美国高中数学2-8 Trig Equations and Applications

美国高中数学2-8 Trig Equations and Applications 8-1 Easier Trigonometric Equations By the nature of the trig functions, solving trig equations will yield an infinite number of solutions. Watch closely to the domain of x as we do each problem. Sometimes we will ask for all possible solutions, called the general solutions of a trig equation. These are generally found by finding particular solutions and adding in a factor for the period length of the particular function. Many times we will restrict the domain of x to the first period of the given function or one time around the unit circle, 2,. Sample Problems 1) Find the values of x between 0 and 2,for which cos x = .4 Solution: To find x, we are looking for the angle measured in rads, one time around the unit circle. Since the cos is positive in quads 1 and 4, we should find two answers in the given domain. -1x = cos .4 Use your calculator to get the answer in the first quadrant. Round to 4 decimal places. x = 1.1593 This is the solution in quadrant 1. The other solution is in quad 4. Remembering our reference angles, the other answer is found by taking 2,,1.1593 = 5.1239 Therefore, the answers are x = 1.1593 and 5.1239 oo2) Solve 4 sin , + 5 = 2 for 0 < , < 360 Solution: Isolate the variable. 4 sin , = -3 sin ,= -3/4 -1,= sin (-3/4) The answers this time are in quadrants III and IV because the sin is negative in those quadrants. Find the reference angle first then apply the correct reference angle formula. -1,= sin(3/4) o,= 48.6 This is rounded to tenths place. Now find the answers in the correct quadrants. oIn three, it's 180 + 48.6 = 228.6 oIn four, it's 360 - 48.6 = 311.4 oTherefore, the answers are ,= 228.6 and o311.4 3) Solve the equation sec x = 4.2 for x between 0 and 2, Solution: Solve the same way as the first example -1x = sec 4.2 x = 1.3304 This is the answer in the first quadrant. Sec is also positive in quadrant IV. Using reference angles, we get 2- 1.3304 = 4.9528 , Therefore the solutions are x = 1.3304 and 4.9528 ( Note: If we wanted all solutions, we would simply add multiples of 2,to each of the above answers. x = 1.3304 + 2nand 4.9528 , + 2n,) 4) Solve the following equation for all solutions. Round answers to the nearest hundredth of a radian. Solve for x: 12 cot x = 35 cot x = 35/12 -1x = cot (35/12) x = .33 Since the cot is positive in Quads I and III, we want the above answer plus this one: 3.14 + .33 = 3.47 Therefore, all answers are: x = .33 + 2n, and x = 3.47 + 2n, Inclination and slope The inclination of a line is the angle ,, where oo0 < ,< 180, that is measured from the positive x-axis to the line. Lines going uphill owill have a slope smaller than 90 and lines going downhill will have slopes greater than o90. Study the two graphs below. For any line with slope m and inclination ,, m = tan , if ,doesn't = 0 oIf ,=90, then the line has undefined slope. Examples 1) To the nearest degree, find the inclination of the line 3x + 4y = 8 Solution: Find the slope of the equation. 3x + 4y = 8 4y = -3x + 8 y = (-3/4)x + 2 m = -3/4 The line is going downhill, so the inclination owill be greater than 90. -1,= Tan (3/4) (reference angle) ,= 37 oSo our inclination is ,= 180 - 37 = 143 2) To the nearest degree, find the inclination of the line -2x + 5y = 15 Solution: Same as above. -2x + 5y = 15 5y = 2x + 15 y = (2/5)x + 3 m = 2/5 The line is going uphill because the slope is positive, so the inclination will be less than o90. -1,= Tan (2/5) o,= 22 oSo our inclination is 22 That's it for this section. On to the next section!! 8-2 Sine and Cosine Curves From previous sections, we worked with horizontal and vertical stretching and shrinking. Recall that vertical stretching/shrinking happens by changing the amplitude of the graph. Horizontal stretching/shrinking happens by changing the period length of the graph. We will apply these concepts to the Sine and Cosine functions. For functions y = A sin Bx and y = A cos Bx (A not = 0 and B > 0) amplitude = | A | and period = 2/B , Demonstration of Graph of y = sin Bx (Manipula Math) Graphing Examples 1) Graph y = 3 sin 2x Solution: Remember that the basic sine curve has zeros at the beginning, middle and end of a cycle. Reaches its maximum at the 1/4 mark and has a minimum value at the 3/4 mark. This graph has an amplitude = 3 and a period length = 2,/2 = , Thus, we have zeros at (0, 0), (,,2, 0) and (,, 0) maximum point at ( ,/4, 3) minimum point at (3,/4, -3) Plot these points and draw a smooth curve to get: 2) Graph y = (1/2)Cos 3x Solution: Remember that the basic cosine graph begins and ends at its maximum point. In the middle, it is at its minimum value, and has zeros at the 1/4 and 3/4 mark. For this graph, the amplitude = 1/2 period = 2,/ 3 The maximums are at the beginning point (0, 1/2) and end point (2,/ 3, 1/2) minimum point at ( ,/3, -1/2) Zeros at ( ,/ 6, 0) and ( ,/ 2, 0) Plot these points and smoothly connect to get: 3) Graph y = -2 Sin (,,,，x Solution: The amplitude = | -2 | = 2 / (/ 2) = 4 Period = 2,, Note that this graph is a reflection about the x-axis. This interchanges the maximum and minimum values. zeros at (0, 0), ( 2, 0), ( 4, 0) minimum ( 1, -2) maximum ( 3, 2) Plot these points and: 4) Graph y = -3 cos (,/ 5)x Solution: This again is a reflection about the x-axis. amplitude = | -3 | = 3 period = 2,/ (,/ 5) = 10 Minimum values at (0, -3), (10, -3) Maximum value at ( 5, 3) Zeros at ( 2.5, 0) and ( 7.5, 0) Connect the points and: Determining the equation from the graph 1) Find the amplitude and period length of each function. Then write an equation. Solution: The graph is a sine curve because it begins at (0, 0). The amplitude = 3 and the period length is (1/2). , Since the amplitude = 3, A = 3. Because period = (1/2),～ B = 2,,(1/2), = 4 Thus, the equation is: y = 3 sin 4x 2) Find the amplitude and period for the function. Write an equation for the graph. Solution: amplitude = 2 and period length = 1 The graph is a cosine graph reflected about the x-axis. The graph starts at a minimum. A = -2 because it is reflected about the x-axis. B = 2,/1 = 2, Thus the equation is: y = -2 cos 2,x Solving an equation algebraically or graphically 1) Solve the equation for 0 < x < 2,. 3 sin 2x = 1 a) Graphically using the TI-82 calculator Set calculator to radian mode. Go to y= screen. Type 3 sin 2x for y and 1 1 for y Press graph. Zoom in. 2 Notice that the graphs cross four times between 0 and 2,. Using your calculator, press the calc button, then the intersect button. Guess first curve, second curve, then best guess. Answer to the point furthest to the left is : (.17, 1) Repeat the process for the other 3 points. Other answers are: (1.40, 1), (3.31, 1) and (4.54, 1). b) Algebraically. 3 sin 2x = 1 sin 2x = 1/3 -1 2x = sin(1/3) The reference angle is .34 (using your calculator) We need to check through 2 periods to find all four answers. sine is also positive in quad II (3.14 - .34) = 2.80 Now add 6.28 to each of these values to get all four answers. 2x = .34, 2,80, 6.62, 9.08 x = .17, 1.40, 3.31, 4.54 In the next section, we will add translations to the graph of sine and cosine!! 8-3 Translations of Sine and Cosine Curves We already know how to translate a graph from our study of functions. A translation is in the form y - k = f(x - h), where it is translated k units vertically and h units horizontally. This fits right into our study of the Sine and Cosine curves. Sine and Cosine Curves If the graphs of y = A Sin Bx and y = A Cos Bx are translated h units horizontally and k units vertically, then the resulting graphs have the equations: y - k = A Sin B(x - h) and y - k = A Cos B(x -h) Demonstration of Graph of y = A sin B(x - C) (Manipula Math) Graphing Examples 1) Graph the function y - 3 = 2 sin,(x - 1) Solution: This graph has the same amplitude and period length without the translation. A = 2 and period length = 2,/ ,= 2 Graph this one. Points on the graph are: zeros (0, 0), (1, 0), (2, 0) maximum at (.5, 2) and minimum at (1.5, -2) Now translate the graph by moving the five points above, three units up and one unit to the right!! This makes the following five points: (1, 3), (2, 3), (3, 3), (1.5, 5), and (2.5, 5) Both graphs are shown on the next grid. The green one is the final graph of the function. 2) Graph the function: y + 2 = 4 cos ,/4(x + 1) Solution: This is the graph of y = cos (/4)x translated 2 units down and one , unit to the left. The amplitude is 4 and the period length is 2,/(,/4) = 8. Here is the graph of the function in blue. Maximum points at: (0, 4), (8, 4) Minimum point at (4, -4) Zeros at: (2, 0) and (6, 0) Now translate the graph by moving each point 2 units down and 1 unit left! The points are translated to: (-1, 2), (7, 2), (3, -6), (1, -2) and (5, -2). These points are marked on the green graph with black dots! The green graph is the final solution! 3) Graph y = 3 cos 4(x - , ) Solution: This has A = 3 with period length 2,/ 4 = ,/ 2. It is shifted ,units to the right. Graph it first without the translation. maximum points at: (0, 3), (,/2, 3) Minimum point at ( ,/4, -3) zeros at: (,/8, 0) and (3,/8, 0) Now shift all the points pi units to the right! Maximum points at: (,～3), (3,/2, 3) minimum point at (5/4, -3) , zeros at: (9/8, 0), (11/8, 0) ,, This is actually the same graph as the one before we translated it. Why? 4) Give an equation for the following graph: This graph has many different solutions including ones for both the sine and cosine functions. In either case, the amplitude and period length are the same. Look at the graph and find the maximum and minimum points. They happen at 1 and -3. Remember the formula to find the amplitude of a graph? That's right, its maximum - minimum divided by two. A = [1 - (-3)]/2 = 2 The period length is 6. Look at the graph and count between maximum points or minimum points. Now solve for B in the equation Period Length = 2,/B 6 = 2,/B 6B = 2, , , ,/3 Our equation has the form y = 2 sin (,/3)x or y = 2 cos (,/3)x before the translation. For the sine curve, the graph starts at a "zero". When you translate a graph, the "zero" becomes the line midway between the max and min values. Look at the graph: The black dot represents a "zero" of the graph before the translation. Now determine where the new point is. It was moved 1 unit right and 1 unit down. You now have h and k!! Put them in the formula and voila!! y + 1 = 2 sin [(,/3)(x - 1)] You can do a similar translation for the cosine curve, but remember, the cosine starts a cycle at a maximum point!! 5) Find an equation for the following graph: Let's find a cosine function for this graph. A = [5-(-3)]/2= 8/2 = 4 counting from the Period length is 4, maximum point to the next maximum point. 4,= 2,/B 4,B = 2, B = 1/2 Before the translation, the equation is y = 4 Cos [(1/2)x] Now determine the translation. Remember, the maximum point is the starting point for the cosine function. Study this graph: Look at the red lines. The maximum point is moved units to the right and one unit up , from the origin. The purple graph is before the translation/ The equation becomes: y - 1 = 4 cos [(1/2)(x - ,)] Hopefully, this will give you a small glimpse into graphing with translations!! 8-4 Relationships between the Functions All of the following relationships need to be memorized. These are the most commonly used functions that appear in many upper level calculus courses. Please make every attempt to really know them!! Reciprocal Relationships csc , = sec , = cot , = 1/sin , 1/cos , 1/tan , Negative Relationships sin(-,) = -sin , cos(-,) = cos , csc (-,) = -csc , sec(- ,) = sec , tan(-, ) = -tan , cot(- ,) = -cot , Pythagorean Relationships 2222sin , + cos 1 + tan , = 1 + cot , = 22, = 1 sec , csc , Cofunction Relationships oosin , = cos(90 - cos , = sin(90 - , ) ,) ootan , = cot(90 - cot , = tan(90 - ,) ,) oosec , = csc(90 - csc , = sec(90 - ,) ,) Any relationship that is true for all values of the variable for which each side is defined is called an identity. Each of the relationships above represent a trigonometric identity. We can use trig identities to simplify trig expressions, prove other identities and solve more complex trig equations. Simplifying trig expressions 1) simplify: sin y cot y = sin y (cos y/sin y) (Basic identity) = cos y (Basic algebra) 222) simplify: csc y(1 - cos y) 22= csc y (sin y) (pythagorean) 22= csc y(1/csc y) (reciprocal) = 1 (basic algebra) 3) simplify: tan x(cot x + tan x) 2= 1 + tan x (distribute) (tan and cot are reciprocals) 2= sec x (pythagorean) 4) simplify: cos x cot x + sin x 5) simplify: 6) simplify: Proving Identities You can use trig identities to prove other statements are identities. You may simplify either the left side of the equation or the right side of the equal sign. Your goal is to make it look like the other side of the equation. You may even work on both sides. It is probably a good idea to simplify the more complicated side. Also, remember your basic algebra skills, like getting a common denominator and factoring. Keep in mind where you are headed. If cosine is on the right side, then when working on the left side, try to write it in terms of cosine. Use the identities. They will help! If all else fails, try writing everything in terms of the sine and cosine. Good luck!! 21) Prove: tan x + cot x = tan x csc x Solution: 2) Prove: Solution: 3) Prove: Solution: 4) Prove: Solution: 5) Prove: Solution: That should give you some techniques in working with identities. Now, let's head onto more difficult trig equations. 8-5 Solving More Difficult Trig Equations Try the quiz at the bottom of the page! go to quiz Most of these type of problems can be solved the same way you solve basic algebraic equations. Also, we can use many of the identities from the previous sections. Remember, we can solve these for a few answers or an infinite number of answers (in rads or degrees)! So watch the directions closely. 1) Solve each of the following for 0 < x < 2, 2a) 4cos x - 1 = 0 Isolate for the square term 24cos x = 1 divide by 4 2cos x = 1/4 take the square root cos x = + 1/2 since we have both positive and negative answers, we will have an answer in each of the four quadrants. Find the reference angle in rads and use the reference angle formulas. Set calculator to rads. -1x = cos (1/2) x = 1.05 This is the reference angle. In quad II, 3.14 - 1.05 = 2.09 In quad III, 3.14 + 1.05 = 4.19 In quad IV, 6.28 - 1.05 = 5.23 Thus, x = 1.05, 2.09, 4.19, 5.23 2b) 2sin x - sin x = 0 factor sin x(2sin x - 1) = 0 write the two solutions sin x = 0 or 2sin x- 1 = 0 sin x = 0 or 2sin x = 1 sin x = 0 or sin x =1/2 -1sin x = 0 at 0 and ,～ x = sin (1/2) --> ,/6 and 5,/6 Thus, the four answers are: x = 0, ,～,,,～ ,,,, 2c) 3cos x + 2cos x - 1 = 0 factor!! (3cos x -1)(cos x + 1) = 0 cos x = 1/3 or cos x = -1 -1-1x = cos (1/3) or x = cos (-1) x = 1.23 or x = , The first solution had another answer in quad IV. (Why?) x = 6.28 - 1.23 = 5.05 Thus, x = 1.23, 5.05, , 2d) 2cos x - sin x - 1 = 0 This is much easier to solve if the function is written in terms of the same trig function. Using the pythagorean relationship, we can write: 22(1 - sin x) - sin x - 1 = 0 distribute 22 - 2sin x - sin x - 1 = 0 Move to the other side to make the quadratic term positive 20 = 2sin x + sin x -1 factor 0 = (2sin x - 1)(sin x + 1) sin x = 1/2 or sin x = -1 -1-1x = sin (1/2) or x = sin (-1) x = ,/6 and 5,/6, or x = 3,/2 (These are values that should have been memorized in previous sections!) Thus, the answers are: x = ,/6 and 5,/6, or x = 3,/2 o2) Solve each of the following for 0 < ,< o360 Make sure your calculator is set to degree mode!! 2a) tan - 5tan + 6 = 0 ,, factor! (tan - 3)(tan -2) = 0 ,, tan , = 3 or tan , = 2 -1-1, = tan 3 or ,= tan 2 oo,= 71.6 or 63.6 These are the reference angles. We also get answers in quad III where the tangent is also positive. o,= 180 + 71.6 = 251.6 o,= 180 + 63.6 = 243.6 oooThus, the answers are:71.6, 63.6, 251.6, o243.6 22b) tan , - 2sec , + 4 = 0 Use pythagorean relationship to get equation in the same trig function. 22(sec , - 1) - 2sec , + 4 = 0 2-sec , + 3 = 0 2-sec = -3 , 2 sec = 3__ , sec , = + \/ 3 o , = 54.7 This is the reference angle. Since it is both positive and negative, we have an answer in all four quadrants. oQuad II: 180 - 54.7 = 125.3 oQuad III: 180 + 54.7 = 234.7 oQuad IV: 360 - 54.7 = 305.3 ooThus, the answers are: 54.7 , 125.3 , oo234.7 , 305.3 c) d) 4sin , + 3cos , = 0 This one would be difficult if we tried to use the pythagorean relationship. Since it deals with sine and cosine rather than their squares, it would introduce radicals in the equation. Not a good choice. It might be smart to try another relationship. How about a third trig function. Sine and cosine are related to Tangent!! If we divide the above equation by cosine, we can change the problem to tangent and have only one trig function!! Well, that does it. Are you ready for the sample test or do you need to review? Pick the appropriate button!! Current quizaroo # 8 1) Solve for 0 < x < 2,. cot x = -2.5 Give answers to the nearest hundredth of a radian. a) .38 and 3.52 b) 2.76 and 5.90 c) 1.19 and 4.33 d) 1.95 and 5.09 e) 1.19 and 1.95 2) Give the amplitude and period for y = -5cos ,x a) amplitude = -5 and period = 2 b) amplitude = 2 and period = 2,/5 c) amplitude = 5 and period = 2 d) amplitude = 5 and period = 1 e) amplitude = -5 and period = , 3) Find the maximum point in the first cycle for the equation: y - 3 = 4sin (,x/2) a) (1, 7) b) (0, 3) c) (2, 8) d) (3, 10) e) (2, 7) 24) Simplify tan x( 1 + cotx)/cot x 2a) sinx 2b) cosx 2c) secx 2d) cscx 2e) tanx 5) Solve for 0 < x < 2,. (sin x)(tan x) = sin x Give answers to the nearest hundredth of a radian. a) ,/4, 5,/4 b) 0, , c) 0, ,～ 2, d) 0, 2, e) 0, ,/4, ,～ 5,/4 click here for answers!! Sample Test 1) Solve each equation either for 0 < , < 360 (round to nearest tenth of degree) or 0 < x < 6.28 radians (round to hundredths of a radian) a) Cos , = -.48 b) Sec , = 1.5 c) Tan x = -1.35 d) Csc x = 2.4 2) Simplify each. 22oa) (sec x - 1)(sec x + 1) b) tan x - sec x c) sin A tan A + sin(90 - A) = = = 3) Let y = -5 sin 3x Give each. a) amplitude = b) period length = c) graph one period and label the appropriate points. 4) Solve each equation for 0 < x < 6.28 (round to nearest hundredth of a radian) a) 3 sin x = 1 b) 3 sin 2x = 1 5) graph one cycle of the equation y = -4 + 4 cos 2( x - ,) Remember to label the appropriate points. 6) simplify each .22 a) sin x sec x = b) (1 - sin x)(1 + tan x) = 7) simplify cos x 1 + sin x + = 1 + sin x cos x 8) solve the equation for 0 < x < 6.28 round to hundredths of a radian. 2 6 cosx + 5 cos x + 1 = 0 9) Solve the equation for 0 < x < 6.28. round to the nearest hundredth of a radian. 5 sin x - cos x = 0 o10) Solve the equation for 0 < , < 360. Round to the nearest hundredth of a radian. 2 2 cos, + 3 sin , - 3 = 0 11) Find an equation for each of the following graphs: This page hosted by Get your own Free Home Page Hope you had great success with this test!! Get ready for some trig of the right triangle!! Answer Page!! -11) a) , , cos .48 Using your calculator, the reference oangle is 61.3 The cosine is negative in the II and III quadrants. Thus, the answers are: o180 - 61.3 = 118.7 and o180 + 61.3 = 241.3 -1-1-1o b) , = sec1.5 = cos1.5 = 48.2 Since the secant is positive in quadrants I and IV the answers are: o48.2 and o360 - 48.2 = 311.8 c) Since tangent is negative, the answers are in quadrants II and IV. Use your calculator to find the reference angle. It is .93 In quadrant II, 3.14 - .93 = 2.21 In quadrant IV, 6.28 - .93 = 5.35 d) The csc is positive in quadrants I and II. Find x using your calculator by typing -1-1sin2.4 = .43 The answers are .43 in the first quadrant and 3.14 - .43 = 2.71 in the second! 222) a) Foil the product to get: secx - 1 = tanx by the pythagorean relationship!! b) Use the pythagorean relationship and the answer is -1 c) (sin A)(sin A/ cos A) + cos A Use the cofunction and basic trig relationship get a common denominator 22(sinA + cosA)/ cos A Use a pythagorean relationship 1/cos A Use a reciprocal relationship sec A 3) amplitude = 5 and period length = 2,/3 zeros at (0, 0), (,/3, 0), (2, /3, 0) Maximum value at ( , /2, 5) Minimum value at (, /6, -5) 4) a) sin x = 1/3 Answers are in quadrant I and II. reference angle is .34 Answers: .34 and 3.14 - .34 = 2.80 b) sin 2x = 1/3 means 2x = .34, 2.80, 6.62, 9.08 because 0 < 2x < 12.56 Thus x = .17, 1.40, 3.31, 4.54 5) amplitude = 4 Period length = 2, /2 = , Translation: 4 down and , right Translated points: zeros translated to (5, /4, -4), (7, /4, -4) Max translated to (, , 0), (2, , 0) Min translated to (3, /2, -8) 6) a) (sin x)(sec x) = (sin x)(1/cos x) = sin x/cos x = tan x 2222 b) (1 - sin x)(1 + tan x) = (cos x)(sec x) pythagorean relationships! = 1 reciprocal relationship! 7) 8) (3cos x + 1)(2cos x + 1) = 0 factor cos x = -1/3 or cos x = -1/2 Reference angles are 1.23 and 1.05 Answers are in quadrant II and III 1.91, 4.37, 2.09, 4.19 9) 10) 11) a) This is not a translation. It is the normal cycle for the cosine graph. Amplitude = 3 with A = 3 because the maximum is at (0, 3) Period length = 4 which makes 4 = 2, /B, 4B = 2, , B = ,/2 Thus the equation is: y = 3cos (,x/2) b) This is a translation. One way to do the problem, certainly not the only way, is to think of it as a sine curved shifted 2 units up and 1 unit to the right. Amplitude = (4 - 0)/2 = 2 with A = 2 Period length = 8 (count from maximum to the next maximum point) To find B, 8 = 2, /B, 8B = 2, , B = ,/4 Putting it all together, gives us the equation y - 2 = 2sin ( ,/4)(x - 3) Hopefully you did well on this test! If not, go back and hit the books a little harder This page hosted by Get your own Free Home Page