Math 142: Trigonometry and Analytic Geometry Practice Final Exam: Fall 2012

Transcription

1 Name: Math 14: Trigonometry and Analytic Geometry 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 f(θ) = sec θ and g(θ) = 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. 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 7

8 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 8

9 Question 7 (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 9

10 Question 8 (10 points). For all in the interval 1 1, verify the following identity: arcsin() + arccos() = π Solution. At first glance, we might try to simplify the epression arcsin() + arccos() by taking the sine of it, but then we have on the left the sine of a sum of two angles. So we use the sum formula: sin(arcsin() + arccos()) = sin(arcsin()) cos(arccos()) + sin(arccos()) cos(arcsin()). Now, we find ourselves trying to simplify two epressions of the form sin(cos 1 ()), so for this we use the following picture describing the angle θ = cos 1 (): 1 θ From the picture we can see that the side opposite to θ has length 1, so we conclude that sin(cos 1 ()) = 1. An identical argument will show that cos(sin 1 ()) = 1 as well. We can now simplify our epression further: sin(arcsin() + arccos()) = sin(arcsin()) cos(arccos()) + sin(arccos()) cos(arcsin()) = = + (1 ) = 1. Now, we see that the angle described by arcsin() + arccos() must be equal to π [ + πk for some integer k. But since arcsin only takes values in the range π, π ], and arccos() takes values in [0, π], it follows that π arcsin() + arccos() π. Therefore arcsin() + arccos() = π for any in the domain of arcsine and arccosine. Page 10

11 Question 9 (10 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 cos(5π/) 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 cos(5π/) = ± 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 11

12 Question 10 (10 points). Use the figure to evaluate g(α) when g() = cos. y α + y = 1 ( 9, y) Solution. We will need to use the Double angle formula: g(α) = cos (α) sin (α). From the image, it is clear that cos(α) =, but we still need to find y. For this, we use 9 the pythagorean theorem to write ( 9) + y = 1, and find that y = 77. Again, from the image it is clear that y should be negative, so we conclude that y = sin(α) = 9, and g(α) = cos (α) sin (α) = = Page 1

13 Question 11 (10 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). Use De Moivre s Theorem to find ( 1 i ) 5. Solution. To use De Moivre s Theorem, we must first put this comple number into polar form. If z = 1 i, then the magnitude of z is: z = (1) + ( ) =, and it s angle is: ( ) tan 1 = 0. 1 So, we can write z = ( cos(0) + i sin(0) ). Now we apply De Moivre s Theorem: z 5 = 5( cos(5 0) + i sin(5 0) ) = ( cos(00) + i sin(00) ) ( ( )) 1 = + i = 1 1 i. Page 14

15 Question 1 (10 points). Consider the quadratic equation = y +. (a) Is the graph of this equation a parabola, circle, ellipse, or hyperbola? (b) Put the equation into the appropriate standard form. (c) Identify any key features of the graph, such as foci, asymptotes, etc. Solution. First, we move everything to one side: + y = 0. We see that there are two square terms, in both and y, and both terms are positive, so this equation describes an ellipse. Net, we put this in standard form by completing the square on (y is already a perfect square): + y = y = 1 4 ( 1 ) + y = ( 1 ) + y = 9 4. Therefore this ellipse is specifically a circle 1, with center. It s four vertices are: ( 1 ) ( ) 1, 0, and constant radius, ( ) 1, (, 0) ( 1, 0), and these lie on the major and minor aes y = 0 and = 1. The two foci will correspond ( ) 1 with the center, and with each other, at, 0, which can be recovered from the formula a = b + c, and the fact that in this case a is equal to b. 1 As a circle, this conic section lacks many of the features of, say, a hyperbola. But this style of problem can be practiced using almost any quadratic equation. Page 15

16 Question 14 (10 points). Consider the diagram that is drawn on the unit circle below. Find the values of each of the si trigonometric functions, for the angle θ. For eample, we can see that cos θ = OA. Find the others. Make sure to eplain your choices with a convincing argument. y E = (0,1) D P C O θ A B = (1,0) Solution. It should be clear that sin θ = AP. To find tan θ, notice that the two right triangles formed by OAP and OBC share the angle θ, and thus are similar. For eample, AP OA = BC OB. But AP = tan θ, and OB is the radius of the circle and is equal to one, so OA tan θ = AP OA = BC 1 = BC. (Think also about the graph of f(θ) = tan θ versus the length BC. When θ = 0 we have BC = 0, but as θ approaches 90, BC will become infinite.) We can use the same argument with the angle 90 θ (for the triangle formed by OED), and some new lines drawn, to show that cot θ = ED. Page 1

17 To find sec θ and csc θ, we could use a similar argument, but we will try a different approach. We know the pythagorean identity sec θ = tan θ + 1. Since tan θ = BC, we might try looking for a right triangle that has sides of length one and BC, and then the length of the hypotenuse will be equal to sec θ. We find the needed triangle to be formed by OBC, so sec θ = OC. Likewise, we can use the other pythagorean identity csc θ = 1 + cot θ to single out the triangle formed by OED, and from its hypotenuse we find that csc θ = OD. Think of the ratios OP OA and OC OB. Page 17

18 Question 15 (10 points). A pendulum 50 centimeters long is moved 0 from the vertical. How far did the tip of the pendulum rise? Solution. We refer to the following picture: y 0 a 50 To find the missing value a, we eamine the reference triangle having angle θ = 0, an adjacent side of length a, and a hypotenuse of length 50. This gives the equation: cos 0 = a 50. Then, a = 50 cos 0 = 50 = 5, so the pendulum rose by (in centimeters): ( 5 ) ( 50) = Page 18

19 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 19

20 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 0