Math 141: Trigonometry Practice Final Exam: Fall 2012

Transcription

1 Name: Math 141: Trigonometry Practice Final Eam: Fall 01 Instructions: Show all work. Answers without work will NOT receive full credit. Clearly indicate your final answers. The maimum possible score is 10 points. Question 1 10 points). Let 0 θ < 90, such that tan θ = 5. Find the eact values of each of the remaining five trigonometric functions. Solution. We use the Pythagorean Theorem and the following triangle to find the missing side: = θ 1 5 Therefore, we have the following: sin θ = opposite hypotenuse = 5, cos θ = adjacent hypotenuse = 1, sec θ = hypotenuse opposite csc θ = hypotenuse opposite cot θ = adjacent opposite = 1 5. =, = 5,

2 Question 10 points). Fill in the missing values on the following diagrams. Be sure to eplicitly state any formulas or previous knowledge that you use to help yourself. 7 B B A 45 0 A C C Solution. For the 45, 45, 90 triangle, we know that if the hypotenuse is equal to 1, then each of the other sides should be equal to. If we scale all distances by a factor of 7, we find that = 7 = 7. For the 0, 0, 90 triangle, if the hypotenuse is equal to 1, then the long side should be equal to, and the short side is equal to 1. We need to find the appropriate factor by which to scale this standard triangle to be the size we have. That is, we want to find a so 10 that a = 5, and so a =. Scaling all distances by a, we find that the hypotenuse is equal to this value: = a = 10. For the circle of radius, we use the formula We first convert 155 to radians: arc length = radius)radians). 155 π 180 = 1π. Page

3 Net, we apply the formula, and find that: arc length = ) ) 1π = 1π. Page

4 Question. Use the Unit Circle to find the eact values of the following epressions: a) 5 points) sec 405 ). Solution. Unit Circle with θ = 405 = 45 = 15 : y, ) So: sec 405 ) = =. b) 5 points) tan ) 5π cot 7π ). Solution. Unit Circles with θ = 5π and φ = 7π : ), 1 y y 0, 1) So: tan ) 5π cot 7π ) = 1 0 = 1. Page 4

5 Question 4 10 points). Graph the following function on the aes provided: f) = tan) Solution. The general graph of f) = tan has its asymptotes precisely where cos = 0: = {..., π, π, π, π },.... Since our function has the argument scaled by ω =, the graph of this function will be compressed by a factor of. The graph is shown in red below π 5π 4π π π 1π -1 1π π π 4π 5π π Page 5

6 Question 5 10 points). Consider the two functions fθ) = sec θ and gθ) = csc θ. In which quadrant are both functions decreasing? Solution. We eamine the graphs of these two functions. The function f) is in red, and g) is in blue, and the asymptotes have been left out. Note that the domain of our graph is [ π, π], which reaches all four quadrants π 5π 4π π π 1π -1 1π π π 4π 5π π The only portion of the graph where both functions are defined and decreasing is which is in Quadrant IV. -5 π, 0 ), Page

7 Question 10 points). Describe, in interval notation, the standard fundamental domain of each of the si trigonometric functions. That is, describe the restricted domains that we use to define the si inverse trigonometric functions. Solution. [ sin θ : π, π ] cos θ : [ 0, π ] tan θ : π, π ) [ sec θ : 0, π ) csc θ : [ π ), 0 cot θ : 0, π) π, π ] 0, π ] Page 7

8 Question 7. Determine the eact values of: a) 5 points) csc 1 ). Solution. We draw the Unit Circle with the two angles/points where csc θ = 1 y =, or where y = 1 : y ) ), 1, 1 However, only θ = 0 = 0 is in the fundamental domain of cosecant, so: csc 1 ) = 0 = π. b) 5 points) sin 1 cos )) 5π. Solution. We draw the Unit Circle to find cos ), 1 y ) 5π : Therefore we have: cos ) 5π =. Now we draw the Unit Circle with the two angles/points where sin θ = y = : Page 8

9 y 1, ) 1, ) Only θ = 00 = 0 is in the fundamental domain of sine, so we conclude that: )) 5π sin cos 1 = 0 = π. Page 9

10 Question 8 10 points). Solve the following equation for all possible values of θ: 8 cos θ = 0. Solution. We begin by solving for cos θ: 8 cos θ = 0 8 cos θ = cos θ = 1 4 cos θ = ± 1. We now have a number of possibilities. If cos θ = 1, then from the Unit Circle we find that: θ = π + πk, or θ = 5π + πk. If cos θ = 1, then from the Unit Circle we find that: θ = π + πk, or θ = 4π + πk. With k allowed to be any integer, this gives all possible solutions. Page 10

11 Question 9 10 points). Verify the following identity: sinα + β) sinα β) = sin α sin β. Solution. We apply the sum and difference formulas to the left hand side and simplify: sinα + β) sinα β) = sin α cos β + cos α sin β) sin α cos β cos α sin β) = sin α cos β + cos α sin β sin α cos β sin α cos β cos α sin β cos α sin β = sin α cos β cos α sin β. Now we apply the Pythagorean Identity to put this all in terms of sine: = sin α1 sin β) 1 sin α) sin β = sin α sin α sin β sin β + sin α sin β = sin α sin β. Page 11

12 Question points). Use any ) trigonometric identities that you know to calculate by 5π hand!) the eact value of sin. 1 Solution. Since our denominator is 1 and 5π is a standard angle that we know, it is best to use a Half Angle Formula: ) 5π 1 cos5π/) sin = ±. 1 ) 5π We can eamine the Unit Circle to find that cos =. Therefore ) 5π 1 cos = 1 + = +, and and we conclude that sin 1 cos ) 5π = +, 4 ) 5π 1 cos5π/) = ± 1 + = ±. To determine the sign of the square root that we should use, we look for the angle 5π 1 on the Unit Circle. We find it in the first quadrant, where sine is positive. Therefore: sin ) 5π = 1 +. Page 1

13 Question points). Solve the following equation for all possible values of θ: cos θ) + cos θ + 1 = 0. Solution. We first use a Double Angle Formula to simplify the left hand side: This leaves us with the possibilities cos θ) + cos θ + 1 = 0 cos θ sin θ ) + cos θ + 1 = 0 cos θ 1 cos θ ) + cos θ + 1 = 0 cos θ 1 + cos θ + 1 = 0 cos θ + cos θ = 0 cos θ cos θ + 1) = 0. cos θ = 0 or cos θ = 1. In the first case, we have that θ = k or θ = k, which might also be written θ = k) and in the second case, we have that k or k. Here k may be any integer. Page 1

14 Question 1 10 points). Two adjacent buildings are separated by an alley. A woman in one of the buildings looks out her window, 0 feet above the ground. She measures the angle of elevation to the top of the second building to be 0, and the angle of depression to the base of the second building to be 75. She finds on the internet that: sin 75 = cos 75 = How tall is the second building? Solution. We begin by drawing a picture of what we know: ft We see that while we don t know much about the top 0, 0, 90) triangle, we do know the length of one side of the bottom 75, 15, 90) triangle. This will allow us to find all relevant information about that triangle. Page 14

15 While we aren t so concerned with the length of the hypotenuse, we would like to know the distance between the buildings, since this information can be transferred up to the top triangle: 75 0 ft Now we have that tan 75 = 0. We can find the eact value of tan 75 from what we know: tan 75 = +1 1 = ) + 1 = = +. So we arrive at: = 0 tan 75 = 0 +. We now know the distance between the buildings, and the length of one side of our 0, 0, 90 triangle: 0 y = 0 + With the same argument from Question, we find that y = = 0 +, and therefore the total height to the building is = 0 ) = feet. Page 15