Periodic and trigonometric functions.

Transcription

1 Periodic and trigonometric functions. We will concentrate our attention on periodic functions, although most of the material will consist on trigonometric functions. There is a good reason for this: periodic functions are approximated by trigonometric functions and so it is essential, from the point of view of constructing models, to understand well the properties of trigonometric functions. Since trigonometry has been a subject of enquiry for thousands of years, we will also present some of its traditional applications. Definition. Let f : R R be a function. We will say that f is periodic of period p if f x p f x,x R. Observation. Iff is periodic of period p, then the graph of f "repeats itself " on,p, p,p, p,p, etc. A graph of a periodic function is shown below. y There are many periodic phenomena in nature, which is induced by he day-night cycle, the seasons, periods of high activity etc. They fall into the category of "circadian rhythms " or, a much abused term, biorhythms..5 x 5 Examples of circadian rhythms are: heart rate, core body temperature, blood pressure, sugar content of the blood, neural activity, amount of certain hormones in the bloodstream, etc. Just to show you an example of how these rhythms are used, we present you with an abstract of research being carried out on Alzheimer s disease (we will explain the special terminology used in the abstract in Section 4.):

2 Circadian changes related to sundowns and sleep-wake disturbances in Alzheimer s disease Ladislav Volicer, David Harper, and Andrew Satlin Abstract Circadian rhythms of core body temperature and motor activity were measured in 5 patients with diagnosis of probable Alzheimer s disease (AD) and 9 healthy controls. AD subjects had lower diurnal motor activity, higher percentage of nocturnal activity, lower interdaily stability of motor activity and delayed acrophase when compared with controls. AD subjects had also higher mesor temperature, increased amplitude of the temperature rhythm and delayed acrophase when compared with controls. AD subjects were divided according to the occurrence of sundowning determined by staff reports. Severity of sundowning was associated with later acrophases of both motor activity and core body temperature, decreased correlation of circadian rhythm of body temperature with 4 hr. cycle, and decreased amplitude of temperature variation. These data indicate that sundowning is related to a disturbance of circadian rhythms caused by AD. The functions they ended with are in the graphed below:

3 In general, if we know that we have a periodic function and have data to try to construct a mathematical model, it still remains a difficult task to complete, and one where the use of the trigonometric functions, to be introduced in the next section, play an essential role. In this course we will only study periodic functions that arise from trigonometric functions (which are essential for many other studies, whenever angles and measurements are needed, as for example, in orthopedics, to determine when a joint is within normal or expected ranges or not, etc.). 4. Radian measure of angles and the sine and cosine functions. This subject is one with a long history, with the first recorded works dating back to the ancient Egypt, works that were influenced by knowledge acquired by the Babylonians (transmitted, supposedly, by oral traditions). The subject matter is the study of angles and measurements that van be done with them. Definition of angle. An angle is generated by the rotation of a closed half-line (ray) about its endpoint. The angle is said to be positive if the rotation is counterclockwise, negative if the rotation is clockwise (ad zero if there is no rotation at all). The endpoint is called the vertex of the angle.

4 To measure angles we will usually do it with the radian measure of an angle, which we describe below: Radian Measure of an Angle. Consider an angle and assume that the closed half line that generates it is the nonnegative x-axis in the coordinatized plane (equal units in both axes). Take the point, and follow its trajectory as the angle is generated; the path it traces lies in the circle centered at the origin with radius one, from now on called the unit circle. The distance traveled by this point is the radian measure of the angle if the angle is positive, zero if the angle is zero, and its additive inverse if the angle is negative. The point x,y where, ends up in the unit circle is called the terminal point of the angle. Two angles are said to be coterminal if they have the same terminal point. Examples.. A positive angle of a full revolution has radian measure. Since we have already been told that an angle of a full revolution measured in degree has measure 6, the it follows that one degree measured in radians has measure 6 8. Suppose that is an angle with terminal point x,y. 4

5 Definition. cos x,sin y. Immediate Identities:. Since the point x,y is in the unit circle, we must have that x y, thus, if is any angle: cos sin.. Since the angle has as terminal point x, y, cos cos sin sin.. Since adding or subtracting any number of full revolutions about the origin yield the same terminal point, we have, for any R: cos k cos,k Z, in radians. sin k sin,k Z, in radians. 4. Define the functions cos : R R,sin : R R in the obvious way (that is cos x, from the terminal point and the same for sin), then both of these functions are periodic of period. 5.Thesignofcos,sin depends on the quadrant where the terminal point lies. If the terminal point lies in the first quadrant, then cos,sin, if it lies in the second quadrant, cos,sin, if it lies in the third quadrant, cos,sin and if it lies in the fourth quadrant, cos,sin. 6. cos k,if k Z is even cos k if k Z is odd sin k 7. cos k,k Z, sin k,k Z sin k,k Z 5

6 Identity, cos sin, is a fundamental fact which is used in many situations. One example is that of its use in proving the so-called trigonometric identities. Examples of simple trigonometric identities:. Show that sin t sin t 4 cos t sin t sin t 4 sin t, whenever cos t sin t. To do this we see that using Identity in the denominator, and factoring the numerator (as we would factor a quadratic polynomial), we get: sin t sin t 4 cos t sin t sin t 4 sin t sin t sin t sin t 4 sin t sin t sin t sin t 4 sin t sin t sin t sin t 4. sin t. Show that sin 4 t sin t cos t sin t sin t. Here we use the fundamental identity to express cos t as sin t. sin 4 t sin t cos t sin 4 t sin t sin t sin 4 t sin t sin 4 t sin t sin t. Special angles. We have seen already that one degree measured in radians has measure By the same token, one radian has degree measure equal to 8. We will always use radian measure whenever dealing with trigonometric functions. However, we will make use of the degree measure of an angle to calculate special values of the trigonometric functions, so that we can use facts learned before about triangles and their properties (for example that the angles in a triangle add up to 8.., this radian measure corresponds to an angle of 4 45 and when considering sin,cos we first observe that the terminal point lies in the first quadrant, so both 4 4 are positive, and that, considering the triangle with vertices,, cos,sin and 4 4 cos,, we have an isosceles right triangle with hypothenuse equal to. Thus

7 cos sin, and : 4 4 cos 4 sin 4 Thus, cos 4 sin 4 cos 4 cos 4,cos 4 cos 4.. This is the way we will state solutions to our problems, so we will not require "rationalizing the denominator ".. (the degree measure is 6 ). In this case we extend our triangle (by adding a side at the point cos,sin to an equilateral triangle with the base in the x axis (draw a picture). Then the height of the triangle is sin and the midpoint of the base is cos. It follows that cos and so: cos sin 4 sin,sin sin 4 sin.. 6 (this corresponds to a measure of ). In this case take the right triangle with vertices at,, cos, and cos,sin and "flipped it about the x axis. You get an equilateral triangle. Since the point cos, is the midpoint of the side 6 opposite the origin, we must have that sin, and by the same type of reasoning 6 we had in, cos (this corresponds to ). Here we observe that, and we use 6 this information to observe that if we consider the angle, which has terminal point in the positive y axis, then we have to add the angle 6 to it, and if we view this using this axis as reference point, then the value of sin cos 6, and that cos cos sin 6. Use of the terminal point information. The position of the terminal point determines if the values of sin,cos of that particular angle are positive or negative. If one value is given, then the value of the other is readily obtainable. 7

8 Examples.. An angle is such that its terminal point lies in the first quadrant and sin 4. Find cos. Solution. Since the terminal point lies in the first quadrant, both of the trigonometric functions of this angle have positive value and cos a sin. We conclude that: cos 6,cos cos An angle has terminal point in the third quadrant and cos. Find sin. 4 Solution. The same type of reasoning tells us that sin, so that: cos sin,sin sin Let be an angle with terminal side in the third quadrant and such that cos.find sin. 5 Since the terminal point is in the Third quadrant, sin. cos sin,sin 9 5 sin sin sin Exercises 4.. Find sin Find cos

9 . Find the terminal point for the angle with measurement of Find the terminal point of. 5. An angle has terminal point in the third quadrant and cos 4. Find sin An angle has terminal point in the fourth quadrant and sin. Find 4 cos. 7. Suppose that an angle is such that cos. Which are the quadrants where the terminal point may lie? 8. Let be an angle such that cos, sin.which are the quadrants where cos the terminal point may lie? 9. Let,,cos sin.which are the quadrants where the terminal point may lie?. If cos 6, which are the possible values for sin? 7. Prove that cos t sin t sin t cos t, whenever cos t. cos t. Show that cos 4 t sin 4 t cos t sin t.. Prove that cos 4 t 5cos t 4 sin t sin t. 4. Find all solutions of the equation cos 4 t 5cos t Find all solutions of the equation sin t sin t. 6. Prove that sin t cos t cos t,t k,k Z. 4. Addition and subtraction formulas. In this section we concentrate in finding the values of the functions sine and cosine for the sum of two angles, say,. What we are after is expressing the value of the desired trigonometric function in terms of the values of these functions evaluated at and. Theorem 4... Let a,b be angles. Then cos a b cos a cos b sin a sin b. Proof. We consider first the case when a b. 9

10 Consider the triangle with vertices at,, cos b,sin b and cos a,sin a and the one with vertices at,,, and cos a b,sin a b. The last triangle is a rotation of the first triangle so that the side joining, and cos a,sin a lies on the x-axis. It follows that the sides opposite to the origin have equal length, concluding that: cos a cos b sin a sin b cos a b sin a b cos a cos b sin a sin b cos a b sin a b cos a cos a cos b cos b sin a sin a sin b sin b cos a b cos a b si cos a cos b sin a sin b cos a b cos a b cos a cos b sin a sin b. This proves the case a b. If b a, then: cos a b cos a b cos a b cos b a cos a b cos b cos a sin b sin a and we get the same result. Consequences:. Let a,b be angles. Then cos a b cos a b cos a cos b sin a sin b cos a cos b sin a sin b.. Let be an angle. Then;

11 cos cos cos sin sin sin. cos cos cos sin sin sin sin cos cos sin cos cos cos cos sin sin cos.. Let a,b be angles. Then sin a b sin a cos b cos a sin b,sin a b sin a cos b cos a sin b. Proof. sin a b cos a b cos a cos b sin a sin b sin a cos b cos a sin b. sin a b sin a b sin a cos b cos a sin b sin a cos b cos a sin b. Examples.. Find cos cos cos cos sin sin. With these tools we can get enough points to graph both sine and cosine. y x - - For the graph of sine we have: graph of cosine

12 y x - graph of sine We have seen that if, are angles, then: cos cos cos sin sin cos cos cos sin sin sin sin cos cos sin sin sin cos cos sin. These formulas are useful in many contexts. The first way we can use them is to find the exact value of trigonometric functions for angles that can be expressed as sums or differences of those which we already know their values. - Example. Find cos 5. Observe that the degree measure of this angle is 5 8 write 5. Thus: 4 6 cos 5 cos so that we can cos 4 cos 6 sin 4 sin 6.

13 Double Angle Formulas. If is an angle then cos cos sin sin sin cos. Example. Find sin. sin sin sin cos. Half Angle Formulas. Let be an angle and observe that. Hence: cos cos cos sin. This, together with the identity cos sin, shows that, for cos : cos cos cos cos cos cos cos cos. To find which of the two roots of the right hand side we must use, we have to consider where that terminal point of the angle lies. Example. Find cos. the terminal side of this angle is in the first quadrant, so 8 cos. Also. Hence: 8 8 4

14 cos 8 cos 4. Therefore, since cos, 8 How about sin? cos 8 cos cos sin sin sin. sin sin cos sin cos Again, the sign of sin will depend on the terminal point of the angle. For example sin 8,sin 8 cos 4. We conclude that: sin 8. Exercises 4.. Find sin 5,cos Let be an angle. Find cos,sin in terms of cos and sin Let be an angle. Find cos,sin in terms of sin and cos. 4. If, find sin and cos in terms of cos. 5. If, find sin and cos in terms of cos. 6. Find sin,cos Find sin 5 5,cos. 8. Find sin,cos Find sin cos

15 . Find sin 5 5 cos.. Find all angles (measured in radians) such that sin.. Find all angles (measured in radians) such that cos.. Find all angles (measured in radians) such that sin. 4. Find all angles (measured in radians) such that cos. 5. Prove that sin 4t 4sin t cos t sin t. 6. Prove that cos t cos t cos t. 7. Prove that cos t cos t 4cos t. cos 8. Show that if then sin Show that if cos t sin t then sin t sin t cos t. sin t cos t. Show that cos t cos t sin t.. Prove that cos 5t sin t sin 5t cos 5t 4sin t. cos x. Show that cos x. cos x cos x 4. Simple harmonic motion and cosinor analysis We have discussed the addition and subtraction identities for the sine and cosine functions. As we have seen, one use of them is to compute the values of these functions for angles that can be written as a sum or difference of angles for which we already know the values of sine and cosine. One very used fact is that: cos cos cos sin sin. This usefulness follows from the following reasoning: Many functions appear in the following form: if B,C are real numbers which are not both zero (in mathematical symbology: if B C ) if, and if D is a real number, consider the function f : R R, f t Bcos t Csin t D. An important question is: how does it graph look like? We re-write the function in the following form: f t B C B B C cos t C sin t D. B C Now we see that the point in the plane B, C B C B C so there exists an angle, such that cos Then we have: is a point in the unit circle, B,sin C. B C B C 5

16 f t B C cos cos t sin sin t C B C cos t C B C cos t C B C cos t C. If we set A B C and t, we get: f t Acos t t C. This proves that the graph of any such function is the graph of a cosine function which is multiplied by a positive constant and translated both horizontally and vertically. The coefficient A is called the amplitude can be regarded as the "amount of variation " between the highest and lowest points in the graph, the period p indicates the length of the intervals over which repetition occurs. The period p is determined in the following way, recalling that is the smallest positive period of the cosine function: f t p f t,t R Acos t p t C Acos t t C,t cos t t p cos t t,t R. From the fact that is the smallest positive period of cosine, it follows that: p p. The number p is called the frequency and indicates the rapidity of "oscillation ", the number t is called the phase shift and shows how far away from zero is the horizontal shift and C indicates how far up or down the function has been shifted. We first will treat some examples to clarify the specific terms. Example. Consider the function f : R R,f t cos t sin t 6. Here we have that A 4 6 4, so our first step is to write it as: f t 4 cos t sin t 6 4 cos 6 cos t sin 6 sin t 6 4cos t 6 6 4cos t 6 6

17 Example. Let f : R R,f t 4cos t 4sin t. A\ f t 4 cos t sin t Since the cosine is negative and the sine is positive the angle has terminal point in the second quadrant, so the angle we choose is 4. f t 4 cos 4 cos t sin 4 sin t 4 cos t 4 4 cos t 4 y x -.5 Graph of last example In biology, more particularly, in the study of circadian rhythms, when the data indicates that the function being studied seems to be of sinusoidal type, that is, it seems to reflect the graph of a shifted cosine function, the term "cosinor analysis " is used. Of course, the first task is to determine if the periodic function being studied seems to be sinusoidal or if it clearly is not. The following graphs illustrate this point: 7

18 y x Non-sinusoidal graph y sinusoidal plot When the phenomenon being studied appears to be sinusoidal, then the biologists usually try to fit a function of the type f t Acos t t M. P This is what is called "cosinor analysis ", a special case of the "time series " technique. In this setting, M is called the "mesor ", an acronym for "midline estimating statistic of rhythm ", A is the amplitude, P is the period and t is called the "acrophase " P which measures the timing of the maximum for the cosine function (or, how far away x 8

19 from zero does the first maximum occur). The plots of periodic functions in the Alzheimer s study presented in the introduction to periodic functions. Of course, once the determination is made that cosinor analysis is the appropriate tool to use, then data (obtained from observations or experiments) is used to determine the mesor, amplitude, period and acrophase. This is not an easy task and researchers work hard to determine them by using statistical and curve fitting techniques that are beyond the scope of this course; the objective in presenting this here is to show how some periodic functions are studied when they arise in applications to biology. We will content ourselves with dealing with functions that are presented to us and interpreting the terms that appear using this terminology. For example, in the article Age-dependent Changes in 4 hour rhythms...by P. Cano, D.P. Cardinali et al, which can be accessed through BioMed Central ( one of their sets of data yields the following information: Mesor: 49, amplitude 69, Acrophase.5 hours, period 4 hours. Thus the function they arrived at is: f t 69cos t y x Plot of Cano et al We have already discussed the setting in which biologists do cosinor analysis, and we have discussed that it is functions of the type f : R R given by: f t Acos t t M. P 9

20 We will always choose t, that is the acrophase is always nonnegative, and always A,P. Notice that in this setting, the maximum value Q attained is Q A M while the minimum value q is q A M Notice that Q q A A Q q. Observation: the mesor is the number that is the midpoint of the interval determined by the minimum value and the maximum value. Example. Consider the function f t cos t / 7 5 (time measured in days). Determine the amplitude, acrophase, period and mesor. Solution. This can be read directly from the function: amplitude, acrophase /, period 7, mesor 5. Example. Data determine that a certain phenomenon is a sinusoidal function of time and cosinor analysis is to be used. The period is one day, the maximum value attained is, the minimum value is and the value at is 5. Determine the function that is searched for. Solution. From the data, A,M,P, so the function is of the form: f t cos t t. We only have to find t. Notice that f cos t 5. Thus cos t cos t. since we are searching for the smallest positive number for which cos t, we conclude that t t 6.

21 The function we are searching for is: f t cos t. 6 The acrophase is 6 of a day. Problems.. Given the function f t cos t sin t 5, write it as f t Acos t t D and state the amplitude and period.. Same for the function f t cos t sin t 5.. Same for the function f t cos t sin t Analysis of data has determined that the cosinor method leads to the function t f t 5cos 6,t in days. 5 Determine the amplitude, period, acrophase and mesor. 5. Same for the function f t 7cos t / Data shows that a certain phenomenon can be modeled using the cosinor method and that the maximum value is, the minimum value is, the period is two hours and the value at zero is 5 6. Find the function, determine the acrophase. 7. Same for: maximum value 4, minimum value is, the period is three hours and the value at zero is. 8. A circadian rhythm is being modeled with cosinor analysis. The period is four hours, the amplitude is 8, the minimum value is 9 and the value at zero is 7. Find the function determined by this method. 9. If there is a function determined by the cosinor method and you know the amplitude and minimum value, how can you find the mesor?. If there is a function determined by the cosinor method and you know the amplitude and maximum value, how can you find the mesor? 4.4 The other trigonometric functions We will enter a brief discussion of the functions that you have called tangent, cotangent, secant, cosecant. All of these functions are obtained using quotients where either sine or cosine in the denominator. Since division by zero is mathematical anathema, we need to find the sets which will be the appropriate domains for these functions.

22 Notice that cos when k,k Z. We define: D x R : there is no k Z such that x k. Also sin when k,k Z. we define: D x R : there is no integer k such that x k. We define:. tangent, tan : D R,tan x sin x. cos x Its graph looks like: y x -5-5 For example: Graph of a portion of tangent tan 4 sin 4 cos 4 tan 6 sin 6 cos 6 tan sin cos.. cotangent, cot : D R,cot x cos x. sin x. secant, sec : D R,sec x. cos x

23 y x -5 Part of the graph of secant 4. cosecant, csc : D R,csc x sin x. Some identities: Recall that sin x cos x sin x cos x cos x tan x sec x. Also recall that sin sin cos cos sin, cos( cos cos sin sin. Let, be angles such that tangent is defined on both of them as well as on. Then: sin tan cos sin cos cos sin cos cos sin sin. tan tan tan tan. Another important observation is that tan x tan x. Exercises 4.4

24 . Find tan 5.. Find sec 5.. Show that if tan and tan are defined, then tan sin cos. Findtan Find sec Show that if 4 k,k an integer, then sec cos 6. Find all angles (measured in radians) such that tan. 7. Find all angles (measured in radians) such that tan. 8. Find all angles (measured in radians) such that tan. 9. Find all angles (measured in radians) such that tan.. Find all angles (measured in radians) such that tan.. An angle has terminal side in the second quadrant and tan. Then sin?. Prove that tan t sec t sec t sec t. Which values of t must be excluded?. Prove that sec t sec t. Which values of t must be excluded? sec t 4. Prove that tan t sec t sec t. tan t cot t 5. Prove that. sec t csc t cos t sin t 6. Prove that tan t sin t sin4 t. cos t 7. Prove that cot t csc t sin t. cos t cos 8. Prove that t sin t sin t. cos t sin t 9. Find sec. 8. Find tan. 4.5 Solving triangles In our incursion into periodic and trigonometric functions, our emphasis has been, so far, in determining functions based on the cosine in order to study periodic functions which often appear in biology. One traditional use of trigonometry is in "solving triangles ", by which we mean given data from a triangle, determine the rest of the characteristics of the triangle. Of course the scope is too broad as stated and we will concentrate on some techniques that will provide answers when the right information is available. These problems also find many uses in biology (and medicine) but the angles that 4

25 appear require, almost always, the use of calculators (or tables), a feature we are consciously avoiding since our interest is in the mathematical thought behind the techniques, not in button pushing on a calculator nor the use of specialized software,something that almost anyone can do without knowing the principles involved. When dealing with triangles, the information we need or seek takes the form of the length of the sides and the measurement of the angles made by two sides that intersect. These triangles are usually drawn without the use of a coordinatized plane but, in order to prove some of our assertions we will make use of the Cartesian coordinates and will position the triangle with one of the vertices at the origin in order to facilitate the proofs. The facts that will be taken for granted in this discussion are: the angles in a triangle are always assumed to be positive, the sum of the angles in any triangle is radians (8 ), similar triangles have the same corresponding angles and all the quotients of corresponding sides are equal. We usually refer to the "side opposite the vertex " or the "side opposite the angle " ; we distinguish between right triangles (one angle measure radians), acute triangles (all angles measure strictly between zero and and obtuse triangles (one angles is greater than. if we do not want to specify the type of a triangle that is not a right triangle we will call it an "oblique " triangle. We assume that you know how to compute areas of triangles. When considering right triangles, the sides that meet in a right angle are called the "legs "of the triangle and the side opposite the right angle is the hypothenuse (and Pythagoras Theorem applies). Given a triangle, we assume it is well known what the height of a vertex is, what the medians are and that all these set of three lines intersect at one point. Suppose that we have a right triangle with vertices A,B,C, C the vertex where the right angle is c is the hypothenuse and a,b are the legs. Let be the angle at vertex A If we set this triangle in the Cartesian plane, with vertex A at the origin, vertex C in 5

26 the x axis and the hypothenuse (or its extension) intersecting at point Q in the unit circle ( so it is the terminal point of the angle ), we see by similarity of triangles, that: cos a c sin b c tan a b. Example. We have a right triangle with one angle measuring radians and with 6 opposite side measuring 5 inches. Find the length of the hypothenuse. Solution. Let b be the length of the other leg (in the same units). Then: tan 6 5 b b 5. Now, we know that the hypothenuse is equal to the square root of the sum of the squares of the legs (Pythagoras Theorem), so if we denote by h the length of the hypothenuse is: h inches. Note. Once you choose the units, the results will always be the same numerical value with that unit system. There are many problems involving right triangles and all of them are dealt with the same ease (be it the height of a tree when its shadow has a given length and the angle of incidence of the sun light is given, if the length of a ladder is given together with the information that it rests on a vertical wall and the distance from the base of the ladder to the wall is also given, etc. etc.) Now we will concentrate on triangles that are not right triangles. Law of Sines. Consider a triangle with vertices A,B,C, corresponding angles (corresponding to vertex A),,, and sides a (opposite to A ),b,c. Then: sin a sin sin b c. Proof. We will illustrate a rigorous proof using a particular triangle, as follows: 6

27 We see that, if h is the length of the perpendicular line from vertex B to the side AB, then: h csin h asin. This implies that csin asin sin c sin a. Now consider the perpendicular (height) from C to the side (or its prolongation) to the side AB, call it h : From this we get that: From this we get: h bsin h csin. 7

28 It follows that: sin sin b c. Putting all our equalities together, we get: sin a sin b sin c. There are several situations where this result allows us to "solve " for the triangle given. We illustrate them with examples. Example. (One side and two angles given) A triangle ABC is such that a 5,,.Find b,c. 4 Solution. Since 7 7 and 5, we need to compute sin sin sin cos cos sin Thus: sin a sin b b b. sin a sin c c c c 5 6. When two sides and an opposite angle are given, then we may have that the "opposite angle " is too big for the remaining side and then no solution exists, as illustrated below: 8

29 On the other hand there could be two triangles that meet our data, as illustrated below: Or, it could happen that there is only one solution. b. Example. For the triangle ABC we are given:,a,c 7. Find, if possible, 6 Solution. From the law of sines we get: sin 6 sin 7 sin 7 6. No triangle will meet ours specifications. There is no solution. 9

30 Example. Solve the triangle ABC with the following specifications:,a,c. 6 Solution. sin 6 sin sin sin. There are two angles between and with this value of sine, namely 4. Consider first. 4 4 Thus 5, 6 4 sin sin 6 b b b b. If we use, then. sin sin sin sin sin sin cos cos sin sin 6 sin b. b b.. and Example 4. Solve the triangle ABC given that a 5,,c 5. Solution.

31 5 5 sin sin 5 5 sin. In this case we have only one solution, with, and so 6. Thus b sin 5 b 5. Law of Cosines. Consider a triangle with vertices A, B, C, corresponding angles,, and opposite sides a,b,c. Then: a b c bccos. Proof. For this proof we will put the triangle in the Cartesian Plane, with vertex A at the origin (coordinates,,c on the x axis (coordinates b, and vertex B in the upper half plane ( coordinates x,y ). Here is a picture of a particular case: We see that:

32 Thus: x ccos y csin. a ccos b csin a c cos cbcos b c sin c b cbcos. This proves the result. Now we only have to familiarize with some of its applications; observe that the result is very dependent on our labeling of vertices, angles and sides. a. Example. Given the triangle ABC where b 5 cm., c cm. and 6,find Solution. From the law of cosines we get that: a 5.cos 6 5 a a 875,a a cm. Exercises 4.5. Suppose that a triangle is such that a,c, 6. Then b?. From the roof of a building, an observer can see the top of a skyscraper with an angle of elevation of radians and the base of the skyscraper with an angle of 4 depression of. if the two buildings are feet apart. Find the height of the 6 skyscraper.. Let ABC be a triangle such that b 5,c,. Find a,sin and sin. 4. Let ABC be a right triangle ( ) such that and a. find,b,c Let ABC be a triangle such that a,b,. Find c. 6. Let ABC be a triangle such that a,,. Find b,c, Solve, if possible, the triangle ABC given that,a 4,c (if there are no solutions, prove it, if there are two solutions, find both). 8. Given a triangle ABC such that,b,c, find a Inverse trigonometric functions

33 We will only consider the functions sin,cos,tan. We know that sin,cos are periodic of period and that the domain of tan is D, and that, D. The conclusion from these preliminary observations is that none of these functions have an inverse function in the domain in which they were defined. It is, however, very useful to find angles with given values of one of their trigonometric functions. In order to do this we will restrict the domain of each of these in order to make this objective possible. Let us first consider the function sin. If we restrict it to the new domain,, then we see that all values of the function (that is, its range) are covered by values of the function on this interval and it is very intuitive, from the unit circle representation of the points cos.sin that the function is one-to-one on this interval. Thus we consider the function sin restricted to the domain,. that is we now consider sin :, with (approximate) graph: y x -.5 Graph of sin Thus, we accept that there is, for this particular case, an inverse function sin :,,. Thus sin 6,sin,sin on ( no addition of multiples of ). -, and so

34 y x Graph if sin Since the restriction we are considering of the function sin is the sine function as defined before (with a different domain), we always have that sin sin x x for x,. However, we must be careful with the reverse composition, as the following examples show: Example. Find sin sin. Solution. Since, we must find an angle whose radian measure is between and that has the same sine. Since, we conclude that : sin sin. Example. Find sin sin. 6 Solution. Since does not belong to,, we must search for one with radian 6 measure in this interval and with the same value of sine. Observe that 6 6 we immediately see that: sin sin 6 6. Example 4. Find sin sin 5. Solution. We must proceed as above. Now we observe that 5, concluding that: sin sin 5. Now we concentrate on the function cosine. For our purposes we restrict it to the 4

35 interval, and consider it as a function from, onto,, and we have the following approximate graph: y x - - Graph of the restriction of cosine We see that it is reasonable to infer that the function is one-to-one, and we will make this assumption (which is valid, as a matter of fact). Thus we get the inverse of this restriction of cosine and write it as cos :,,. y Graph of cos We observe that cos 4,cos 4,cos,cos, 6 and so on. x 5

36 Again we always have that: cos cos x x. However we must be careful with cos cos, and remember that the angle we get belongs to the interval,. Example 5. Find cos cos. Solution. Since cos cos, we have: cos cos cos cos. Example 6. Find cos cos 5. Solution. 5 so we conclude that: cos cos 5 cos cos cos cos. Example 7. Find sin cos. 5 Solution. Since cos 5,,sin cos. 5 sin cos 5 cos cos 5 sin cos sin cos 5 5 sin cos 5 5. As it regards the function tangent, we restrict it to the interval, that its range is R. Thus tan : R,. and recall 6

37 y x tan 4,tan Graph of tan and so on. Example 8. Find tan tan. Solution. The terminal point of is in the second quadrant, so the value of tangent is negative. We conclude that: tan tan tan tan tan tan. Example 9. Find sin tan 6. Solution. We know that tan 6, (why?), so both cosine and sine of this angle are positive. 7

38 tan tan 6 6 sin tan 6 cos tan 6 6 sin tan 6 6 sin tan 6 sin tan 6 6 sin tan 6 7sin tan 6 6 sin tan sin tan There are many other examples we could show you, and some will be included in the exercise set that follows. We now concentrate on some of the uses of these functions when dealing with cosinor analysis and combinations of sine and cosine. Example. A circadian rhythm of period 4 hours has been determined to fit the cosinor analysis method. The maximum value is 75, the minimum value is 5 and the value at zero is. Find the amplitude, mesor and acrophase and show the function obtained. Solution. f t Acos t t M 4 Acos t t M A ,M 4, 8

39 f t 5cos t t 4 f 5cos t 4 5cos t 8 cos t 8 5 t cos 8 5 t cos 8 5 f t 5cos t cos Example. Consider the function f t cos t 5sin t 5. write it in the form f t Acos t t 5. Solution. f t 4 4 cos t 5 4 sin t 5 The point, 5 is in the fourth quadrant so we choose the angle 4 4 cos,cos,sin 5, concluding that f t 4 cos t cos 4 5 f t 4 cos t cos 4 5. Exercises.. Find sin sin 6.. Find cos cos 8.. Find sin cos. 4. Find cos tan Find sin cos Find cos sin. 7. A given circadian rhythm can be represented using the cosinor method, it has period of 4 hours a maximum value of 9, a minimum value of and the value at time zero is. Find the amplitude, mesor and acrophase. 8. A given circadian rhythm can be represented using the cosinor method, it has a 9

40 period of hours, maximum value of, minimum value and at zero its value is. Find the amplitude, mesor and acrophase. 9. Write the function f t 4cos t 4sin t 6 in the form f t Acos t t 6.. Write the function f t cos t sin t 5 in the form f t Acos t t 5.. Find tan sin. 5. Find sin sin cos 5.. Prove that if t, then sin t cos t. 4. Prove that if t R then tan t sin t. t 5. Solve if possible, the triangle ABC given that,a,b (if there are no solutions, prove it, if there are two solutions, find both). 6. Solve, if possible, the triangle ABC given that,a 5,c (if there are 6 no solutions, prove it, if there are two solutions, find both). 6. Solve, if possible, the triangle ABC given that,b 5,c (if there are 6 no solutions, prove it, if there are two solutions, find both). 7. Consider the function f : R R,f t cos 5t 6sin 5t 5 in the form f t Acos w t t C,t o,. 8. Consider the function f : R R,f t 4cos 5t 6sin 5t in the form f t Acos w t t C,t o,. 9. Consider the function f : R R,f t cos t sin t in the form f t Acos w t t C,t o,.. Consider the function f : R R,f t 5cos t sin t 5 in the form f t Acos w t t C,t o, 4