(0) + πn = π + πn Now put all these together. Solutions to sin(5x) + sin(x) = 0 are all of the form: x = 0 + π n or x = π + π n or x = arccos(0) + πn = π + πn or x = arccos(0) + πn = π + πn No new questions beyond this point. 7

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