Chapter 8A - Angles and Circles

Transcription

1 - Chapter 8A Chapter 8A - Angles and Circles Man applications of calculus use trigonometr, which is the stud of angles and functions of angles and their application to circles, polgons, and science. We start with the definition of angles and their measures. Angles Roughl, an angle is the space between two ras or line segments with a common endpoint. The ras or line segments are called the sides and the common endpoint is called the verte. More precisel, if a ra or line segment rotates about an endpoint from some initial position, called the initial side, to some final position, called the final side, then the angle between the sides is the space swept out. Within a plane, we sa the angle is positive if the rotation is counterclockwise, and is negative if the rotation is clockwise. An angle is in standard position if the verte is at the origin and the initial side is along the positive -ais. θ θ A positive angle An angle in standard θ A negative angle Measures of Angle The size of an angle ma be measured in revolutions (rev), in degrees ( ) or in radians (rad). An angle is called a full rotation if the ra rotates from the initial side all the wa around so that the final side coincides with the initial side. A full rotation is measured as rev 6 rad. An angle is called a half rotation or a straight angle if the ra rotates from the initial side to a final side which is directl opposite to the initial side. A half rotation is measured as rev 8 rad. An angle is called a quarter rotation or a right angle if the ra rotates from the initial side to a final side which is perpendicular to the initial side. A quarter rotation is measured as rev 9 rad. An angle is called a null rotation if the ra never rotates so that the final side coincides with the initial side. A null rotation is measured as rev rad.

2 - Chapter 8A Several other important angles are: rev 7 rad rev 6 rad rev 6 6 rad rev 5 8 rad Eample : Find the angular measure of one time zone on the surface of the earth. Solution: The earth rotates once a da or b rev in hours. Thus a hour time zone has an angular measure of rev 5 rad. An angle will be bigger than a full rotation if the ra rotates from the initial side all the wa around and past the initial side again. There is no limit to the size of an angle either positive or negative. Two angles with the same initial and final sides are called coterminal and their measures must differ b an integral multiple of rev 6 rad. Below are some pictures of various angles. In each figure, the initial sides of the angles coincide. 75 o 8 o 5 o Positive angles in degrees 5π π π Positive angles in radians 57 o -5 o - o Other angles in degrees 7π 6 π 5π Other angles in radians Below are some pictures of various angles in standard position. 5 o 7π 6 5 o π 8 o o 5π 5 o o Positive angles in degrees Positive angles in radians Other angles in degrees

3 - Chapter 8A 8π π Other angles in radians π 6 You must learn to identif angles in standard position in radians and degrees. These are the angular measurements most often used in trigonometr and calculus, with radians predominating in calculus. In the following either degrees or radians will be randoml used. Definition: An angle is called acute if its measure is between and 9, and an angle is called obtuse if its measure is between rad and rad. Definition: The complement of an angle whose measure is is an angle whose measure is (in radians), and the supplement of an angle is an angle whose measure is 8 (in degrees). The Parts of a Circle Definition: The circle with center P and radius r is the set of all points X in the plane whose distance from P is r. If the center is P a,b and the general point on the circle is X, then the equation of the circle is a b r We often take the center to be the origin O,. Then its equation is r The region inside of a circle is called a disk. Definition: A radial line (or a radius) is an straight line segment from the center of the circle to a point on the circle. The word radius can refer to either a radial line or its length r. Definition: A diameter line (or a diameter) is an line segment between two points on the circle which passes through the center. The word diameter can refer to either a diameter line or its length d r. Definition: The circumference of the circle is the distance around the circle. B the definition of the circumference is C d r. Definition: The area of the circle (actuall of the disk) is A r. Definition: An arc is an piece of the circle between two points on the circle. A chord is an line segment between two points on the circle. A sector is an piece of the disk between two radial lines.

4 - Chapter 8A Two radial lines are shown in black.. Anarcisshowninblue. A chord is shown in green. A sector is shaded in can. Radial lines Sector Definition: An angle whose verte is at the center of a circle is called a central angle. The sides of a central angle are radial lines which intersect the circle at two points. The arc between these two points, the chord between these two points and the sector between the two sides of the angle are called the arc, chord and sector subtended b the central angle. We use the following notations: C Circumference of the circle L Length of an arc of the circle A Area of the whole circle A Area of a sector of the circle In the previous figure the arc, chord and sector are subtended b the central angle. The fraction of the circle or disk subtended b a central angle is 6 for in degrees, or rad for in radians. So the length of an arc (the arc length) is this fraction of the circumference and the area of a sector (the sector area) is this fraction of the area of disk. Hence the arclength is L 6 C r 6 for in degrees rad C r r for in radians and the sector area is A 6 A r 6 rad A r r for in degrees for in radians Notice that the formulas for arc length and the area of a sector are much simpler when written in terms of radians than in terms of degrees. The formulas for angular velocit and angular acceleration used in phsics are also simplier using radians. This is one of the reasons for using radians rather than degrees, and when ou learn about derivatives of the trig functions in calculus, ou will appreciate radians even more. Eample : In the figure above, suppose the radius is r 6cm and the central angle is 5 7 rad. Find the circumference and area of the circle. Find the fraction subtended as well as the arc length and sector area of the arc and sector subtended b the central angle. Solution: The circumference is C 6cm cm. The area is A 6cm 6cm. Using degrees: The fraction subtended is cm5. The arc length is L 6 7 cm. And the sector area is A 6cm 5 6 cm. 7rad Using radians: The fraction subtended is rad 7. The arc length is L 6cm 7 7 cm. And the sector area is A 7 6cm cm. Notice how much simplier the computations are in terms of radians.

5 - Chapter 8A Secant line, Tangent line Definition: chord. Definition: of tangenc. A secant line is a line which intersects the circle twice. The part inside the circle is a A tangent line is a line which intersects the circle at eactl one point called the point A secant line is shown in purple. Its chord is shown in green. A tangent line is shown in blue. Its point of tangenc is in ellow.

6 - Chapter 8A Eercises for Chapter 8A - Angles and Circles. If a point on a disk rotates about the disk s ais 5 times, how man degrees will it rotate, how man radians?. An angle of 5 equals how man radians?. An angle of radians is how man degrees?. If a disk is rotating at 5 rotations per minute, how man radians will the disk rotate in minutes? 5. A circular race track is being built. The outside diameter of the track is ft. and the inside diameter is 8 feet. If two runners run around the track once with one of them running on the outside edge of the track, while the other is running on the inside edge, how much further will the outside runner have to run? 6. Referring to the runners in the preceding problem, if the runners both run at a speed of 6 miles per hour, how much time will elapse between the first runner finishing and the second runner finishing. Assume the start at the same time. 7. A child runs half wa around a circle. If the radius of the circle is meters, how far did the child run? 8. A dog is tied to a centimeter diameter pole, with a meter long leash. If the dog runs around the pole until his collar is tight to the pole, how man degrees has the dog run around the pole? 9. The apparent diameter of the planet Venus is, kilometers. Assuming that Venus is a sphere, what is its circumference?. The average distance of the earth to the sun is approimatel 5 million kilometers. Assuming it takes the earth 65 das to orbit the sun, that there are hours in a da, that its orbit is a circle, and that its speed is constant, what is the speed of the earth in kilometers per hour about the sun?. If a circle has a meter diameter, find the eact arc length and eact area of the sector subtended b a central angle of 9 degrees.. If a circle has a radius of 5 centimeters, find the eact arc length and eact area of the sector subtended b a central angle of 7 radians.

7 - Chapter 8A Answers to Eercises for Chapter 8A - Angles and Circles. Since one rotation is equivalent to 6 degrees or radians, 5 rotations equals 56 8 degrees or radians.. If is the measure of the angle in radians, then we have the following ratio: Thus, If is the measure of the angle in degrees, then Thus, In two minutes the disk will have rotated 9 times, or 9 8 radians. 5. The outer circumference is feet while the inner circumference is 8 feet. The difference between these two distances, feet, is how much further the first runner has to travel. 6. Since both runners are running at the same speed, we just need to determine how much time it will take to run feet at a speed of 6 miles per hour. First we convert the speed from miles per hour into feet per second: 6 miles/hour ft/sec 6 Since, see preceding problem, one of the runners has to run an additional feet, it will take him longer to run once around the track seconds 7. distance r meters. 8. Each turn of the dog around the pole uses cm or. meters of the dogs leash. Thus, number of revolutions 9.59 revolutions or degrees. 9. It s circumference 8,6 kilometers.. The distance the earth travels in one orbit about the sun is approimatel kilometers. The amount of time it takes for one orbit is hours. Thus, the speed of the earth in its orbit is

8 - Chapter 8A kilometers/hour.. Since the diameter is meters, here we have r and we are given that 9. Thus, the arc length is r meters, and the area of the sector is: r square meters. Here r 5 and 7 radians. Thus, the arc length is r r and the area of the sector is: centimeters r r square centimeters

9 - Chapter 8B Chapter 8B - Trig Functions Triangle Definition We are now read to introduce the basic trig functions. There are two consistent was to define them. The first uses a right triangle and is valid onl for angles strictl between and 9, while the second uses a circle and is valid for all angles. Consider a right triangle with one angle. The sides are: the leg adjacent to : adj the leg opposite to : opp and the hpotenuse: hp In terms of these sides, the trig functions, sine, cosine, tangent, cotangent, secant and cosecant, of the angle are given b sin opp hp cos adj hp tan opp adj cot adj opp θ hp adj sec hp adj csc hp opp opp In terms of sin and cos, the other trig functions are tan cos sin cot cos sin tan sec cos csc sin B using similar triangles it is eas to see that these definitions do not depend upon an particular right triangle. Once again we note that this definition of the trig functions is limited to angles which are acute, that is, less than a right angle and greater than. Below we give a definition which works for an angle and show that if we have an acute angle then both definitions give the same values. Circle Definition Consider a circle of radius r centered at the origin in which a radial line has been drawn at an angle measured counterclockwise from the positive -ais. The radial line intersects the circle at a point,. Note: and/or ma be negative. (,) r θ In terms of, and r, the trig functions, sine, cosine, tangent, cotangent, secant and cosecant, of the angle are given b.

10 - Chapter 8B sin r tan, sec r, cos r cot, csc r, Note that sin and cos are defined for all values of, but this is not true for the other trig functions. tan and sec are not defined when, that is, when is an odd integer multiple of /. cot and csc are not defined when. That is, the are not defined when is an integer multiple of. Notice that this sas the coordinates of a point on the circle are, rcos,rsin. In terms of sin and cos, the other trig functions are tan cos sin cot cos sin tan sec cos csc sin Using similar triangles it is clear that the definition of these trig functions does not depend upon the radius of the circle. That is, no matter what the circle s radius is, the values of these functions do not depend upon the value of r. Consider a small red circle of radius r and a big magenta circle of radius R. The radial line at an angle intersects the red circle at, and intersects the magenta circle at a point X,Y. (,) ( X,Y) R r θ Since the blue and green triangles are similar, the ratios of corresponding sides are equal. Hence: r R Y r X R But these equations sa the sine and cosine are the same for both triangles: sin r R Y cos r X R From the equalit of the sine and cosine functions, the equalities of the rest of the trig functions follow. Notice that this circle definition of the trig functions works for an angle not just acute angles. However, to make sense, the definitions in terms of circles and triangles must agree for acute angles and the do, as one sees b comparing the definitions when the angle is acute. Trig Functions on a Unit Circle It is often convenient to use a unit circle. Then the trig functions are given b sin tan sec cos cot csc and the coordinates of a point on the circle are, cos,sin.

11 - Chapter 8B Special Angles You need to know (or be able to figure out) the values of the trig functions for specific angles. That is those angles which are multiples of 6 rad or 5 rad. Below is a table of the values of the trig functions for some special angles. You should absolutel not memorize this table. Rather, in each case, ou should figure out the values of the trig functions using the circle definition and our knowledge of 6 9 and 5 right triangles. Trig Functions for Special Angles Deg Rad sin cos tan cot sec csc rad 6 rad 5 rad 6 rad 9 rad 6 9 triangle To determine the values of the trig functions for either or 6, draw an equilateral triangle with each side of length. Remember that an equilateral triangle is also equiangular, which means each of its angles is / of 8 or 6. Now draw a perpendicular from an verte to its opposite side. This gives ou the triangle show below, and the Pthagorean theorem tells us that the vertical side must have length. o = π 6 Notice that the shortest side is opposite the smallest angle 6 rad, the middle length side.7 is opposite the middle angle 6 rad 6 o = π and the longest side is opposite the biggest angle 9 rad. From this triangle we get sin 6, cos6,tan6,csc6,sec6, cot6. sin, cos,tan,csc, sec, cot.

12 - Chapter 8B 5 triangle 5 o = π 5 o = π A 5 right triangle must be an isosceles right triangle, which means the legs of such a triangle have the same length. In the picture to the left the are of length. Computing the values of the trig functions we get sin 5, cos5,tan5, csc5,sec5, cot5. Eample : Find the trig functions at 5 rad. Solution: Draw a circle of radius, draw the radial line at 5 rad and drop (raise) a perpendicular to the -ais. The hpotenuse of the resulting right triangle is since this is the radius of the circle. Moreover 5/ / 6. Since this is the IV th quadrant, the coordinate is positive and the coordinate is negative and we label the two legs of the triangle with and. 5π - Then the trig functions are - sin 5 r tan 5 cos 5 r cot 5 sec 5 r csc 5 r

13 - Chapter 8B Eample : Find the trig functions at rad. Solution: We draw a circle of radius, draw the radial line at rad and drop a perpendicular to the -ais. The hpotenuse of the resulting right triangle is since this is the radius of the circle, and we note that this must be an isosceles right triangle since / / 5. Since we are in the II nd quadrant, we label the two legs of the triangle with and. Then the trig functions are - π - sin r cos r tan cot sec r csc r Eample : Find the trig functions at rad. Solution: We draw a circle of radius and draw the radial line at endpoint of the radial line are and. rad. The coordinates at the π - So the values of the trig functions are: sin r cos r tan cot sec r csc r -

14 - Chapter 8B Eercises for Chapter 8B - Trig Functions. If sin. 55 and the angle is acute what are the values of cos and tan?. If sin. 55 and the angle is obtuse what are the values of cos and tan?. If cos. 5 and the angle is acute, what are the values of sin and tan?. If cos. 5 and the angle is not acute, what are the values of sin and tan? 5. If tan. 5 and is acute what are the values of sin and cos? 6. If tan. 5 and is not acute what are the values of sin and cos? 7. Using the Pthagorean theorem deduce that sin cos for all values of. 8. If cos and is in Quadrant IV, eactl find the value of sin, tan, csc, sec, and cot. 9. Give eact answers to each of these: a. cos b. tan 5 6 c. csc 5 d. sin e. sin 9 f. tan. Find the eact value of all the trig functions of the angle Find the eact value of all the trig functions of the angle.

15 - Chapter 8B Answers to Eercises for Chapter 8B - Trig Functions. Since the angle is acute, the value of its cosine must be positive. If, is that point on the unit circle, which corresponds to the angle, then we have sin.55. Moreover we have Thus, cos.85 and tan Since the value of the sine function is positive, if, is that point on the unit circle, which corresponds to the angle, then, which means that the point (, must lie in the second quadrant. That is,. Thus, using the value from the preceding problem, we have cos.85 and tan If, is that point on the unit circle, which corresponds to the angle, then.5 and Thus, sin.968 and tan Since the angle is not acute and cos is positive then the point (, must lie in the fourth quadrant. That, is with. Using the value computed in the previous problem, we have: sin.968 and tan Since is acute, the point, must lie in the first quadrant. That is both and are positive, and we know that tan.5 and. Thus, /, which implies that /. Thus, / 5,and / 5. Thus, sin 5.7 and cos Since tan is positive, and if, is that point on the unit circle determined b the angle, then and must have the same sign. Since is not acute, this means that both and are negative. Solving the same sstem of equations as in the preceding problem, we now have / 5 and / 5. Thus, sin 5.7 and cos 5.89

16 - Chapter 8B 7. Let, be that point on the unit circle determined b the angle, then, as and arethe lengths of the legs of a right triangle whose hpotenuse has length, we have cos sin. 8. We recall that if we are looking at a right triangle, cos adj hp Thus, in our right triangle, the length of the side adjacent to the angle is, the length of the hpotenuse is, and we can use the pthagorean theorem to find the length of the side opposite the angle: opp But since we are in Quadrant IV, we know that opp. Thus, we can use this triangle to find the values of our other trig functions: sin opp hp,tan opp adj,csc sin sec cos, and cot tan 9. a. cos b. tan 5 6 tan 6 tan 6 c. csc 5 csc csc sin / d. sin sin sin e. sin 9 sin sin f. tan sin/ cos/. We see here that 7. Thus, 6 6 cos 7 cos 6 6 sin 7 6 sin 6 tan 7 6 tan 6

17 - Chapter 8B sec 7 6 cos 7 6 csc 7 6 sin 7 6 cot 7 6 tan 7 6. We see here that. Thus, cos cos sin sin tan tan sec cos csc sin cot cos sin

18 - Chapter 8C Chapter 8C - Graphs of the Trig Functions If we compute and plot the values of the si trig functions, for those values of in radians for which the computations are eas, we get the following si plots. Remember that is in radians. Asmptotes π 5 6 π θ π 5 6 π θ 5 6 π π θ sin cos tan Asmptotes Asmptotes Asmptotes π π θ π π θ π π θ cot sec csc If we add more points, or use a graphing calculator, we get the following si plots. Asmptotes π 5 6 π θ π 5 6 π θ π π 5 6 θ sin cos tan

19 - Chapter 8C Asmptotes Asmptotes Asmptotes π π θ π π θ π π θ cot General Graphs sec csc There are three was we need to clarif our graphs of the trig functions. First, although calculators will compute trig functions using either degrees or radians, mathematicians almost alwas use radians when working with the trig functions. The reasons for this will become clear after ou learn about the derivatives of the trig functions. Second, in the circle definition of the trig functions, there is no reason that the angle is restricted to revolution. So the trig functions are also defined for angles less than rad or greater than rad 6.8rad. Since the coordinates of the point on the circle repeat with each revolution, the graphs of sine, cosine, cosecant, and secant repeat ever rad. We denote this b saing that these four trig functions are periodic with period. Third, mathematicians like to use the letter for the independent variable and the letter for the dependent variable of a function. So we often write sin, or cos, or tan, oretc. Here is the angle (in radians) and is the value of the particular trig function. Caution: The and in a formula such as sin have absolutel nothing to do with the and in the circle definition of the trig functions. With these modifications, the graphs of the si trig functions become: sin cos Notice that sin and cos have period, while tan and cot (below) have period. In the net four trig functions note the vertical asmptotes. Those for tan and sec are at odd multiples of ; those for cot and csc are at multiples of tan cot

20 - Chapter 8C sec csc Notice that sin, cos, sec and csc all have period, while tan and cot have period. Also recall that tan and sec have vertical asmptotes at the odd multiples of while cot and csc have vertical asmptotes at the multiples of. Shifting and Rescaling Graphs The trig functions ma also be shifted and/or rescaled. We demonstrate using the sin function. We start with the graph of sin : Shifting As an eample, if the sine function is shifted to the right b.785andupb graph is then the shifted which has the equation sin. In general, if the sin function is shifted to the right b andupb then the shifted graph is which has the equation sin. If the horizontal shift is positive, the shift is to the right and we sa that the graph of sin leads sin b. If is negative, the shift is to the left and we sa that the graph of sin lags behind sin b. If the vertical shift is positive, the shift is up. If is negative, the shift is down.

21 - Chapter 8C Rescaling As an eample, if the sine function is epanded verticall b and contracted horizontall b then the rescaled graph is which has the equation sin. In general, if the sin function is epanded verticall b A and contracted horizontall b c then the rescaled graph is which has the equation Asinc. If the horizontal scale factor c, the graph is contracted horizontall. If c, the graph is epanded horizontall. If c, the graph is reflected horizontall and contracted or epanded b a factor of c. If the vertical scale factor A, the graph is epanded verticall. If A, the graph is contracted verticall. If A, the graph is reflected verticall and contracted or epanded b a factor of A. Shifting and Rescaling The function sin ma be shifted and rescaled at the same time. For eample, the graph of sin is: which is epanded verticall b, contracted horizontall b, shifted to thr right b b. and shifted up In general, the graph of Asinc is: which is epanded verticall b A, contracted horizontall b c, shifted to the right b and shifted up b.

22 - Chapter 8C Interpretation The vertical shift gives the average value of the function Asinc and the line is called the center line of this function. The absolute value of the vertical scale factor A is called the amplitude and gives the maimum distance the function goes above and below the centerline. The horizontal shift gives the distance the function Asinc leads or lags behind sinc. The absolute value of the horizontal scale factor c gives the angular frequenc which is related to the period P and the frequenc f b the equations f P c. Finall, the quantit c is called the phase of the wave and the product c is called the phase shift of the wave. So the phase shift c gives the amount the phase of Asinc leads or lags behind the phase of sinc. Using these various definitions, if c (for simplicit), the general shifted and rescaled sin curve can be rewritten in several other was: Asin Asin Asin P Asin P

23 - Chapter 8C Eercises for Chapter 8C - Graphs of the Trig Functions. Graph sin and sin on the same aes. If ou could pick up the plot of sin and move it around, how should ou move this plot so that it las right on top of the plot of sin?. Compute the value of sin for, /, /6, /, /8, /, and /. Then, knowing what the graph of sin looks like, graph the function sin on the interval,.. The graph of a function f looks like the graph of the function sin ecept that it takes on the value when. What could f equal?. The graphs of sin and cos look like each other ecept that the are out of phase. Conjecture a relationship between the sine and cosine functions. 5. Sketch the graphs of the functions sin, sin /, and sin / on the same plot. Pa attention to the correspondence between the plots and the sign of the shift /. 6. Sketch the graphs of cos and cos. Conjecture a relationship between these two functions. 7. A scientist recorded data from an eperiment. She noticed that the data appeared to be periodic with period and had a maimum value of 5, which occurred at. After staring at the data for some time, she decided to use a cosine function to model the data. What was the form of the function she used? What would the form be if she had decided to use a sine function? 8. If f 8sin 6, what is the amplitude, period, phase shift, and vertical shift? 9. If f cos 9, what is the amplitude, period, phase shift, and vertical shift?. Write a function of the form f asink b c whose graph is given, where a, k, andb are positive and b is as small as possible

24 - Chapter 8C. Match each function with its graph below: a. sin b. sin / c. sin / d. sin e. sin / () () () () (5)

25 - Chapter 8C Answers to Eercises for Chapter 8C - Graphs of the Trig Functions. Moving the plot of sin one unit to the right or units to the left will place it on top of the plot of sin.. sin, sin 6,sin,sin 8, sin, sin sin. f sin. cos sin 5. sin black, sin / red, sin / green 6. cos cos - -

26 - Chapter 8C 7. f 5cos or f 5sin 8. The amplitude is 8, the period is, the phase shift is, and the vertical shift is The amplitude is, the period is, the phase shift is, and the vertical shift is 9.. 5sin 8. a. () b. () c. () d. (5) e. ()

27 - Chapter 8D Chapter 8D - Trigonometric Identities There are man identities of an algebraic nature between the trigonometric functions. Some of them involve relationships between the different functions and some involve the same function but with sums or differences of their arguments. Two eamples of this are: sin cos andsin sin cos sincos, respectivel. In the following pages we will list some of the more useful trig identities and give their proofs. Man more identities can be found in the eercises. The first series of identities we will list involve negative angles and complementar angles. Negative Angle Identities: sin sin tan tan sec sec cos cos cot cot csc csc These identities follow from the fact that if, is that point on the unit circle, which corresponds to the angle, then the point which corresponds to the angle is (,. So, for eample, we have sin sin tan tan Note that sin, tan, cot, and csc are odd functions, while cos and sec are even functions. Complementar Angle Identities: sin cos tan cot sec csc cos sin cot tan csc sec Thus, the trig function of the complementar angle is equal to the complementar trig function of the original angle. To prove these complementar angle identities it suffices to verif that cos sin is true. The corresponding identit for sin follows from the one for cosine, and the other four follow from these two. To see that cos sin is valid, we first observe that the identit is valid if or/. So let s assume that /. Upon eamining the figure below we see that sin and that this is the value of cos/. Similar arguments verif this identit for

28 - Chapter 8D values of larger than /. The corresponding identit for sine follows from: sin/ cos/ / cos. Pthagorean Identities: The trig identit sin cos is called a Pthagorean identit as it follows from the Pthagorean theorem. To see this, let (, denote that point on the unit circle determined b the angle. Then sin and cos. Moreover, and are the lengths of the legs of a right triangle whose hpotenuse has length. Thus, sin cos. There are two other such identities, and the are obtained from this one b dividing this equation b cos or b sin. Thus, sin cos sin cos cos sec tan sec. Now divide b sin. cos sin sin csc cot csc Supplementar Angle Identities: The following identities are eas to prove as the follow from the corresponding complementar angle identities, and this is demonstrated below for one of the identities. sin sin tan tan sec sec cos cos cot cot csc csc sin sin cos cos sin. Be sure ou can eplain wh each of the equalities in the above lines is correct. Sum of Two Angles Formulas: sin sin cos cossin cos coscos sin sin tan tan tan tan tan The proofs of these identities are more complicated, but as usual once we have the identit for sine and cosine, then the two angle sum formula for the tangent function follows.

29 - Chapter 8D We give a proof for these formulas for the special case where and are acute and so is their sum. That is, we assume,, and /. The figure below was constructed as follows: Lines, which are here denoted b BP, BQ, andbr were drawn so that angle PBQ and angle QBR, which means that angle PBR. Point A on line BR was chosen arbitraril, and a line from A perpendicular to line BP was drawn with point G being its intersection with line BP. A similar line was drawn from D to line BP with C its point of intersection. Line DE was then drawn perpendicular to DC with E being its point of intersection with AG. Note: triangles DAE and DBC are similar, EG DC, andgc DE. We now have sin AG AB AE AE EG AB AB DC AE AB EG AB AE AB AD AD DC AB BD BD AD AD BD BD AB sin sin sincos sin cos sin cos. cos BG BA BC BC CG BA BD DE BA BD BA AD coscos sinsin. BC BA DE BA AD BD BC BD BA DE AD AD BA The other cases for various values of and can be shown b using the above identities for these restricted values of and and the complementar trig identities we ve alread verified. See the eercises. Difference of Two Angles Formulas: These identities follow from the Sum of Two Angles Formulas as we show for the sine function. sin sin cos cossin cos coscos sin sin tan tan tan tan tan sin sin sin cos sin cos sincos sincos.

30 - Chapter 8D Double Angle Formulas: The proofs of these formulas, as are all remaining trig identities in this section, are assigned as eercises. sin sin cos cos cos sin cos sin tan tan tan Square Formulas: The proofs of these formulas, as are all remaining trig identities in this section, are assigned as eercises. sin cos cos cos Half Angle Formulas: The proofs of these formulas, as are all remaining trig identities in this section, are assigned as eercises. sin cos with if if is in Quadrants I or II is in Quadrants III or IV cos cos with if if is in Quadrants I or IV is in Quadrants II or III tan cos cos with if if is in Quadrants I or III is in Quadrants II or IV tan cos sin sin cos Product to Sum Formulas: The proofs of these formulas, as are all remaining trig identities in this section, are assigned as eercises. coscos cos cos sin sin cos cos sin cos sin sin cos sin sin sin Sum to Product Formulas: The proofs of these formulas are assigned as eercises. cos cos cos cos cos cos sin sin sin sin sin cos sin sin cos sin

31 - Chapter 8D Eercises for Chapter 8D - Trigonometric Identities. Using the fact that sin sin, and cos cos, show that tan is an odd function of.. If lies in the third quadrant, i.e., /, and cos /, what is the value of sin?. Epress sin5 7 in terms of sines and cosines of 5 and 7 degrees.. Using the formulas for sin and cos, show that tan tan tan tan tan tan tan tan tan tan 5. Show that sin cos and cos cos. 6. If is an acute angle and sin.6, compute cos, sin/, and cos/. 7. Verif the following identities coscos cos cos sincos sin sin 8. Given that cos7, find sin5 and cos5. 9. Verif the following trig identit sin tan tan coscos. The two points cos,sin and cos,sin are on the unit circle. a. What is the angle formed b the radial lines from the center of the unit circle to these points? b. What is the length of the chord joining these two points? c. Rotate the unit circle so that one of the two points now lies on the point,. So one of the original points has coordinates,. What are the coordinates of the second point, and what is the length of the chord joining these two points? d. Show how the fact that the lengths of the chords in parts b. and c. are equal implies the formula cos coscos sinsin. Eactl evaluate sin where sin 8 and is in Quadrant II. 9. Ifcot 6,where is in Quadrant IV, what is the eact value of sin? 5. Eactl evaluate cos where cos and is in Quadrant IV.

32 - Chapter 8D Answers to Eercises for Chapter 8D - Trigonometric Identities. tan sin cos sin cos tan. Since is in the third quadrant, sin is negative. Thus, sin cos /9.9. sin5 7 sin5 cos7 sin7 cos5. tan sin cos sin cos sin cos sin sin coscos sin cos sincos coscos sinsin tan tan tan tan tan tan tan tan tan tan tan tan tan tan 5. Start with the sum of angles formula for cosine. cos cos coscos sinsin cos sin sin. Solving for sin, we have sin cos. The second formula follows from the identit cos cos sin cos. 6. cos sin sin/ cos/ cos cos

33 - Chapter 8D 7. For the first identit, start with the sum of two angles formula for cosine. cos coscos sinsin cos coscos sinsin Now add these two equations to get the first identit. To get the second identit do the same thing with the sum of two angles formula for sine. 8. Note that 5 is / of 7, and that 5 is in the second quadrant, which means that sin5 is positive, and cos5 is negative. sin5 cos7 /.77 cos5 cos7 / sin coscos sin cos sincos coscos sin cos sin cos coscos coscos cos sin sin cos tan tan. a. Either or. b. length cos cos sin sin cos coscos cos sin sinsin sin coscos sinsin c. Assume that the angle formed b the radial lines joining the original two points is. Then after rotating the unit circle, the coordinates of the second point are cos,sin, and the length of the rotated chord is length cos sin cos cos sin cos d. Since the lengths of both chords are equal we have coscos sinsin cos coscos sinsin cos cos coscos sinsin

34 - Chapter 8D. Since sin 8/9 and is in Quadrant II, we can determine that cos 7 /9. Using the sum of two angles formula for sine we have sin / sin cos/ cossin/ 8/9 / 7 /9 / Since cot 6 6 and is in Quadrant IV, we can determine that cos 5 8 sin 5. Using the double angle formula for sine we have 8 and sin sincos Since cos / and is in Quadrant IV, we can determine that sin 7 /. Using the sum of two angles formula for cosine we have cos / coscos/ sinsin/ // 7 / / 8 8 8

35 - Chapter 8E Chapter 8E - InverseSineFunction In addition to the original 6 trig functions there are several more that are etremel useful in mathematics, science and engineering. One of these functions is defined in this section and a few of its properties are also discussed. The others will be defined in the net section. In order for a function f to have an inverse function it must be one-to-one. None of the trig functions as previousl defined satisf this condition. So our first item is to address this issue, which we do in some detail with the sine function, whose graph is shown below sin It s clear from the horizontal line test that sin is not one-to-one. However, if we restrict the domain of this function from all real numbers to just the interval /,/, this new function is one-to-one. See the plot below sin, The range of this function is the closed interval,. We define the arcsin function and denote it as follows: for an,, the range of sin, arcsin sin if and onl if sin. Note that the domain of sin is the range of the arcsin function and the domain of arcsin is the restricted domain of the sine function.

36 - Chapter 8E The tables below ma help to clarif this definition sin sin # Since sin/ / we must have sin / /. Note: sin does not mean the reciprocal of the sine function. It is standard notation for the inverse function. Geometricall if the point, is on the graph of sin, i.e.sin, then the point, is on the graph of the arcsin function, whose graph we show below arcsin sin, The following plot shows both sin, sin, and the line. Note that the graph of the arcsin function is obtained b reflecting the graph of sin through the line sin (blue) and sin (red)

37 - Chapter 8E Before restating the definition of the arcsin function we point out that the domain of the sine function does not have to be the interval /,/ in order to force the function to be one-to-one. We could have taken man different intervals. For eample sin is one-to-one on each of the following intervals:, and,. In fact the sine function is one-to-one on an interval of the form k,k for an integer k. Setting k,, and gives the three intervals we ve mentioned. The reason for picking the interval, is that this is more useful than other possible choices. We summarize below the above discussion. sin :,, sin :,, where sin if and onl if sin. Another wa to write this last line is the following: sinsin for, and sin sin for,. Eample : The sine of 6 equals. Thus, sin /. Note that the value of arcsin is a number with out units. So sin / 6 would not be correct as this has units, and sin / 6 is also not correct, as 6 is not between / and /. Eample : The sine of 5. What does the arcsin of / equal? Solution: 5 is equivalent to 5/ radians. However, 5/ is greater than /, so the answer cannot be sin / 5/. We need a number,, between / and / such that sin /. Let s start noting that 5/ /, and use the sum of two angles formula for the sine function. sin 5 sin sincos sin cos sin sin. Thus,wehavesin.

38 - Chapter 8E Eercises for Chapter 8E - InverseSineFunction. sin.? Give our answer in degrees as well as radians. (A calculator ma be useful.). Ifsin. and / what is? (You should use a calculator to determine sin.). Ifsin.5, what does cos equal?. tansin /? 5. sin? 6. Eactl evaluate sin. 7. sin sin 5? 8. sin sin 7?

39 - Chapter 8E Answers to Eercises for Chapter 8E - Inverse Sine Function. Using a calculator we have: sin..5. Since m calculator is set to radian mode the answer given is in radians. The answer in degrees is degrees.. Looking at a plot of the sine function, see below, sin we see that sin is smmetric about the line /. So if sin. and /,and if is such that sin sin and / then we must have. Thus,.5.7. Draw a right triangle with hpotenuse of length one, and an angle labeled. Then the side opposite will have length.5. This implies that the side adjacent to will have length Thus, cos.968.

40 - Chapter 8E. As in the preceding problem, draw a right triangle with hpotenuse of length three, and an angle labeled. See the figure below. Thus, tan Since sin /, we must have sin. 6. Since sin/ / and / /, /, we conclude that sin. 7. Since 5, we know that sin 5 sin. Thus, sin sin 5 sin sin 8. sin sin 7 7

41 - Chapter 8F Chapter 8F - Inverse Trig Functions In this section we define the inverse cosine and inverse tangent functions. The last three inverse trig functions: arcsec, arccot, and arccsc are not discussed. Inverse Cosine The cosine function is also not one-to-one until we restrict its domain. The graph of this function shows that there are man possible restrictions cos We could use an of the following intervals:,,,, or,. The usual choice is to use the interval, for the domain of the cosine function. When this choice is made the arccos function, written as cos is defined for, b cos, if and onl if, and cos. Note that the range of the arccos function is the restricted domain of cos and the domain of arccos is the range of the cos function. The table below lists some values of the arccos function. cos not defined The graph of the inverse cosine function is shown below. You should sketch the graph of the cosine function with restricted to lie in the interval,, draw the line and reflect the graph of cosine about this line. You should get the plot below.

42 - Chapter 8F arccos Summarizing the above discussion we have: For, such that,, cos if and onl if cos, and.thatis,for and, we have cos cos coscos. Inverse Tangent Function The graph of the tangent function below shows, that if the domain of tan is restricted to / /, then the tangent function is one-to-one, with range the set of all real numbers tan The inverse tangent function is defined as follows. For an real number tan if and onl if tan and. A plot of arctan is shown below

43 - Chapter 8F The table below lists some values for the arctan function. tan To better help our understanding of this function sketch the graph of tan for between / and /, draw the line, and then sketch the reflection of the graph of tan through the line. You should get the graph of the arctan function. Eample : If /, compute cos sin. Solution: Let cos sin. Then, and cos sin. We know that sin cos/. Since both and / are in the interval, and cos is one-to-one on this interval we ma conclude that /. Thatis, cos sin. Eample : Compute cossin. Solution: Since cos is an even function of, we ma as well assume that sin is positive, and set sin.thensin. Draw a right triangle with angle, and label the sides as shown below. It is clear from this picture that cossin cos.

44 - Chapter 8F Eercises for Chapter 8F - Inverse Trig Functions. cos /?. tan /?. tan?. sintan /? 5. costan? 6. sincos tan? 7. tansin? 8. cossin cos? 9. Show that sin sin sin.. Eactl evaluate sincos if.

45 - Chapter 8F Answers to Eercises for Chapter 8F - Inverse Trig Functions. Looking at the /6 degree right triangle we see that cos6 /, and since 6 corresponds to radians, which is between and, we have cos /.. None of the standard right triangles have an angle whose tangent is /, so we are forced to resort to a calculator, which gives tan.6 radians.. Since tan / and tan is an odd function we know that tan /. Thus, tan.. Sketch the right triangle with legs of length and. Call the angle opposite the leg of length,. See the sketch below. The Pthagorean theorem implies that the hpotenuse of this triangle has length sin /,and. Thus, sintan / sin. 5. As in the previous problem draw the appropriate right triangle. See sketch below. From this sketch we have costan cos.

46 - Chapter 8F 6. Be sure to draw the appropriate right triangles as we did in the previous eercises. sincos tan sincos costan sintan coscos 7. tansin. 8. cossin cos cossin coscos sinsin sincos 9. Show that the sine of both sides is equal, and since the sine function is one-to-one, the two sides must be equal. sinsin sin sinsin cossin sinsin cossin. Since sinsin, we see that the sine of the right hand side of the original equation equals the sine of its left hand side.. Draw a right triangle with one leg and an hpotenuse of length. B the pthagorean theorem, the length of the other leg is 9. Label the angle opposite this side. Thus, we have sincos sin 9

47 - Chapter 8G Chapter 8G - Law of Sines and Law of Cosines A classic series of problems in geometr and trigonometr fall under the heading "to solve the triangle". What this means is that some data about a triangle is known, and from this information all of the lengths of the triangle s sides and all of angles are to be determined. There are two formulas that are ver helpful in this matter, and the are Law of Sines: sina a sinb b sinc c Law of Cosines: c a b abcosc The notation is eplained in the figure below. Law of Sines, Fig. The Law of Sines: This law states, for an triangle, that the ratio of the sine of an angle to the length of the side opposite the angle does not depend on which verte of the triangle we use. That is, see the figure above: sina a sinb b sinc c A proof of this formula is relativel eas and proceeds as follows. Draw a perpendicular from the verte with angle C to the side opposite this verte. Call the length of this line segment h. We then have Law of Sines, Fig. sina h b and sinb h a. Solving both equations for h and setting the resulting two epressions for h equal to each other we have the equation bsina asinb, which leads to sina a sinb b To derive the other equaltit(s) drop a perpendicular from a different verte. While the picture changes if angle B is obtuse ( /) the argument is the same as in this case, where angle B is acute ( /)..

48 - Chapter 8G Eample : A triangle has two sides of length inches and 5 inches with the angle opposite the side of length 5 inches equal to 5. Determine the length of the third side. Solution: The first step is to sketch and label a triangle, which models this data. Eample The ratio of the sine to the length of the opposite side is R sin Knowledge of this ratio enables us to determine angle B. sinb sin There are two angles between and whose sine equals this number, one of which is less than / and the second, which is greater than /. Since b is less than c, angle B cannot be greater than /. Thus, B arcsin radians, or degrees, and angle A equals , and its sine equals Finall we are able to determine a. sina a sinc c a sina sinc/c inches It s clear from the answer that angle A is greater than 9, and that the picture drawn is not a good representation of the triangle in question. Note that in the process of determining the length of the unknown side we have solved the triangle. The Law of Cosines: This law is a generalization of the Pthagorean Theorem for right triangles. If a, b, c are the lengths of the sides of an triangle and C is the angle opposite the side of length c (equivalentl C is the angle included b the sides with lengths a and b), we have c a b abcosc. Note that if C is 9 degrees, then cosc, and we have the Pthagorean Theorem. The following figure will be used in the proof of the Law of Cosines. Note that the line of length h from verte C is perpendicular to the side of length c. Verif the following equations. Law of Cosines

49 - Chapter 8G c c c, C h acos bcos c bsin, c asin The following proof of the Law of Cosines is a bit more complicated than it needs to be (see the eercises for a simpler derivation of this law), and it depends upon the fact that angle B is an acute angle. However, our goal here is two fold: to use previous identities and definitions, and to prove the Law of Cosines. See the eercises for a proof of the case when B is obtuse. c c c c c c c b sin absin sin a sin b cos absin sin a cos a b a cos b cos absin sin a b a cos b cos abcos cos cosc a b abcosc a cos abcos cos b cos a b abcosc acos bcos a b abcosc. Eample : A triangle has two sides of length inches and 5 inches with the angle included between the sides of known length equal to 5. Determine the length of the third side. Note, the difference between this eample and the previous eample is the relation of the known angle to the known sides. Solution: Thus, Eample Using the law of cosines we have: a b c bccosa a

50 - Chapter 8G Eample : Solve the triangle given in Eample. Solution: We know the following about this triangle a. 697, b, c 5 A 5, B?, C? To determine angles B and C we ll use the Law of Sines. To this end we first compute the ratio sina a. 69 sinb b sina a sinc c sina a Thus, we have B arcsin C arcsin As a check let s make sure that the sum of the angles equals 8. A B C This is not good. So where did we go wrong? See the eercises.

51 - Chapter 8G Eercises for Chapter 8G - Law of Sines and Law of Cosines. Prove the Law of Sines for the case when angle B is not less than /. There are two cases here: B / and B /.. Find a simpler derivation of the law of cosines. Hint, place verte C at the origin, verte A on the -ais, with verte B in the first quadrant, then use the distance formula to calculate c.. We did not show in the Law of Sines proof that sina sinc. Using the picture in the proof, drop a sinc c. a c perpendicular from B to the opposite side and show that sina a. Let ABC be such that a, b 5, c 5. Find angle B. 5. Let ABC be such that a, A 6,andB. Solve this triangle. 6. Let ABC be such that a, b 5, c. Find angle B. 7. Let ABC be such that b, c 5, and B 5. Solve this triangle. (Be careful) 8. Let ABC be such that a, b, and c 6. Solve this triangle. 9. Find where we went wrong in Eample, and fi it.. Let ABC be such that a 5, b 5,andA. Solve this triangle.. Let ABC be such that a 5, b 5,andC. Solve this triangle.

52 - Chapter 8G Answers to Eercises for Chapter 8G - Law of Sines and Law of Cosines. Prove the Law of Sines for the case when angle B is not less than /. ThefirstcaseiswhenB /, and we have (See the figure below) sina a sinc c sinb b a/b a cosa c b b c/b c b The second case occurs if B /. From the picture below we have: h bsina h asin B asinb Setting these two epression for h equal to one another we have sina a perpendicular from B to the opposite side we have h csina asinc. These equations give us sina a sinc c. sinb b.ifwealsodrawa. Find a simpler derivation of the law of cosines. Hint, place place verte C at the origin, verte A on the -ais, with verte B in the first quadrant, then use the distance formula to calculate c.