Trigonometry Exam II Review Problem Selected Answers and Solutions

Transcription

1 Trigonometry Exam II Review Problem Selected Answers and Solutions 1. Solve the following trigonometric equations: (a) sin(t) = 0.2: Answer: Write y = sin(t) = 0.2. Then, use the picture to get an idea of how the answers are related. Find sin 1 (t) = πn Then, the final answer is: t = (π ) + 2πm (b) 3 cos(5t) = 1: Answer: Get the trig function on one side by itself: x = cos(5t) = 1 3. Write BLOB = 5t. Then, use the picture to see how the answers are related. Find cos 1 (1/3) = Then the

2 final answer is t = πm πn 5 (c) 2 sin(5t + 4) = t 3 t: Answer: Since we have trig functions and non-trig function in the same equation, we can only solve this graphically. The graphs, with the points we re looking for: Answers: { , , , , }. (d) tan(3t) sin(5t + 1) 3 sin(5t + 1) = 0. solution: Start by taking out what each term has in common: tan(3t) sin(5t + 1) 3 sin(5t + 1) = sin(5t + 1)(tan(3t) 3) = 0. We now set each factor equal to 0: sin(5t + 1) = 0: We write BLOB for 5t + 1, and get y = sin(blob) = 0. Drawing the line y = 0 and the unit circle gives us the two points A and B: We see, then, that every distance BLOB is of the form BLOB = πm

3 Putting BLOB = 5t + 1 = πm, and solving for t gives us t = πm 1. 5 tan(3t) 3 = 0. Getting the trig function on one side by itself, we get We write BLOB for 3t, and get tan(3t) = 3. m = tan(blob) = 3. Drawing the line through the origin whose slope is 3, we get the points C and D: We see, then, that the every distance BLOB is of the form BLOB = arctan(3) + πl. Finding arctan(3) = , and putting BLOB = 3t, and solving for t gives us πl t =. 3 Final Answer: t = πm πl and t = 5 3 (e) 10 cos 2 (4t) cos(4t) = 2. solution: This is an equation of quadratic type, so I need to get everything on one side and try to factor: 10 cos 2 (4t) cos(4t) 2 = (5 cos(4t) + 2)(2 cos(4t) 1) = 0. Setting each factor equal to 0, we get two equations to solve: 5 cos(4t) + 2 = 0: Getting the trig function by itself on one side gives us x = cos(4t) = 2 5.

4 Letting BLOB = 4t, we have to solve x = cos(blob) = 2 5, which means we have to graph the unit circle and the line x = 2 5 : Using our calculator, we see that cos 1 ( 2.5) = , which gives one of our values of BLOB. Using our picture we see that the other distance is This gives us kπ 4t = BLOB = mπ Solving for t now gives us t = kπ mπ 4 2 cos(4t) 1 = 0: Getting the trig function by itself on one side gives us: Letting BLOB = 4t, we have to solve cos(4t) = 1 2. x = cos(blob) = 1 2, which means we have to graph the unit circle and the line x = 1 2 :

5 Using our calculator, we that BLOB = , to 5 decimal places. Using our picture we see that the other BLOB = , to 5 decimal places. This gives us Solving for t now gives us lπ 4t = BLOB = nπ t = lπ nπ 4 Final answer: Here are all the solutions: kπ t = mπ t = lπ t = nπ t = 4 (f) 2 sin(θ) = 2 cos(θ) solution: The problem we have is that (a) there is nothing to factor and (b) there are two trig functions. In this case, dividing both sides by cos(θ) solves the biggest problem, giving us 2 sin(θ) cos(θ) = 2 tan(θ) = 2. Getting the trig function by itself on one side gives us 2 m = tan(θ) = 2 =

6 With no BLOB to worry about, this just means that we need to find where the 2 line through the origin whose slope is m = intersects the origin, as shown 2 below: ( ) 2 Our calculator says one of the answers is t = tan 1 = Using 2 our picture, we see that all of the answers, then are t = πm. (g) 3 sin 2 (t) = 5 5 cos(t) solution: We would like to get something to factor, but a bigger problem is that both sin(t) and cos(t) are in the equation. We do have the Pythagorean identity (remember, this just means a formula that came from using the Pythagorean theorem) sin 2 (t) + cos 2 (t) = 1. If we solve this for cos 2 (t) and plug into our equation, we would still have to deal with the cos(t). (YUCK). So, we re going to solve for sin 2 (t), and get then plug this into our equation to get sin 2 (t) = 1 cos 2 (t), 3(1 cos 2 (t)) = 5 5 cos(t). Moving everything to the right hand side and combining like terms gives us Factoring the right hand side gives us 0 = 3 cos 2 (t) 5 cos(t) = (3 cos(t) 2)(cos(t) 1). Setting each factor equal to 0 (Why does that work again?), we get two trig equations to solve:

7 3 cos(t) 2 = 0: Getting the trig function by itself on one side gives us x = cos(t) = 2 3. Without a BLOB to worry about, this means finding the intersection of the line x = 2 with the unit circle, like this: 3 ( ) 2 Using our calculator, we see that one of the answer is t = cos 1 = Using our picture, we see that the other answer is t = This gives us kπ t = mπ cos(t) = 1: This is already set up for us to use the unit circle: We need to look for the intersection of the line x = 1 and the unit circle, like this: There s only one point, and it happens when = 0, so all the solutions to this equation are t = 0 + 2nπ. Conclusion: Here are all the solutions: t = kπ

8 t = mπ t = 2nπ 2. Graph the following function by framing. Check your answer using your calculator, and explain how you checked. ( ) π P (t) = cos 12 (t + 3). Answer: You should have found all of the following: Middle line height: 40 Amplitude (remember, this is just the distance from the middle to the top): 15 Top: = middle + amplitude = = 55 Bottom: = middle - amplitude = 25 Left side of frame: t = 3 2π Width of frame: π/12 = 24 Right side of frame: = = 21. remember: You have to graph an upside-down cosine graph, because of the -15. So, including the frame, you should see something like: Now, use the above and your graphing calculator to check your answer! 3. Find a possible equation for the graph below. Check your answer using your calculator and explain how you checked.

9 Answer: There are many correct answers! Just for practice, I m going to draw a frame that has one cycle of sine inside of it; I m adding the middle line so you can see that, too: Now we go through and find all the various elements: The middle line is at about The top is at about 4.25 So, the distance between the middle and the top is about (-2.75) = 7 The left hand side is at about The right side is at about 9.25 The width of the frame is then (-5.75) = 15. ( ) 2π The graph we then get is y = sin( 15 (t ). If you decided to draw a ( ) 2π cosine frame, your answer might be more like y = cos 15 (t + 2)

10 4. Find a possible trig equation that matches the data below: t h(t) (See next page)

11 solution: If we plot the points, we get something like: We can draw a frame from top point to top point and get roughly the following: Top = 60 Bottom = -30 Left = 1 Right = 14 We re trying to find an equation of the form y = a + b cos(c(t d)). This leads to our calculations: 60 + ( 30) Middle line = = 15 2 Amplitude = Distance from middle line to top = = 45 Left = 1 Width of box = Right - Left = 13 (so, the coefficient c is c = 2π 13 ) ( ) 2π Putting these all together gives us the answer: y = cos 13 (t 1).

12 5. At midnight, the power requirement of City A is at a minimum of 40 megawatts. By noon the city has reached its maximum powre consumption of 90 megawatts and by midnight it once again requires only 40 megawatts. This pattern repeats every day. Find a possible formula for f(t), the power, in megawatts, required by City A as a function of t, the number of hours since midnight. solution: This is still an ungraphing question! We need to come up with a graph, so, we need to come up with some points, to graph, etc. First, I make a very simple table of values that start at 40 when t = 0, hit its high point at 90 when t = 12, and get back to its low point of 40 when t = 24. Here is the little table of values, the points, the graph I sketched through the points, and the frame it made: We then do the usual stuff: The middle line A is = 65 The amplitude B is the top - middle = = 25. The left side of my graph D hasn t moved, so is just 0. The width of my box is the right side minus the left side = 24-0 = 24. This tells me that C = 2π 24. Putting these all together, ( we) get: π Answer: y = cos 12 t 6. Bearing problem (Extra credit) 7. A surveyor measures the angle of elevation to the top of a building to be She then walks 100 feet away, and then measures the angle of elevation to be 8.3. To the nearest foot, how far was she originally standing from the building, and how tall is the building?

13 solution: A note about this problem: This is an angle of elevation/depression problem, but harder than the one that might appear on the exam (but not harder than might appear on the final exam): Here s the picture: We actually have two different angle of elevation equations, and we need them both: Solving equation for y, we get Let s just work on the second part: Get all the x s on the left hand side, Factor out the x s: tan(10.1 ) = y x tan(8.3 y ) = x y = x tan(10.1 ) = (x + 100) tan(8.3 ). x tan(10.1 ) = x tan(8.3 ) tan(8.3 ) x tan(10.1 ) x tan(8.3 ) = 100 tan(8.3 ) x(tan(10.1 ) tan(8.3 )) = 100 tan(8.3 ) Then divide both sides by the stuff next to x: Then from above we have x = 100 tan(8.3 ) tan(10.1 ) tan(8.3 ). y = x tan(10.1 ) = 100 tan(8.3 ) tan(10.1 ) tan(10.1 ) tan(8.3 ) Answer: The building was 81 ft high, and she was standing 452 feet away.

14 8. Give approximate values of each of the following by estimating the corresponding distance around the unit circle, and estimating the requested value: (a) sin 1 ( 0.8): solution: Recall how sin(t) works: You plug in a distance t, and the answer is a y-value on the unit circle. Inverse sine works the other way. You plug in a y-value, and the answer is the shortest distance you can go around the unit circle to get to that point. So, here, they re telling us y = 0.8. Drawing a circle and the correct line gives us something like Answer: t 1.1 (b) cos 1 (1.4) solution: We see that we re plugging in x = 1.4, and we re asking for the shortest distance you can go around the unit circle to get to such a point. Here s the picture: Answer: The picture shows us that we can never find such a distance, so that cos 1 (1.4) is undefined. (c) tan 1 (2)

15 solution: We re being told that m = 2, and we have to find the shortest distance around the unit circle to get to such a point. Here s the picture: Answer: t 1.2-ish. : 9. Evaluate each of the following exactly, using both the unit circle and a separate reference triangle: ( (a) sin (tan 1 7 )) 3 solution: We need to figure out which quadrant to draw a reference triangle, then we need to draw and label the reference triangle, then we need to use it. Starting from the inside, we see that we need to make sense of tan ( 1 3) 7. As we did in the previous problem, we see that we are plugging in the slope m = 7, 3 (which is about -2.something), and are being asked for the shortest distance to get to such a point. Here s our first picture: The point with the shortest such distance is in the fourth quadrant; now, we draw and label a really good reference triangle:

16 We now know exactly what tan 1 ( 7/3) looks like: It s the reference angle t in the picture. And, we re being asked for the sine of that angle, which according to the picture is just ( ( )) 1 (b) csc cos 1 9 solution: Following the same procedure as in part (a), we first draw the unit circle along with the line x = 1 9 : Now, we see that the point we need to look it is in the first quadrant, and we see about where we need to draw the reference triangle. Here s the labeled reference triangle:

17 We can now see that we re being asked for the cosecant of the angle, which is one 9 over the sine. This then gives us Without the use of a calculator, sketch all basic six trig functions on the same set of axes (in fact, the sheet of axes we ve been using in class) (For solution, see handouts from class) 11. Using the graphs of sin(t), cos(t), and tan(t), sketch the graphs of each of their inverse functions, indicating the domain and range of each function. (For solution, see handouts from class)

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