#### Transcript Ex 1

Objectives: • Convert between degree and radian measure • Pythagorean identities • Find the trigonometric values of any angle measure • Unit circle • Graph trigonometric functions • Solve for lengths and angle measures of any type of triangle. • Verify identities • Application of trigonometry Uses of trigonometry: Used to describe the motion of any object that behaves in a circular, oscillating or periodic manner. Angle: Consists of two rays or half lines that originate at a common point called the vertex. These two rays have names: • Terminal side: The ray that rotates to generate the angle. • Initial side: The ray that does not move. Angles are commonly denoted using lower case Greek letters. , , , alpha, beta, theta, gamma (respectively) To better describe the formation of angles we superimpose an xy plane on the angle with the vertex at the origin. y Angles are generated by the terminal side rotating counterclockwise or clockwise. B A 0 x If angles are generated by the terminal side moving counterclockwise the angle is positive. If angles are generated by the terminal side moving clockwise the angle is negative. The direction of the arrow inside the angle will tell you if the terminal side is generating a positive or negative angle. y B If angles are generated by the terminal side moving clockwise the angle is negative. A 0 x If we have an angle of 390°, this is one revolution of the unit circle plus 30°. So, we say that 390° is coterminal with 30° Coterminal angles differ by multiples of 360° Ex 1: Determine if the following angles are coterminal. a.) 90° and 1170 ° YES b.) 123 ° and 844 ° NO Solution: 1170 90 3 360 If you get a whole number they are coterminal. If the result is not a whole number then they are not coterminal. 844 123 2.002777778 360 To define the measure of an angle, we first add the unit circle centered at the origin to the coordinate system. y This is called a unit circle because the length of the radius is 1. 1 -1 1 -1 x The equation of the unit circle is: 2 2 x y 1 The domain is: [-1, 1] To generate angles we must consider the terminal and initial sides. The initial side is aligned with the x-axis and the terminal side starts at the x-axis and rotates to generate the angle. y This point is P(t). P is the function and t is the input value. x t For every P(t) on the unit circle we can define its measure by using degrees or radians. Radian Measure: For any real number t, the angle generated by rotating counterclockwise from the positive x-axis to the point P(t) on the unit circle is said to have radian measure t. t 3 2 2 y (0,1) …and so on… (-1,0) (0,-1) 1 (1,0) -1 -2 -3 x We can see that an angle that measures 90° is the same measure as 2 We can also see that 180° is We will use this fact to convert between degrees and radian measure. To convert degree measure to radian measure you multiply the degree measure by: 180 Ex: Convert the following degree measure to radian measure. a.) 150° b.) 225° 150 5 a.) 150 6 180 180 c.) -72° 225 45 5 b.) 225 36 3 180 180 72 2 c.) 72 180 180 5 To convert radian measure to degree measure you multiply the radian measure by: 180 Ex 2: Convert the following radian measures to degree measure. 7 a.) 6 12 b.) 5 a.) 7 180 1260 210 6 6 b.) 12 180 2160 432 5 5 11 c.) 6 11 180 1980 c.) 330 6 6 3.2 The Sine and Cosine Functions Recall: The terminal side rotates to generate the angles. There are infinitely many points on the unit circle that the terminal side “could” generate. We will only “memorize” a few of them. We will memorize all angles on the unit circle that are in increments of 30° and 45°. y Unit Circle 90° 120° 60° 45° 135° 150° 30° 180° 0° 330° 210° 315° 225° 240° 270° 300° x An ordered pair has the form: (x, y) The Sine and Cosine Functions: Suppose that the coordinates of the point P(t) on the unit circle are (x(t), y(t)). Then the sine of t, written sin t, and the cosine of t, written cos t, are defined by sint t = y(t) and cos t = x(t) Our new ordered pairs are of the form: P(t) = (cos t, sin t) Finding the cosine and sine values y of the common angles on the unit circle. 120° 90° 60° 45° 135° 150° 30° 180° 0° x 330° 210° 315° 225° 240° 270° 300° We create right triangles by drawing lines perpendicular to the x or y axis. It does not matter which axis. To find the cosine and sine of 30° we must use the triangle we created. 1 60° 30° 1st: We know the third angle is 60° by the triangle sum theorem. 2nd: We know the length of the hypotenuse since this is a unit circle: r = 1 3rd: Using the properties of a 30-60-90 right triangle we can find the other two sides. Since the base of this triangle is on the x-axis this side would represent the cos t. The height would represent the sin t. Recall: * The side opposite the 30° angle is half of the hypotenuse. * The side opposite the 60° angle is the product of the short leg and the square root of 3. If the hypotenuse has length 1, then the side opposite 30° is ½ . The side opposite the 60° angle is So, P(t) = 3 1 P , 6 2 2 1 3 3 2 2 This is how you will solve for all sine and cosine on the unit circle dealing with 30 – 60 – 90 right triangles. Finding the coordinates of a point using a 45 – 45 – 90 right triangle. We know: 45° • The third angle is 45° • The hypotenuse has length 1. 1 • In a 45 – 45 – 90 right triangle the legs are the same length. We can call both legs the same variable since they are equal. 45° a2 a2 1 Now solve for a. 2a 2 1 1 a 2 1 a 2 1 a 2 1 a 2 2 1 2 2 2 2 2 4 2 We rationalized the denominator in this step because we do not leave radicals in the denominator. Since the legs have the same length, the cosine and sine values are the same. 2 2 Pt P , 4 2 2 Pythagorean Identity: For all real numbers t, (sin t)2 + (cos t)2 = 1 Because of the Pythagorean Identity, sine and cosine have bounds: For all real numbers t, 1 sin t 1 The cosine function is even For all real numbers t, cos(-t) = cos t and 1 cos t 1 The sine function is odd For all real numbers t, sin(-t) = - sin t A Periodic Function is a function that repeats the same thing over and over again. Trigonometric Functions are periodic because they repeat. Sine and Cosine functions have a period of 2 From your unit circle you can see that the ordered pairs on your unit circle do not begin to repeat until after one complete revolution of the unit circle 360° sin(t 2 ) sin t cos(t 2 ) cos t and Reference Number For any real number t, the reference number r associated with t is the shortest distance along the unit circle from t to the x-axis. The reference number r is always in the interval Ex 2: What is the reference number of 0, 2 11 6 Since the answer is always in the above interval, it is 6 Ex 3: Determine the values of t in [0,2 ] that satisfy 2(sin t ) 2 3 cos t 0 This is simply a quadratic function. Use what you know. You can view this quadratic as… 2 y 2 3x 0 We cannot work with it this way; we can only have one unknown. We need to turn both variables into y (sine) or x (cosine). Pythagorean Identity: (sin t)2 + (cos t)2 = 1 Manipulate this identity so that it is something that you can use. (sin t ) 2 (cos t ) 2 1 (sin t ) 2 1 (cos t ) 2 We can replace (sin t)2 with (1 – (cos t)2) 2(1 (cos t ) 2 ) 3 cos t 0 Now distribute 2 2(cos t ) 2 3 cos t 0 Now factor and set each factor equal to zero. 2(cos t ) 2 3 cos t 2 0 To make things easy, and we don’t like our leading coefficient being negative, multiply both sides by -1. 2(cos t ) 2 3 cos t 2 0 Now factor 2 x 2 3x 2 0 It may help you to view it like this to factor. (2x + 1)(x – 2) = 0 Set each factor equal to zero and solve. 2x + 1 = 0 x 1 2 and and x–2=0 x2 1 cos t and cos t 2 2 1 2 4 cos t at : t and t 2 3 3 Replace x with cos t since x represents cos t. Where on the unit circle is this true? cos t = 2 will never happen; it is outside the bounds of cosine. Lesson 3.3: Graphs of the Sine and Cosine Functions Sine Curve We will use the values that we memorized from the unit circle to graph one period of the sine function. Sin t 1 4 4 -1 2 3 4 5 4 3 2 7 4 2 The more points you plot the more precise your graph will be. I will plot only the 45° increments of the angles. t Cosine Curve cos t 1 4 4 -1 2 3 4 5 4 3 2 t 7 4 2 Ex 1: Use the graphs of y = sin x and y = cos x to sketch the graphs of y sin x 2 y sin x 2 and Trigonometric functions are no different than any other function when “shifts” are involved… minus to the right …plus to the left. Sin t 1 graph of y = sin x 4 4 -1 2 3 4 5 4 3 2 7 4 2 Again, we must be sure the leading coefficient is 1 before we try and see the shifts. t Note: Your shift will help guide you as to what you should count by on the x-axis. sin x 1 -1 2 y sin x 2 3 2 2 x 5 2 Q: What is the last value going to be that we will write on the x-axis? 5 A: 2 NOW YOU TRY THE SECOND ONE!!! y sin x 2 sin x 1 2 2 -1 You try… x y cos x and 2 3 2 2 y cos x 2 5 2 We can see from the previous 4 graphs that when we shift one graph to the right or left that we obtain the graph of a different equation, such as: The graph of: sin x cos x 2 cos x sin x 2 sin x cos x 2 cos x sin x 2 These are all true only because we are shifting by 2 When you shift by other increments different equalities occur. When shifting by increments of pi… y sin( x ) sin( x ) y cos( x ) cos( x ) Ex 1: Sketch the graph of f ( x) 2 cos x 2 All trig functions are of the form: y = A sin(Bx + C) The name of the trig function changes only. |A| is the amplitude Period is Shift is 2 B C B Don’t forget!!! The leading coefficient must be 1 before you can see the shift!!! f ( x) 2 cos x 2 Amplitude is: 2; so the height will go to +2 and -2 on y-axis. B = 1, so the period has not changed. Shift: right 2 units First, I will graph y = 2cosx, then shift this graph to the right 2 units. Since the period did not change and the shift is units, we will count by on the x-axis. 2 4 cos x y = 2cos x Notice that the zeros of this function do 2 not change. Amplitude is just a vertical elongation or compression of the graph. x -2 2 3 2 2 Now, we shift this graph! cos x 2 -2 2 Ex 2: Sketch the graph of f ( x) 2 cos x 2 3 2 2 x 5 2 f ( x) 2 cos 3x 2 B = 3, not 1 so we must find the length of the new period. Recall: period is found by: 2 B 2 3 Always reduce if necessary. Instead of this graph being graphed from 0 to 2π, the entire cosine curve will be graphed between 0 and 2 3 Recall: To see a horizontal shift the leading coefficient must be 1 and it is not. We must factor out the 3 from the quantity. f ( x) 2 cos 3 x 2 f ( x) 2 cos 3 x 6 We will shift this graph to the right units. 6 The main points of the cosine curve (y = cos x) are: cos x 2 -2 (0,1) x 2 5 6 3 2 3 6 7 4 3 5 11 2 6 3 2 3 6 2 ,2 3 New points: (0,2), 6 ,0 , 3 ,2 , 2 ,0 , with shift ,0 2 ( ,1) 3 ,0 2 ( 2 ,1) Divide all of these orginal x-coordinates by 3 and multiply the y-coordinates by 2, these will be the locations of your new points. 2 5 ,2 , ,0 , ,2 , ,0 , ,2 6 3 2 3 6 Ex 3: Sketch the graph of y 3 cos(2 x ) cos x 2 x 2 4 -2 4 2 3 4 3.4 Other Trigonometric Functions The Tangent, Cotangent, Secant, and Cosecant Functions: The tangent, cotangent, secant and cosecant functions, written respectively as tan x, cot x, sec x, and csc x are defined by the quotients tan x sin x cos x 1 sec x cos x cot x cos x sin x 1 csc x sin x Note: Tangent and secant are only defined when cos x ≠ 0 cotangent is cosecant are only defined when sin x ≠ 0 Trig. function Sin x Cos x Tan x Csc x Sec x Cot x Sign in quadrant… I II III IV + + - - + - - + + - + - + + - - + - - + + - + - Ex: Suppose sin x 23 and that 2 x . Determine the values of the other trigonometric functions. Solution: Since we know that csc x is the reciprocal of sin x, write the reciprocal of sin x. csc x 1 3 sin x 2 Next we must find cos x because the remaining trig. functions contain it. What do you know that involves both sine and cosine that will help you find cos x? Pythagorean Identity. (sin x) 2 (cos x) 21 2 2 2 (cos x) 1 3 4 2 (cos x) 1 9 4 (cos x) 2 1 9 9 4 (cos x) 2 9 9 5 (cos x) 2 9 5 cos x 9 5 cos x 3 The interval for sin x is given. Quadrant II Cosine is negative in QII, so… cos x 5 3 2 sin x 2 3 6 2 5 2 5 tan x 3 5 cos x 3 3 5 3 5 5 5 5 cot x = 5 2 or you could have started at the beginning to find this solution. sec x is the reciprocal of cos x. Write it and simplify: sec x 3 3 5 3 5 5 5 5 5 Note: You are now ready to write your answers. Make sure they have the correct sign for quadrant II. Given : sin x 2 3 5 3 2 5 tan x 5 3 csc x 2 3 5 sec x 5 5 cot x 2 cos x Ex 2: Given that cos t 3 3 and that t 2 2 Find the values of the other trigonometric functions. Solution: We can see that we are in QIII. The given is the cosine value for 7 6 7 1 sin x sin 6 2 1 1 2 1 3 3 tan x 2 3 3 3 3 3 2 2 csc t 2 sec x 2 2 3 3 3 cot x 3 3 The graph of the tangent function. The tangent function is zero when the sine function is zero because sine is in the numerator of the tangent function. The tangent function is undefined when the cosine function is zero because the cosine function is in the denominator of the tangent function. The tangent function will be zero at: The tangent function is undefined at: tan t 1 0, , 2 3 2 2 3 6 -1 2 , 2 5 6 3 2 3 6 t y = tan x Sine and cosine have periods of 2 , therefore, tangent will also repeat itself on that same period. Ex 3: Sketch the following graphs: a. y tan 2 x b. y tan 2 x 3 c. y tan 2 x 1 3 Solution: a.) B = 2; set up your inequality: 0 2 x 2 0 x The length of your new period is . Since we divided the period by 2 we will also have to divide the restrictions by 2: Original restrictions: 2 and 3 2 New Restrictions: 4 and 3 4 a.) y = tan 2x b.) y tan 2 x 3 c.) y tan 2 x 3 This graph is the graph from part b reflected over the x-axis. c. contd.) y tan 2 x 1 3 Now we shift it up one unit. Note: The zeros from the graph are not obvious. To find them we would set the function equal to zero. 0 tan 2 x 1 3 Which implies… 1 tan 2 x 3 If we let x = 2 x 3 We have… tan x = 1; where does this occur, when x = ? We will find this later this chapter. Now that we have the graphs of sine, cosine and tangent we can graph the remaining trig. Functions using the reciprocal technique. Do you remember those properties of graphing reciprocals? … as f(x) increases, its reciprocal… decreases! The Cosecant function: Graph: y = sin x. y= csc x is undefined at: Where sin x = 0. There, we have vertical asymptotes. Now, use the fact that as y = sin x increases, y = csc x decreases and vice versa. Ex: Graph y = sec x Ex: Graph y = cot x Be careful with y = cot x. The restrictions for y = tan x are different from y = cot x. They have different vertical asymptotes. 1 Ex: Sketch the graph of y csc 3x 2 2 Note: There are several strategies for graphing: 1. You could start with the parent function and build on it one change at a time. 2. You can find the most important pieces of information of this new graph and make the changes to the pieces, and then plot your new points and new vertical asymptotes. This time, lets make the changes to the parent function: y = csc x y = csc x has vertical asymptotes at: 0, , 2 2 4 5 , , , y = csc 3x has vertical asymptotes at: 0, , 3 3 3 3 Now, the amplitude changes to ½ Now, we shift the graph to the left 6 units. The vertical asymptotes go with it!!! They get shifted too! 3.5 Trigonometric Identities Using the first Pythagorean Identity we can manipulate it and obtain two more. Pythagorean Identities: (tan x) 2 1 (sec x) 2 (cot x) 2 1 (csc x) 2 Sum and Difference Formulas for Sine and Cosine For every pair of real numbers x1 and x 2 , we have sin( x1 x2 ) sin x1 cos x2 cos x1 sin x2 and cos(x1 x2 ) cos x1 cos x2 sin x1 sin x2 Ex 1: Determine the following: a.) sin 12 7 b.) cos 12 Using the values from the unit circle that we have committed to memory we will find a combination of two that equal the function we are trying to evaluate. ? 12 What equals We now label these two 12 3 4 x1 and x2 Using the difference formula (we subtracted above and for sine you do the operation of the combination that you used to equal your function), we will evaluate sin 12 sin sin cos cos sin 3 4 3 4 3 4 sin sin cos cos sin 3 4 3 4 3 4 3 2 1 2 sin 3 4 2 2 2 2 2 3 2 sin 3 4 4 4 Both terms have a common factor: 2 4 Factor it out! 2 3 1 sin 3 4 4 You try part b. Hint: For cosine you use the opposite formula compared to the sign you used to obtain your function. 7 If you subtracted to get cos then you will use the sum formula, and 12 visa versa. 7 cos cos b.) 12 3 4 cos cos cos sin sin 3 4 3 4 3 4 3 2 1 2 cos 2 2 3 4 2 2 2 1 3 cos 3 4 4 We can also find the tangent of functions on the unit circle… tan( x1 x2 ) sin( x1 x2 ) cos( x1 x2 ) sin x1 cos x2 cos x1 sin x2 cos x1 cos x2 sin x1 sin x2 We can find tangent of t by using this formula or we can continue to manipulate the formula to obtain a smaller one. sin x1 cos x2 cos x1 sin x2 tan( x1 x2 ) cos x1 cos x2 sin x1 sin x2 We maintain equality as long as we do the same thing to both the numerator and the denominator. Lets divide the numerator and denominator by cos x1 cos x2 tan( x1 x2 ) tan( x1 x2 ) sin x1 cos x2 cos x1 sin x2 cos x1 cos x2 cos x1 cos x2 cos x1 cos x2 sin x1 sin x2 cos x1 cos x2 cos x1 cos x2 tan x1 tan x2 1 tan x1 tan x2 Now simplify Ex 2: Determine tan 12 12 3 4 tan tan 3 4 tan 3 4 1 tan tan 3 4 To use this formula it is a must that you can easily find the tangent values of the measures you have committed to memory. 3 1 tan 3 4 1 3 1 3 1 tan 3 1 3 4 YOU pick which is easier for you!!! Double angle formulas for sine and cosine sin 2x = sin(x + x) = sin x cos x + sin x cos x = 2sin x cos x cos 2 x (cos x) 2 (sin x) 2 1 (sin x) 2 (sin x) 2 1 2(sin x) 2 Ex 3: Determine all the values of x in [0, 2π] that satisfy each equation. a.) cos x = sin 2x Solution: b.) sin x = cos 2x c.) 1 = sin x + cos x a.) We need to get everything into sin x and cos x because that is what we know on the unit circle. cos x = sin 2x = 2sin x cos x cos x = 2sin x cos x 0 = 2sin x cos x – cos x = cos x(2sin x – 1) 0 = cos x(2sin x – 1) cos x = 0 2sin x – 1 = 0 and sin x = ½ cos x 0 when x sin x 2 3 2 and 1 5 when x and 2 6 6 b.) sin x = cos 2x sin x 1 2(sin x) 2 2(sin x) 2 sin x 1 0 (2 sin x 1)(sin x 1) 0 sin x 1 2 when x 6 or 5 6 sin x 1 when x 3 2 c.) 1 = sin x + cos x Square both sides. 1 (sin x cos x) 2 1 (sin x) 2 2 sin x cos x (cos x) 2 Re-order expression 1 (sin x) 2 (cos x) 2 2 sin x cos x 1 1 2 sin xcos x 0 2 sin xcos x 0 sin 2x In order for sine to ever be equal to 0, 2x must be a multiple of pi. x could be any of the following to make this statement true: x 0, x 2 , x , x 3 , x 2 2 CHECK!!! x 0, x 2 , x , x 3 , x 2 2 NOTE: We squared the equation earlier, so this could introduce extraneous solutions. This must have happened because the original expression, 1 = sin x + cos x is only satisfied if the following is true: x 0, x 2 When x = , x 2 3 2 the statement does not equal 1. Half Angle Formulas Half angle formulas come from double angle formulas involving “2x.” Ex: Double Angle formula half angle formula x cos 2 cos 2 x Instead of doubling the angle it is cut in half. Half Angle Formula: For any real number x we have x 1 cos x x 1 cos x sin and cos 2 2 2 2 2 2 If we replace x with 2x we have sin x 1 cos 2 x and (cos x) 1 cos 2 x 2 2 7 7 7 , cos , and tan . 8 8 8 Ex 4: Determine sin Solution: 7 1 cos 2 7 8 sin 2 8 7 sin 8 7 1 cos 2 7 8 sin 2 8 7 1 cos 7 4 sin 2 8 7 sin 8 1 2 2 2 Substitution for x Reduce fraction Now take cosine Now use algebra and simplify expression 7 sin 8 1 2 2 Now we must determine the sign. 2 2 2 7 sin 2 2 2 8 2 2 7 2 sin 2 8 2 2 1 7 sin 2 2 8 2 2 7 sin 4 8 2 2 7 sin 2 8 This function is in quadrant IIsine is positive here. 2 2 7 sin 2 8 You try the next one!!! 7 cos 8 7 2 1 4 2 2 2 1 cos Since 7 8 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 is in the second quadrant, cosine is negative. 2 2 7 cos 2 8 7 sin 7 8 tan 8 cos 7 8 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Verifying Identities Using all of the trigonometry we’ve learned so far we can use this information to prove identities. (make one side of the equation look like the other) Ex 5: verify the identity (1 (cos x) 2 )(sec x) 2 (tan x) 2 It does not matter which side you start with to make it look like the other side. Many times we only have to manipulate one side to look like the other. Sometimes we work both sides together and as we go along they will be the same. Solution: I will start with the left side. I notice that I can use the Pythagorean Theorem. (sin x) 2 (sec x) 2 (tan x) 2 1 2 (sin x) (tan x ) (cos x) 2 2 (tan x) 2 (tan x) 2 Remember, your goal is to turn it into tangent. (sin x ) 2 2 (tan x ) (cos x) 2 Now you try one!!! Ex 6: Verify: cot x + tan x = sec x csc x cos x sin x sec x csc x sin x cos x (cos x) 2 (sin x) 2 sec x csc x sin x cos x (cos x) 2 (sin x) 2 sec x csc x sin x cos x 1 sec x csc x sin x cos x 1 1 sec x csc x sin x cos x sec x csc x sec x csc x 3.6 Right Triangle Trigonometry Trigonometric Functions of an Angle in a Right Triangle For the angle θ in the right triangle shown, we have b sin , c c csc , b a cos , c c sec , a b tan , a a cot b Recall: SohCahToa c b θ a opp adj opp , cos , tan , hyp hyp adj hyp hyp adj csc , sec , cot opp adj opp sin 3 Ex 1: Suppose that an acute angle θ is known to satisfy sin . Determine the 5 other trigonometric functions of this angle. Solution: Lets use a right triangle and label what we know. 5 3 We must find a before we can find the others, except for csc θ θ a =4 The Pythagorean Theorem will help us find the value of a. a2 b2 c2 a 2 32 5 2 a 9 25 2 a 2 16 a4 Now that we know all of the values for a, b, and c, we can write all of the solutions: 3 sin , 5 5 csc , 3 4 cos , 5 5 sec , 4 3 tan , 4 4 cot 3 Note: The last example said that the angle was acute which means it lies in the first quadrant. If the angle was in the second, third, or fourth, we would have solved the problem like we just did, and then change the signs as necessary. Ex 2: Find the value of the six trigonometric functions if the cos 3 5 Solution: Using the Pythagorean theorem we find that the opposite side equals 4. 4 sin , 5 5 csc , 4 3 cos , 5 5 sec , 3 4 tan , 3 3 cot 4 Ex 3: A climber who wants to measure the height of a cliff is standing 35 feet from the base of the cliff. An angle of approximately 60° is formed by the lines joining the climber’s feet with the top and bottom of the cliff, as shown. Use this information of approximate the height of the cliff. cliff x tan 60 opp x adj 35 tan 60 x 35 1.732050808 60° 35 ft Solution: Solve for x. x 35 x ≈ 60.6 ft. When you have a right triangle situation and at least one angle and one side is known, follow these steps to find the missing piece. 1. Ask yourself: What side do I have and what side do I want? We have the adjacent side and we want the opposite side. 2. Which trig function involves the two answers from question #1? The tangent function will be used. Ex 4: Two balls are against the rail at opposite ends of a 10 foot billiard table. The player must hit the ball on the left with the cue ball on the right without touching any of the other balls on the table. This is done by banking the cue ball off the bottom cushion, as shown. Where should the cue ball hit the bottom cushion, and what is the angle that its path makes with the bottom cushion. 3 θ x 2 2 3 tan x 10 x 20 – 2x = 3x 2 3 x 10 x x=4 2 Since tan ; x θ 10 - x 2(10 – x) = 3x tan 2 1 4 2 Unfortunately, we do not know a value on our unit circle where this is true, but this is the best we can do for now, even though we need the angle measure, not the tangent of that angle. We can answer where the cue ball should hit the bottom cushion: The cue ball should hit the bottom cushion 2 feet from the bottom left or 8 feet from the bottom right. Ex 5: An engineer is designing a drainage canal that has a trapezoidal cross section, as shown. The bottom and sides of the canal are each L feet long, and the side makes an angle θ with the horizontal. a.) Find an expression for the cross-sectional area of the canal in terms of the angle θ with the horizontal. b.) If the canal is S feet long, approximate the angle θ that will maximize the capacity of the canal. L Lsin θ L Lcos θ a.) Area of a trapezoid: 1 A (b1 b2 ) h 2 L Lcos θ We know the length of one base (L) and can find the height by making a right triangle. We have angle θ and the hypotenuse; we want the opposite side: 1 A (b1 b2 ) h 2 A 1 2 L 2 L cos L sin 2 A ( L L cos ) L sin A L(1 cos ) L sin A L2 (1 cos ) sin b.) Capacity means volume, so a canal that is S feet long has the capacity… V A S V SL2 (1 cos ) sin feet 3 To find the exact value of θ will have to wait until the next section! 3.7 Inverse Trigonometric Functions Recall properties of inverse functions: Properties of Inverse Functions Suppose that f is a one-to-one function. Then the inverse f-1 is unique, and … 1. The domain of f-1 is the range of f. 2. The range of f-1 is the domain of f. 3. If x is in the domain of f-1 and y is in the domain of f, then f-1(x) = y if and only if f(y) = x. 4. f(f-1(x)) = x when x is in the domain of f-1. 5. f-1(f(x)) = x when x is in the domain of f. 6. The graph of y = f-1(x) is the graph of y = f(x) about the line y = x. Sine The sine function is not one to one because it is periodic. If fails the horizontal line test for every horizontal line between -1 and 1 on the y-axis. However, we can restrict the domain of the sine function so that it is one-to-one. The sine function is one to one on the interval: 2 , 2 On this interval, the sine function does have an inverse and is denoted: Arcsine function or simply: y = arcsin x Note: csc x = (sin x)-1 The Arcsine Function , The arcsine function, denoted arcsin, has domain (-1,1) and range 2 2 and is defined by arcsin x = y if and only if sin y = x Arcsine Properties sin(arcsin x) = x when x is in [-1, 1] and arcsin(sin x) = x when x is in 2 , 2 The graph of y = arcsin x is the reflection of the restricted function of y = sin x reflected about the line y = x. Notice how steep the ends of the curve are on the inverse. This corresponds to the flatness on the sine curve. Ex 1: Find a.) arcsin Solution: 1 2 a.) arcsin b.) arcsin sin 3 1 2 6 b.) arcsin sin 3 3 3 c.) arcsin sin 4 3 c.) arcsin sin 4 Remember that we have a restricted interval now for sine. So we ask ourselves: Where in the restricted interval does sin = ½ ? Recall the arcsine properties: arcsin(sin x) = x. Here we have to be careful – our answer must lie in the restricted interval. 3 is not in the restricted interval. 4 3 sin Where in the restricted interval is the the same value? 4 3 sin sin So... 4 4 3 c.) arcsin sin 4 4 All of the other trigonometric functions are defined by making domain restrictions, too. Cosine The Arccosine Function The arccosine function, denoted arccos, has domain [-1, 1] and range [0, π]. and is defined by arccos x = y if and only if cos y = x Arccosine Properties cos(arccos x) = x when x is in [-1, 1] in [0, π]. and arccos(cos x) = x when x is Ex 2: Find 1 a.) cos arccos 2 1 b.) sin arccos 2 c.) arccos cos 3 d .) arccos cos 4 Solution: 1 1 a.) cos arccos 2 2 2 3 1 b.) sin arccos sin 3 2 2 c.) arccos cos 3 3 d .) arccos cos except this is not in the restricted domain. 4 4 2 Where in the restricted domain does cos ? 4 2 At 4 Tangent The Arctangent Function The arctangent function, denoted arctan, has domain (-∞, ∞) and range , , and is defined by 2 2 arctan x = y if and only if tan y = x Arctangent Properties tan(arctan x) = x for x in (-∞, ∞) and arctan(tan x) = x for x in , . 2 2 The graph of y = arctan x is the graph of y = tan x reflected over the line y = x. Graph together with students on board 12 3 Ex 3: Find cos arctan arcsin 5 5 We can see that this is of the form: cos( x1 x2 ) Solution: We will use the difference formula for cosine. 12 3 12 3 12 3 cos arctan arcsin cos arctan cos arcsin sin arctan sin arcsin 5 5 5 5 5 5 We know the value of the very last identity. 12 3 12 3 12 3 cos arctan arcsin cos arctan cos arcsin sin arctan 5 5 5 5 5 5 The remaining identities must be found using a right triangle: tan 13 52 122 12 12 5 But this is not one of our identities… To get θ by itself we take the inverse of both sides. arctan(tan ) arctan θ 5 12 5 arctan 12 5 Now, using this triangle find values of identities involving arctan 12 5 12 3 12 3 12 3 cos arctan arcsin cos arctan cos arcsin sin arctan 5 5 5 5 5 5 12 3 12 3 12 3 cos arctan arcsin cos arctan cos arcsin sin arctan 5 5 5 5 5 5 These values are: 5 13 12 13 12 3 5 3 12 3 cos arctan arcsin cos arcsin 5 5 13 5 13 5 3 cos arcsin All we have to do now is draw another triangle to help us find 5 3 3 sin arcsin 5 5 5 3 3 4 Now, find the value of your identity: cos arcsin θ 5 5 4 12 3 5 4 12 3 cos arctan arcsin 5 5 13 5 13 5 Now evaluate 12 3 16 cos arctan arcsin 5 5 65 Arcsecant The arcsecant function has the same restricted interval as the cosine function. Recall that y = sec x has a vertical asymptote at 2 The Arcsecant Function The arcsecant function, denoted arcsec, has domain (-∞, -1] U [ 1, ∞) and range 0, , and is defined by 2 2 arcsec x = y if and only if sec y = x Arcsecant Properties Sec(arcsec x) = x when x is in (-∞, -1] U [1, ∞) and arcsec(sec x) = x when , 2 2 x is in 0, Ex: Graph y = arcsec x Graph with students on board. 3.8 Applications of Trigonometric Functions A Cessna Citation III business jet flying at 520 miles per hour is directly over Logan, Utah, and heading due south toward Phoenix. Fifteen minutes later an F-15 Fighting Eagle passes over Logan traveling westward at 1535 miles per hour. We would like to Determine a function that describes the distance between the planes in terms of the time after the F-15 passes over Logan until it reaches the California border 20 minutes later. d = 1535t Distance = rate x time d (t ) (1535t ) 2 (520t 130) 2 d(t) d = 520(t + 0.25) = 520t + 130 Recall: the distance formula here involves time in hours: 20 minutes is 1/3 of an hour t = 1/3 d (t ) (1535t ) 2 (520t 130) 2 1 1 1 d (1535 ) 2 (520 130) 2 3 3 3 1 d 595 miles 3 It is more likely that the planes will not be traveling in paths that are perpendicular. Lets change the problem a little. A Cessna Citation III business jet flying at 520 miles per hour is directly over Logan, Utah, and heading due south toward Phoenix. Fifteen minutes later an F-15 Fighting Eagle passes over Logan traveling 24° west of south at 1535 miles per hour toward Nellis Airforce base 395 miles away. Since this situation does not include a right angle, we can not use the Pythagorean Theorem…directly. 24° 1535t 520t + 130 d(t) Law of Cosines Suppose that a triangle has sides of length a, b, and c and corresponding opposite angles α, β, and γ as shown. Then a 2 b 2 c 2 2bc cos A c α B β a γ b We also get… b 2 a 2 c 2 2ac cos c 2 a 2 b 2 2ab cos C Ex 1: A triangle has sides of length 6 and 8, and the angle between these sides is 60°. What is the length of the third side? 6 x 60° 8 Solution: Using the Law of Cosines… x 2 6 2 82 2(6)(8) cos 60 1 x 100 96 2 2 x 2 52 x 2 13 Ex 2: If two sides of a non right triangle are of lengths 15 and 25 and the included angle measures 35°, find the missing side and one of the other angles. 15 x γ 35° 25 x 15 2 25 2 2(15)(25) cos 35 x 15.35 -----------------------------------------------------------------152 15.352 252 2(15.35)(25) cos 225 235.6225 625 767.5 cos 225 860.6225 767.5 cos 635.6225 767.5 cos 0.8281726384 cos 34.0085 34 We can now go back and solve the aircraft problem… (d (t )) 2 (1535t ) 2 (520t 130) 2 2(1535t )(520t 130) cos 24 Lets let t = 0.333 (d (0.333)) 2 (511.155) 2 (173.16 130) 2 2(511.155)(173.16 130) cos 24 1 d 265 miles 3 Ex 3: A picture in an art museum is 5 feet high and hung so that its base is 8 feet above the ground. Find the viewing angle θ(x) of a 6-foot tall viewer who is standing x feet from the wall. APB and APC are both right triangles. We can Use the Pythagorean Theorem. AC 49 x 2 AB 4 x 2 Using the Law of Cosines… 25 (4 x 2 ) (49 x 2 ) 2 4 x 2 49 x 2 cos ( x) 25 53 2x 2 2 4 x 2 49 x 2 cos ( x) 25 53 2 x 2 cos( ( x)) 2 4 x 2 49 x 2 2 14 x ( x) arccos 2 2 (4 x )(49 x ) Using a graphing calculator we can see that the best viewing of the painting is where the maximum occurs which is when x 3.7 feet. This is the distance the viewer should stand from the wall. When the angles and one side of a triangle are known we can use the Law of Sines to find the other missing parts. The Law of Sines Suppose that a triangle has sides of length a, b and c with corresponding opposite angles α, β and γ as shown. Then B β sin sin sin c a a A α γ b C b c Ex 4: Find the missing angles. α 6 2 13 β 60° If we knew α we could find β. β = 180 – 60 – α 8 Solution: We could use the Law of Cosines to find α… 82 6 2 (2 13) 2 2(6)(2 13) cos but it is easier to use the Law of Sines. sin sin 60 8 2 13 2 3 arcsin 13 8 sin 60 2 13 2 39 arcsin 13 sin arcsin 8 sin 60 2 13 73.9; 46.1 Ex 5: The aircraft carrier Carl Vinson leaves the Pearl Harbor naval shipyard and heads due west at 28 knots. A helicopter is 175 nautical miles from the carrier 35° south of west. a.) On what course should the helicopter travel at its cruising speed of 130 knots to intercept the aircraft carrier? b.) How long will it take. Solution: Draw a picture of the given information. First, find θ which will give the course the helicopter should fly. sin sin 35 28t 130t sin 28t sin 35 0.1235 130t arcsin 0.1235 7 The helicopter should fly 42° north of east. Since we know θ = 7° we also know the third angle is 138°. sin 138 sin 35 175 130t 130t sin 138 175sin 35 t 175 sin 35 130 sin 138 t 1.154 hours It will take the helicopter about 1 hour and 9 minutes to intercept the aircraft carrier. Ex 6: Think Pair Share A campground lies at the west end of an east-west road in a relatively flat, but dense, forest. The starting point for a hike lies 30 kilometers to the northeast of the campground. A hiker begins at the starting point and travels in the general direction of the campground, reaching the road after 25 kilometers. Approximately how far is the campground from the road? B 30 25 camp A 45° γ C First, we need to find γ: γ' C' sin sin 45 30 25 Lets assume triangle ABC gives the correct solution. If the hiker traveled along the line BC' would mean the hiker was way off course – BUT IT COULD HAPPEN! This gives us an isosceles triangle. This could be the value of γ or γ' sin sin 45 30 25 or 58 Since γ is an obtuse angle it cannot be 58°, this must be the value of γ'. γ is part of a linear pair and the other angle is the same as γ' since they are the base angles of an isosceles triangle. γ = 122° The angle at B is: 180 – 45 – 122 = 13° sin 13 sin 45 AC 25 AC = distance of hiker from camp 7.95 km. If the hiker was lost, but measured the distance from the road correctly, B is 77°, therefore the distance would be… sin 77 sin 45 AC 25 AC 25 sin 77 34.5 km sin 45 If the hiker was badly off course he is about 34.5 km from camp. Heron’s Formula Heron’s formula is used to find the area of a triangle when only the lengths of the sides of the triangle are known. Heron’s Formula A triangle with sides of length a, b and c has area given by 1 P( P 2a)( P 2b)( P 2c) 4 where P is the perimeter of the triangle, P = a + b + c Ex 6: Find the area of the triangle. 7 9 12 P = 7 + 9 + 12 = 28 A 0.25 28(28 14)(28 18)(28 24) A 14 5