Trigonometric Functions

Transcription

1 Similar Right-Angled Triangles Trigonometric Functions We lan to introduce trigonometric functions as certain ratios determined by similar right-angled triangles. By de nition, two given geometric gures are similar if they have the same shae and di er only in size. Pairs of similar right-angled triangles are shown in the examles below. Examle R C A B Two similar right-angled triangles P Q Because the two triangles have the same shae, the angle at A is equal to the angle at P, the angle at B is equal to the angle at Q, (they are both right angles), and the angle at C is equal to the angle at R. AB and P Q are called corresonding sides. BC and P Q are also corresonding sides as are AC and P Q. Examle W Z U V X Y Another air of similar right-angled triangles In this air of similar right-angled triangles, the angle at U is equal to the angle at X, the angle at V is equal to the angle at Y and the angle at W is equal to the angle at Z. UV and XY are corresonding sides as are V W and Y Z. W U and ZX are also corresonding sides. As we can see from the above examles, corresonding sides in a air of similar triangles need not have the same length. What is the same are their ratios. Thus in Examle () Likewise Length of AB Length of BC = Length of P Q Length of QR. We may also write this as Length of AB Length of BC = Length of P Q Length of QR Length of BC Length of CA = Length of QR Length of RP which we may also write as Length of BC Length of QR = Length of CA Length of RP

2 The last one is Length of CA Length of RP = Length of AB Length of P Q which may be written as Length of CA Length of RP = Length of AB Length of P Q In Examle (), Length of UV Length of XY which is a short form for the following three: = Length of V W Length of Y Z = Length of W U Length of ZX Length of UV Length of XY = Length of V W Length of Y Z Length of V W Length of Y Z Length of W U Length of ZX = = Length of W U Length of ZX Length of UV Length of XY Of course they may be re-arranged at will. The above ratios may be used to calculate unknown lengths in similar triangles if su cient information is furnished about some of the lengths. Examle Say we are given that the lengths, in centimeters, of the sides AB, BC and P Q in Examle Length of P Q are 8, 5 and 0 resectively. Since we know the ratio Length of AB, (it is 0 ), we may use it to calculate 8 the length of QR. We simly use the fact fact that Length of QR Length of BC = Length of P Q Length of AB This translates into Exercise 4 Length of QR = Therefore the length of QR is = centimeters.. Suose, in Examle, the lengths, in inches, of QR, RP and BC are 8, 0:4 and 5 resectively. Calculate the length of CA.. Suose, in Examle, the lengths, in feet, of ZY, XY and U V are, and resectively. Calculate the length of V W.. A 6 feet tall man walks away from the base of a house as shown in the gure below. When he is 8 feet away from the house, his shadow starts forming 8 feet ahead of him. How tall is the building? house house shadow

3 Trigonometric Functions As a relude to trigonometric functions, consider the following roblem: A student was assigned the task of estimating the height of an uright tree in her home back yard. She roceeded as follows: She measured the length of the tree s shadow and found it to be feet. She then erected an uright stick and measured its height and the length of its shadow. She found them to be 4 feet and.7 feet resectively. She then argued that the two gures above are similar. Therefore The height of the tree The length of the tree s shadow = The height of the stick The length of the stick s shadow () Substituting the known lengths gives Therefore the height of the tree is 4 :7 The height of the tree = 4.7 = 5:8 feet, rounded to decimal lace. As we have said before, we will de ne trigonometric functions as ratios of right-angled triangles like the ones used by the student to determine the height of the tree in her backyard. There are six of them, denoted by tan x, sin x, cos x, cot x, csc x and cot x. We start with tan x. To introduce it, consider the angle x shown below in the rst quadrant. We denoted the origin by A and took the ositive horizontal axis as the initial ray. We then then droed a erendicular to the horizontal axis from a oint C on the terminal ray for x. Thus ABC is a right-angled triangle and the right angle is at B.

4 The longest side AC is called the hyotenuse of the triangle. It is common ractice to call BC the "side oosite angle x" and AB the "side adjacent to the angle x", then de ne tan x = Length of oosite side Length of adjacent side However, because oosite sides and adjacent sides are not so obvious for angles in the other quadrants, it is more ractical to use the coordinates of the oint C. Clearly, the length of the oosite side BC is the vertical coordinate of C and the length of the adjacent side AB is the horizontal coordinate of C, therefore Vertical coordinate of C tan x = Horizontal coordinate of C Of course this ratio deends on the value of x, not on the size of the triangle, (since ratios of corresonding sides in similar gures are equal). It turns out to be a very useful comutational tool, therefore it has been evaluated accurately for di erent angles x and recorded in tables and the common calculators. For some angles x, it can be calculated using the geometry of triangles. Among such angles are 0, 45, and 60 : Of these three, tan 45 is the easiest one to calculate because when x is 45, the oosite side and the adjacent side are equal in length, (see the left gure below). This imlies that the vertical and horizontal comonents of C are equal, therefore tan 45 = : C A 45 B An angle of 45 An angle of 0 To determine tan 0, consider the equilateral triangle ACD, (i.e. a triangle whose sides are all equal in length) shown in the right gure above. Actually, all we need is the the uer half of the triangle. There is no harm assuming that each of the sides AC; CD and AD has length. Then BC has length, (since it is half the length of CD). If we now use the Pythagorean theorem we easily obtain the length of AB to be =. Now the coordinates of C can be read o the gure. The horizontal one is and the vertical one is. Therefore tan 0 = Vertical coordinate of C Horizontal coordinate of C = = 4

5 To determine tan 60, consider the same equilateral triangle ACD but drawn di erently as shown below. All we need is the left half of the triangle. If we assume that each of the sides AC; AD, and CD has length then AB must have length and BC must have length =. Therefore C has horizontal coordinate and vertical coordinate, so that Angles Bigger Than 90 tan 60 = Vertical coordinate of C Horizontal coordinate of C = = The same formula is used to de ne the tangent of an angle bigger than 90. For an examle, take the angle 5, shown in the gure below, which the line segment AC makes with the horizontal axis. Dro a erendicular from C to the horizontal axis as shown in the next gure. We have denoted the oint where it intersects with the horizontal axis by B. Since angle BAC is 45, (i.e ), angle BCA must also be 45, therefore the line segments BA and BC must have the same length. If, for ure convenience, we assume that the length of BC is one unit then so must be the length of BA. Therefore the oint C has coordinates ( ; ) and so tan 5 = Vertical coordinate of C Horizontal coordinate of C = = 5

6 For another examle, we calculate tan 0. The 0 is drawn in the gure below. The coordinates of B are ( ; 0) and BC has length. Therefore the coordinates of C are ; and so tan 0 Vertical coordinate of C = Horizontal coordinate of C = = Use the gure below to calculate the exact value of tan 50 : For most angles x, we use a calculator to evaluate tan x. Say you want to nd tan. Make sure the calculator is in degree mode. Press the button labelled tan then enter the number. If you now ress the = symbol, a number which we may round o to 0:44 47, should aear. Use a calculator to comlete the table below. Round o the values to decimal laces. x tan x Soon you will be required to draw an angle whose tangent is a given number. Here are two examles: Examle 5 To draw an angle u that satis es the following conditions: (i) it is in standard osition, (ii) it is in the rst quadrant and (iii) its tangent is 5 6, (i.e. tan u = 5 6 ). Solution We have to take the ositive horizontal axis as the initial ray. To draw a terminal ray for u, it su ces to draw a line segment that originates from (0; 0) and asses through a oint C with vertical coordinate 5 and horizontal coordinate 6 so that tan u = Vertical coordinate of C Horizontal coordinate of C = 5 6 6

7 . 6 C 4 u 4 6 Examle 6 To draw an angle w which, (i) is in standard osition, (ii) is in the third quadrant, and (ii) has tangent 4, (i.e. tan w = 4 ). Solution We have to take the ositive horizontal axis as the initial ray. We have to draw a terminal ray with the roerty that if C is a oint on the ray then Vertical coordinate of C Horizontal coordinate of C = 4 Since the horizontal and vertical coordinates of oints in the third quadrant are negative, it must be the case that Vertical coordinate of C Horizontal coordinate of C = 4 The two negative signs haen to cancel out to give the result tan w = 4. Therefore, it su ces to draw a line segment that originates from (0; 0) and asses through ( ; 4). The angle is drawn in the gure below. Examle 7 In the given triangle, angle A is 9, the side AB has length cm and angle ABC is a right angle. Calculate the length of the side BC and round o to decimal lace. Solution We use the de nition tan 9 = Length of oosite side Length of BC = Length of adjacent side Therefore Length of BC = tan 9. We use a calculator to determine tan 9 then multily it by to get 5.9 cm, to decimal lace. 7

8 Exercise 8. Draw an angle u, in the second quadrant, whose tangent is 8, (i.e. tan u = 8 ) and an angle w in the third quadrant whose tangent is 4 5. Angle u Angle w. In triangle ABC, the angle at B is a right angle, the angle at A is 4 and the side AB has length 8 cm. The length of the side BC is: A) 5:88 cm B) 6: cm. C) 6:74 cm D) 7:56 cm Practice Problems Set 5, v. The shadow of an electric ole is found to be 8 feet long. At the same time and lace, the shadow of 8

9 a 6 foot man is found to be.7 feet long. What is the height of the electric ole, to the nearest foot? 5 5. Draw an angle u, in the fourth quadrant, whose tangent is, (i.e. tan u = ) and an angle w in the third quadrant whose tangent is 7 5. Angle u Angle w 9

10 . In triangle ABC, the angle at B is a right angle, the angle at A is and the side BC has length 4 ft. Calculate the length of the side AB. 4. To determine the height of a building and the communication tower on to of the building, the angles of elevation of the bottom and to of the tower were measured from a oint 47 meters from the bottom of the building and the results are as shown in Figure. If we remove the un-necessary details, we get gure. The angle at A is 90. Use triangle OAB to calculate the height AB, then use triangle OAC to calculate the height AC. Tower Building Figure Figure Figure Practice Problems Set 5, v. The shadow of an electric ole is found to be 8 feet long. At the same time and lace, the shadow of 0

11 a 6 foot man is found to be.7 feet long. What is the height of the electric ole, to the nearest foot? 5. Draw an angle u, in the fourth quadrant, whose tangent is 9, (i.e. tan u = 5 9 ) and an angle w in the third quadrant whose tangent is 8 7. Angle u Angle w. In triangle ABC, the angle at B is a right angle, the angle at A is 5 and the side BC has length 48

12 ft. Calculate the length of the side AB. C 48 ft A 5 B 4. To determine the height of a building and the communication tower on to of the building, the angles of elevation of the bottom and to of the tower were measured from a oint 47 meters from the bottom of the building and the results are as shown in Figure. If we remove the un-necessary details, we get gure. The angle at A is 90. Use triangle OAB to calculate the height AB, then use triangle OAC to calculate the height AC. Tower Building Figure Figure Figure

13 General De nition of the Basic Trigonometric Functions. Consider an angle x in standard osition. As we have done above, we the origin by A and take ositive horizontal axis, (starting at A), as the initial ray. Take any C on the terminal ray for x. Then the line segment AC makes an angle x with the ositive horizontal axis. Dro a erendicular CB from C to meet the horizontal axis at B and use it to determine the coordinates of C. In the examle shown below, we considered an angle x in the third quadrant, but that need not be the case. The line segment AC is called the hyotenuse of the right triangle ABC. The trigonometric functions tan x, sin x and cos x are de ned as follows: tan x = Vertical coordinate of C Vertical coordinate of C, sin x = Horizontal coordinate of C Length of hyotenuse AC, Horizontal coordinate of C cos x = Length of hyotenuse AC For angles x in the rst quadrant and in standard osition, we may de ne sin x and cos x by sin x = Length of oosite side Length of hyotenuse cos x = Length of adjacent side Length of hyotenuse It turns out that tan x = sin x. Verifying this is easy: cos x Vertical coordinate of C sin x cos x = = Length of hyotenuse AC Vertical coordinate of C Length of hyotenuse AC = Horizontal coordinate of C Length of hyotenuse AC Horizontal coordinate of C Length of hyotenuse AC Vertical coordinate of C Horizontal coordinate of C = tan x The following table gives a few values of these functions. The angles x may be given in degrees or radians.

14 We have given them in both units. x in degrees x in radians rad tan x 0 sin x 0 cos x x in degrees x in radians tan x sin x cos x 6 rad 4 rad rad rad rad 4 rad rad rad rad 5 4 rad 4 rad rad 5 rad 7 4 rad 6 More Trigonometric Functions Unde ned Unde ned There are three more trigonometric functions denoted by cot x, csc x and sec x. 0 rad rad 6 Using the same gure above, they are de ned by cot x = Horizontal coordinate of C = tan x Vertical coordinate of C csc x = Length of hyotenuse AC = sin x Vertical coordinate of C sec x = Length of hyotenuse AC = cos x Horizontal coordinate of C Most calculators give only the values of sin x, cos x and tan x. The other functions are evaluated using the above formulas. For examle, sec 60 = cos 60 =. 4

15 Examle 9 ( 5; ) is a oint on the terminal ray of an angle x. To nd the exact value of: sin x, cos x, tan x, cot x, csc x and sec x. Solution: The angle is shown in the gure below The horizontal and vertical coordinates of the oint ( 5; ) on the terminal ray are 5 and resectively. The hyotenuse has length 5 + = 69 = Therefore sin x =, cos x = 5, tan x = 5, cot x = 5, csc x = and sec x = 5. Examle 0 To nd the exact values of tan x, sin x and cos x given that sec x = and x is in quadrant IV. Solution: The angle is shown in the gure below. Since sec x =, we may take the horizontal coordinate of a oint on the terminal ray to be and the corresonding hyotenuse to have length. Then the corresonding vertical coordinate should be = 5. It has to be negative because the vertical coordinates of oints in the fourth quadrant cannot be ositive. Thus ; 5 is a oint on the terminal ray for the given angle x. Therefore tan x = 5, sin x = 5 and cos x =. Exercise. ( ; 4) is a oint on the terminal ray of an angle. Find the exact value of sec. A) 5 4 B) 5 C) 5 D) 4 5. You are given that is an acute angle and sin =. Find the exact value of sec. A) sec = 5 5 B) sec = 5 5 C) sec = D) sec = 5 5

16 . The angles x, y and z are shown in the gure below. Determine the exact value of: (a) sin z (b) cos z (c) tan z (d) sin y (e) cos y (f) tan y (g) sin x (h) cos x (i) tan x (j) sec z (k) csc z (l) cot z 4. You are given that u is an angle in the second quadrant with sin u = 5 and v is an angle in the fourth quadrant with tan v = 4 (a) Draw the angles in standard osition on the axes below. (b) Determine the exact value of: (a) tan u (b) cos u (c) cot u (d) sec u (e) csc u (g) cos v (h) cot v (i) sec v (j) csc v (k) sin v 6

17 5. You are given that x is an angle in the second quadrant and sin x = 4. Draw the angle in standard osition then calculate cos x, tan x, cot x, sec x, and csc x. 6. You are given that is an angle in the third quadrant and tan = 5. Draw the angle in standard osition then calculate cos, sin, cot, sec, and csc. The signs of the Sine, Cosine and Tangent functions Since the horizontal and vertical coordinates of the oints in the rst quadrant are ositive, it follows that ALL the three functions sin x, tan x and cos x are ositive in the rst quadrant. The vertical coordinates of oints in the second quadrant are ositive but the horizontal coordinates are negative. It follows that, of the three, it is the SINE function that is ositive in the second quadrant. The other two are negative. The vertical and horizontal coordinates of oints in the third quadrant are negative. It follows that of the three, it is the TANGENT function that is ositive in the third quadrant, (because it is a quotient of two negative numbers). The other two are negative. The horizontal coordinates of oints in the fourth quadrant are ositive but the vertical coordinates are negative. It follows that of the three, it is the COSINE function that is ositive in the fourth quadrant. The other two are negative. The above observations are summarized in the following table We may condense this further to Quadrant 4 Positive function All Sine Tangent Cosine 4 A S T C Or simly ASTC Reference Angles Consider the angles 0, 50, 0, 0, 90. The non-acute angles 50, 0, 0, 90, 50,... are all related to the single acute angle 0 as follows: 50 is short of 80 by 0 therefore its terminal ray makes an angle of 0 with the x-axis. Terminal ray makes an angle of 0 with the x-axis 7

18 0 exceeds 80 by 0 therefore its terminal ray makes an angle of 0 with the x-axis. Terminal ray makes an angle of 0 with the x-axis 0 is short of 60 by 0 therefore its terminal ray makes an angle of 0 with the x-axis. Terminal ray makes an angle of 0 with the x-axis The attern should be clear. We say that 0 is the reference angle for the 50, 0, 0, etc. In general, the reference angle of a given angle x is the acute angle that the terminal ray for x makes with the x-axis, assuming that one has taken the ositive x-axis as the initial ray. Note that sin 50 = sin 0, cos 50 = cos 0, tan 50 = tan 0, thus the values of the trigonometric functions at 50 are either the same or they di er from the corresonding values at 0 by just a sign. Before calculators were introduced, the values of sine, cosine and tangent functions were tabulated for acute angles only. If, say, one wanted to determine sin 90, which is not an acute angle, one would (a) Determine the reference angle for 90, which is 70 ; (b) Look u sin 70 from a table of the sine function. This would be a number rounded to 4 decimal laces, most robably 0:997; (c) Note that sin x is negative in the fourth quadrant, therefore sin 90 = 0:997: Exercise. Find the reference angle for 5. Find the reference angle for 7. Find the reference angle for 44 A) 5 B) 5 C) 65 D) 75 8

19 Practice Problems Set 6, v. You are given that is an angle in the second quadrant and cos =. Draw the angle, form an aroriate right triangle, calculate its unknown length then determine the exact values of sin, tan, csc and cot. Where alicable, give your answer as a radical with a rational denominator. Calculate the unknown length in this sace sin = tan = csc = cot =. You are given that with sin = 4 and tan > 0, (i.e. the tangent of is ositive). Determine the quadrant in which lies, draw it then calculate the exact value of cos 9

20 . A guy wire is attached to a oint feet below the to of a vertical electric ole. The ole is 8 feet tall and the guy wire makes an angle of 78 with the horizontal ground. Draw a diagram dislaying this information then calculate the length of the guy wire. 0

21 Practice Problems Set 6, v. You are given that is an angle in the second quadrant and sin = 4. Draw the angle, form an aroriate right triangle, calculate its unknown length then determine the exact values of cos, tan, sec and cot. Where alicable, give your answer as a radical with a rational denominator. Calculate the unknown length in this sace sin = tan = csc = cot =. You are given that with cos = and tan < 0, (i.e. the tangent of is negative). Determine the quadrant in which lies, draw it then calculate the exact value of sin

22 . A guy wire is attached to a oint feet below the to of a vertical electric ole. The ole is 7 feet tall and the guy wire makes an angle of 75 with the horizontal ground. Draw a diagram dislaying this information then calculate the length of the guy wire.

23 Grahs of sin x and cos x Samle values of cos x are given below. x in degrees cos x to dec. l. x in degrees cos x to dec. l :87 0:7 0:5 0 0:5 0:7 0: :87 0:7 0:5 0 0:5 0:7 0:87 They are lotted on the axes below. Joining them with a smooth curve gives the grah below As shown in the gure below, we also get this same shae between any two consecutive multiles of 60.

24 For this reason, we say that cos x is eriodic with eriod 60. Its values lie strictly between and. We say that it has amlitude. One cycle of sin x is given in the gure below It is this same shae that one gets between 60 and 70 and between 60 and 0, (see the grah 4

25 below). In general, we get the same shae between any two consecutive multiles of 60. For this reason, we say that sin x is eriodic with eriod 60. Its values are also strictly between and, therefore its amlitude is also. Magnifying the grahs of sin x and cos x For an examle, consider the grah of sin x. We get it by simly doubling the values of sin x. The result is a grah with amlitude drawn below on the same axes as the grah of sin x. grah of sinx In general, if b is a ositive number then the grah of b sin x will have the familiar shae of a sine function but with amlitude b. A similar statement with sine relaced by cosine is also true. In the gure below, the 5

26 grah of :8 cos x is drawn on the same axes as the grah of cos x..8 grah of.8cosx The grah of sin x is obtained by drawing the re ection, in the horizontal axis, of the grah of sin x, (the ositive values are made negative and the negative values are made ositive) Grah of sin x Grah of sin x The grah of sin x is obtained by rst magnifying the grah of sin x by a factor then draw a re ection, 6

27 in the horizontal axis, of the magni ed grah. grah of sinx grah of sinx The same rocedure is followed to sketch a cycle of cos x or b cos x, (when b is ositive). For examle, to sketch a cycle of :8 cos x, rst magnify the grah of cos x by a factor of :8 then draw a re ection, in the horizontal axis, of the magni ed grah..8 grah of.8cosx.8 grah of.8cosx In general, to sketch a cycle of a sin x or a cos x where a is a given negative number, rst magnify the grah of sin x or cos x by a factor jaj then draw a re ection, in the horizontal axis, of the magni ed grah. Translating the grahs of sin x and cos x horizontally Say you are asked to sketch one cycle of the grah of cos (x 60 ). You simly take a cycle of cos x and slide it, (i.e. translate it), to the right through 60 degrees. The result is drawn below on the same axes as one cycle of cos x, (shown dotted). We say that the grah of cos x is shifted through 60 to the right. The 7

28 60 angle is called the horizontal shift. Grahs of cos x, (dotted), and cos (x 60 ) Here is a way of visualizing what is going on: Take the largest value of cos x. You get it when x is 0 or 60, (assuming you are restricting yourself to one cycle of cos x). It follows that cos(x 60 ) has value when the angle x 60 is 0 or 60, i.e. when x = 60 or when x = 40 as shown on the above grah. If you are asked to sketch the grah of cos (x + 60 ), you would shift the grah of cos x to the left, (NOT TO THE RIGHT), through 60. Grahs of cos x, (dotted), and cos (x + 60 ) In general, if b is a ositive number then the grah of cos (x b) is obtained by shifting the grah of cos x through b degrees to the RIGHT and the grah of cos (x + b) is obtained by shifting the grah of cos x through b degrees to the LEFT. Changing the Period of sin x and cos x We change the eriod of sin x or cos x when we multily the variable x by a constant. For an examle, consider the function sin x. We note that x is zero when x = 0 and it is 60 when x = 60 = 60 = 40 : 8

29 It follows that one cycle of sin x is between 0 and 40 degrees, suggesting that its eriod is 40. This is indeed the case as the grah below shows. In general, the function sin bx has eriod 60. The same alies to cos bx. b A Combination of All Three Consider sketching one cycle of the grah of sin 4 (x about it from its formula. 40 ). The rst ste is to deduce useful information It has amlitude. Its eriod is = 60 = 480 degrees. It has a horizontal shift of 40 to the right. Now draw coordinate axes and introduce an interval from 0 to 480 on the horizontal axis, (because the function has eriod 480 ). Divide the interval into four equal arts corresonding to angles 0, 0, 40 ; 60 and 480. The vertical axis should extend at least to and because the function has amlitude

30 Without the shift, you have the function sin 4x whose grah is shown dotted in the gure below. The grah of sin 4 (x 40 ) is obtained by shifting the dotted grah through 40 to the right. Another Examle To sketch one cycle of y = cos (x + 5 ) Solution We start by extracting useful information from the given formula. (i) Its amlitude is. (ii) 60 Its eriod is = 0. (iii) To get its shift we rst write it in the form cos (x + 5). Now the shift is clear; it is 5 to the left. Turning to the required sketch, we rst sketch a cycle of cos x, shift it 5 to the left, magnify it by a factor then re ect, in the horizontal axis, the magni ed grah. The various stages are shown in the gures below Sketch of cos x Sketch of cos (x + 5) 0

31 Sketch of cos (x + 5) Sketch of cos (x + 5) Exercise. Determine the amlitude of y = 4 sin x. Determine the eriod of y = 5 cos 4x A) 4 B) 4 C) 4 D) 4. Determine the hase shift of y = 5 sin 4 x 4 A) B) 4 C) D) A) 6 units to the left B) units to the right C) 4 units to the left D) 4 units to the right 4. The current I, in ameres, owing through a articular ac (alternating current) circuit at time t seconds is I = 0 cos 4t (a) What is the eriod of the current? A) 0 seconds B) 4 seconds C) seconds D) 44 seconds (b) What is the hase shift of the current? A) 6 seconds B) 6 seconds C) 44 seconds D) 44 seconds 5. You are given one cycle of the grah of f(x) = sin x. 6

32 Extend the coordinate axes as necessary then sketch the grah of g(x) = sin (x + 0 ). 6. You are given a sketch of the grah of h(x) = cos x: Extend the coordinate axes as necessary then sketch the grah of u(x) = cos (x 45 ) : 7. You are given a sketch of the grah of h(x) = cos x: Extend the coordinate axes as necessary then sketch the grah of v(x) = cos x 8. Sketch, on the same axes, one cycle each for the grah of (a) sin x and :5 sin x (b) cos x and = cos x (c) sin x and 4 sin x (d) cos x and 4 5 cos x (e) sin x and sin (x 0 ) (f) sin x and sin (x + 40 ) (g) cos x and cos (x + 45 ) (h) cos x and cos (x 8 ) (i) sin x and sin (x + 54 )

33 Practice Problems Set 7, v. Sketch one cycle of the curve y = cos x and the curve y = cos x on the same axes below.. Sketch one cycle of the curve y = sin x and the curve y = sin(x 0 ) on the same axes below.

34 . Sketch one cycle of the curve y = sin 4 x 4. Determine the amlitude, eriod and shift of the curve y = sin (x 0 ) Amlitude = Period = Shift = (Secify whether it is to the left or right.) Sketch one cycle of the curve y = sin (x 0 ) 4

35 Practice Problems Set 7, v. Sketch one cycle of the curve y = sin x and the curve y = cos x on the same axes below.. Sketch one cycle of the curve y = cos x and the curve y = cos(x + 40 ) on the same axes below. 5

36 . Sketch one cycle of the curve y = cos x 4. Determine the amlitude, eriod and shift of the curve y = 4 sin 4 5 (x + 0 ) Amlitude = Period = Shift = (Secify whether it is to the left or right.) Sketch one cycle of the curve y = sin (x 0 ) 6

37 Exercise 4. Determine the eriod of the given function then sketch one cycle of its grah: (a) sin x (b) cos x (c) sin 4 x (d) cos 5 x (e) sin 5 x (f) sin 4 x (g) cos 5 6 x (h) cos 4 5 x. Give the eriod and amlitude of f(x) = sin x then sketch one cycle of its grah on the given axes.. Determine the eriod, amlitude and shift of the trigonometric function f(x) then sketch one cycle of its grah on the given axes. 4. Determine the eriod, amlitude and shift of the given function then sketch one cycle of its grah: (a) sin (x 0 ) (b) cos (x + 45 ) (c) 4 sin (x 0 ) (d) :5 cos 4 5 (x + 5 ) (e) :8 cos 5 (x 50 ) (f) :5 sin 4 (x + 48 ) (g) 4 sin (x + 5 ) (h) sin (x 5 ) (i) 5 cos 5 (x + 90 ) 5. Determine the eriod, amlitude and shift of f(x) = 4 sin 4 x + 8 grah. The angles are in radians. then sketch one cycle of its 6. You are given that x is in degrees. Determine the eriod, amlitude and shift of y = cos x 0 then sketch one cycle of the curve. 7. Sketch one cycle of the curve y = sin 4 x 5. Assume that x is in degrees. 7

38 Inverse Trigonometric Functions Here is a roblem that may be solved using an inverse trigonometric function: A rectangle has sides of length inches and inches. What is the acute angle x, shown in gure (i), between its diagonals? 6 x y Figure (i) Figure (ii) One aroach is to to calculate angle y shown in gure (ii) then double it to get x. Clearly, tan y = 6. Therefore we have to nd an acute angle y whose tangent is 6. This is the oosite of what we have done so far, which was to calculate the tangent or sine or cosine of a given angle. Now we have to nd an angle whose tangent is a given number. This requires the use of an inverse trigonometric function. The inverses of tan x, sin x and cos x are introduced below. It is convenient to view a function as a table with two rows, (or two columns), that shows how two variable quantities are related. The table must satisfy one crucial condition: Every number in the rst row, (column), is aired with exactly one number in the second row, (column). Take the function tan x: The table below shows a samle of its values. x tan x 5:67 :9 0:6 0 0:6 :8 :08 The decimals are aroximate values. We have restricted the angles x: they are between 90 and 90. When we do this then di erent angles in the rst row are aired with di erent numbers in the second row. Under these circumstances, we can de ne an inverse function. To this end, swa the two rows. The result is a table that gives samle values for the inverse of tan x which is denoted by arctan y or tan (y). (The use of y instead of x is urely for convenience.) y 5:67 :9 0:6 0 0:6 :8 :08 tan y Thus tan ( 5:67) = 80 ; tan () = 45, tan (:8) = 54, etc. Clearly, if y is a given number then arctan y, or tan (y), is the angle between 90 and 90 whose tangent is y. This function is given on calculators. Set your calculator to degree mode then use it to con rm that, (to decimal lace), tan (0:7) = 5:8, tan ( 0:4) = :8, then determine the following: a) tan ( 0:567) b) tan (0:) c) tan (0:57) d) tan (4:7) e) tan ( 6) 8

39 To determine the acute angle between the diagonals of the above rectangles, we had to nd an angle y whose tangent is 6 6. Now we know that we have to nd tan. A calculator gives 8:6 to decimal laces. Therefore x = 57: or 57 to the nearest degree. The inverse of sin x is denoted by sin y or arcsin y and it is obtained in a similar way. We rst restrict the angles x to values between 90 and 90, (so that di erent angles have di erent sines). A samle of values of sin x is given below x sin x 0:98 0:77 0 0:8 0:95 When we swa the two rows we get samle values for sin y or arcsin y: y 0:98 0:77 0 0:8 0:95 sin y As you would exect, sin (y) is the angle between 90 and 90 whose sine is y. The values of sin y are also given on calculators. Use it to con rm that, to one decimal lace, sin (0:) = 7:5, sin (0:87) = 60:8 and sin ( 0:7) = 46:. To get an inverse for cos x, we restrict the angle to values between 0 and 80 to guarantee that di erent angles have di erent cosines. A samle of values of cos x is given in the table below. x cos x 0:64 0: 0 0:77 0:97 Samle values for its inverse are obtained by swaing the two rows. y 0:64 0: 0 0:77 0:97 cos y If y is a number between and then cos (y) is the angle between 0 and 80 whose cosine is y. Set your calculator to degree mode then use it to determine cos (0:6), sin ( 0:4), cos ( 0:567) ; sin (0:), cos (0:57). Examle 5 To nd all the angles x such that 4 sin x = : Solution 6 Divide both sides by 4 to get sin x = 0:75. Therefore x is an angle whose sine is 0:75. In other words, x = arcsin 0:75. A calculator gives x = 48:6 to dec. l. But this is not the only angle whose sine is Another one is 80 48:6 = :4. Since the sine function is eriodic with eriod 60, we may add any multile of 60 to 48:6 or :4 and the result will still be a solution of the equation 4 sin x =. Thus 48:6 + 60, : ; 48:6 + 70, : ; and many others are solutions of the given equation. The exressions (48:6 + 60n), (:4 + 60n), n = ; ; : : : cature all the solutions. 9

40 Examle 7 To calculate the exact value of cos sin ( 5 ) : Solution 8 sin ( 5 ) is an angle in the rst quadrant whose sine is 5. We may name that angle u. Thus u = sin ( 5 ) and so sin u = 5. The gure below shows the angle. 5 u We use the Pythagorean theorem to calculate the length of the third side of the triangle. If its length is a, then a + = 5 When we solve for a we get a = 4. We are required to calculate cos sin ( 5 ) which is actually cos(u). Since the horizontal side of the triangle has length 4, it follows that cos u = 4 5, therefore cos sin ( 5 ) = 4 5 Exercise 9. Solve the following equations for x between 0 and 60. When necessary, round o your answers to decimal lace. (a) 5 sin x = 0 (b) 5 tan x + 9 = 0 (c) cos x + = (d) sin x + = (e) 4 cos x + = 7 tan x + 6 = 4. Clearly, tan tan = tan (60 ) =. Exlain why tan tan y = y for any given number y. What is tan (tan x)?. A student was asked to determine tan sin ( ) and he roceeded as follows: We know from a table of samle values of the sine function that sin ( ) = 0. Therefore the roblem requires one to nd tan (0 ) which is. It follows that tan sin ( ) = tan (0 ) =. Find the exact value of cos tan () and sin(cos ( )) in a similar way. 4. Follow the stes of Examle 7 to calculate the exact value of: (a) sin(tan ( 5 )): (b) tan sin ( 4 5 ): (c) cos(tan ( 7 4 )): (d) cos(sin (x)), (hint: write x as x ). 5. Verify that sin cos ( 4 ) = Use a sketch to nd the exact value of cos sin 4 5 A) 5 B) 4 5 C) 5 D) 5 7. Assume that x is ositive. Use a sketch to nd sin tan x A) x x + x + B) x x x + C) x x D) x + x + 40

41 8. The exact value of cos tan x in terms of x is (A) 4 x x (B) 4 + x x (C) x 4 + x (D) x 4 x 9. Evaluate the following exressions (a) sin (arccos x) (b) tan arcsin x (c) cos arctan x (d) sec arctan x 0. Find all the angles x such that 6 sin x 7 sin x + = 0 :. Find all the angles x such that tan x 6 tan x + 5 = 0. 4