小透明数学老师投稿：5粉丝：200分享--dynmicweibozoneqqbaidu将视频贴到博客或论坛视频地址复制Flash代码复制Html代码复制微信扫一扫分享收藏0硬币--稍后看马克一下~用手机看转移阵地~用或其他应用扫描二维码手机下视频请使用扫码若未安装客户端，可直接扫此码下载应用看过该视频的还喜欢miniOFF高中高一数学课件：三角函数的周期性_课件_无忧考网
高中高一数学课件：三角函数的周期性
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课件网权威发布高中高一数学课件：三角函数的周期性，更多高中高一数学课件相关信息请访问课件网。
【导语】课件中对每个课题或每个课时的教学内容，教学步骤的安排，教学方法的选择，板书设计，教具或现代化教学手段的应用，各个教学步骤教学环节的时间分配等等，下面是无忧考网整理的高一数学课件：三角函数的周期性，欢迎阅读与借鉴。　　一、学习目标与自我评估　　1掌握利用单位圆的几何方法作函数的图象　　2结合的图象及函数周期性的定义了解三角函数的周期性，及最小正周期　　3会用代数方法求等函数的周期　　4理解周期性的几何意义　　二、学习重点与难点　　“周期函数的概念”，周期的求解。　　三、学法指导　　1、是周期函数是指对定义域中所有都有　　，即应是恒等式。　　2、周期函数一定会有周期，但不一定存在最小正周期。　　四、学习活动与意义建构　　五、重点与难点探究　　例1、若钟摆的高度与时间之间的函数关系如图所示　　(1)求该函数的周期;　　(2)求时钟摆的高度。　　例2、求下列函数的周期。　　(1)(2)　　总结：(1)函数(其中均为常数，且　　的周期T=。　　(2)函数(其中均为常数，且　　的周期T=。　　例3、求证：的周期为。　　例4、(1)研究和函数的图象，分析其周期性。(2)求证：的周期为(其中均为常数，　　且　　总结：函数(其中均为常数，且　　的周期T=。　　例5、(1)求的周期。　　(2)已知满足，求证：是周期函数　　课后思考：能否利用单位圆作函数的图象。　　六、作业：　　七、自主体验与运用　　1、函数的周期为()　　A、B、C、D、　　2、函数的最小正周期是()　　A、B、C、D、　　3、函数的最小正周期是()　　A、B、C、D、　　4、函数的周期是()　　A、B、C、D、　　5、设是定义域为R,最小正周期为的函数，　　若，则的值等于()　　A、1B、C、0D、　　6、函数的最小正周期是，则　　7、已知函数的最小正周期不大于2，则正整数的最小值是　　8、求函数的最小正周期为T，且，则正整数的最大值是　　9、已知函数是周期为6的奇函数，且则　　10、若函数，则　　11、用周期的定义分析的周期。　　12、已知函数，如果使的周期在内，求正整数的值　　13、一机械振动中，某质子离开平衡位置的位移与时间之间的函数关系如图所示：　　(1)求该函数的周期;　　(2)求时，该质点离开平衡位置的位移。　　14、已知是定义在R上的函数，且对任意有成立，　　(1)证明：是周期函数;　　(2)若求的值。三角函数(1)(2)(3)(4)_新东方高一数学精讲（必修一）_腾讯视频高一数学三角函数定义_百度文库
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你可能喜欢CHAPTER 3TRIGONOMETRIC MEASUREMENTSLEARNING OBJECTIVES Upon completion of this chapter, you should be able to do the following: 1. Measure angles in degrees, radians, and mils. 2. Find angular velocity and the area of a sector using radians. 3. Apply the Pythagorean theorem and properties of similar right triangles to problem solving. 4. Apply trigonometric ratios, functions, and tables to problem solving.INTRODUCTION However, trigonometry is not restricted to solving problems it also forms a foundation for some advanced mathematical concepts and subject areas. Trigonometry is both algebraic and geometric in nature, and in this course both of these qualities will be applied. MEASURING ANGLES In this course, an angle is considered to be generated when a line having a set direction is rotated about a point, as depicted in figure 3-1.o A figure3-1._Genenretion of an angle.In figure 3-1, line OA is laid out as a reference line having a set direction. One end of the line is used as a pivot point and the line is rotated from its initial position (line OA) to another position (line OB), as in opening a door. As the line turns on its pivot point, it generates the angle AOB. The following terminology is used in this and subsequent chapters: 1. Radius vector -The line that is rotated to generate an angle. 2. Initial position -The original position
corresponds to line OA in figure 3-1. 3. Terminal position -The final position
corresponds to line OB in figure 3-1. 4. Positive angle -The angle generated by rotating the radius vector counterclockwise from the initial position. 5. Negative angle -The angle generated by rotating the radius vector clockwise from the initial position. The convention of identifying angles by use of Greek letters is followed in this text. When only one angle is involved, it will be symbolized by θ (theta). Other Greek letters will be used when more than one angle is involved. The additional symbols used will be φ (phi), α (alpha), and β (beta). DEGREES The degree system is the most common system of angular measurement. In this system a complete revolution is divided into 360 equal
so, 1 revolution = 360 ° For accuracy, each degree is divided into 60 so, 1° =60' Each minute is divided into 60 so, 1'=60& For convenience in working with angles, the 360 ° are divided into four equal parts of 90 °each, similar to the rectangular coordinate system. The 90 °sectors, called quadrants, are numbered according to the 3-2180° QUADRANT III 90° QUADRANT II QUADRANT I 0° QUADRANT Iv270° Figure 3-2.-Quadrant positions.convention shown in figure 3-2.3-3If the angle generated by rotating the radius vector in a positive (counterclockwise) direction is between 0 ° and 90 ° , then the angle is in the first quadrant. If the angle is between 90 °and 180 ° , then the angle is in the second quadrant. If the angle is between 180°and 270° , then the angle is in the third quadrant. And if the angle is between 270 ° and 360 ° , then the angle is in the fourth quadrant. If the angle generated by rotating the radius vector in a positive direction is more than 360 ° , then the quadrant in which the angle lies is found by subtracting from the angle the largest multiple of 360°that the angle contains. The quadrantin which the remainder angle lies is determined as described in the previous paragraph. The original angle lies in the samequadrant as the remainder angle. E X A M P L E 1 : In which quadrant is the angle 130 0?SOLUTION: Since 130 °is between 90 ° and 180° , it is in the second quadrant. (See fig. 3-3, view A). quadrant is the angle 850 0?E X A M P L E 2 : In whichSOLUTION: The largest multiple of 360 ° contained in 850 °is 720 ° ; so, 850 °- 720 °= 130° . Since 130° is in the second quadrant, then 850 ° also lies in the second quad-rant. This relationship is shown in figure 3-3, view B.Figure 3-3.-Angie generation.PRACTICE PROBLEMS: Determine the quadrant in which each of the following angles lies: 1. 260° 2. 290°3-43. 800 ° 4. 1,930 °ANSWERS: 1. 3rd 2. 4th 3. 1st 4. 2ndRADIANS Another even more fundamental method of angular measurement involves the radian . It has certain advantages over the degree method. Radian measurement greatly simplifies work with trigonometric functions in calculus. Radian measurement also relates the length of arc generated to the size of an angle. A radian is defined as an angle that, if its vertex is placed at the center of a circle, intercepts an arc equal in length to the radius vector of the circle. Assume that an angle is generated, as shown in figure 3-4, view A. If we impose the condition that the length of the arc, s, described by the extremity of the line segment generating the angle, mustABCDFigure 3-4.-Radian measure.3-5equal the length of the radius vector, r, then we would describe an angle exactl that is, for 1 radian,s =rIn a broader sense, the radian measure of an angle, θ , is the ratioof the length of the arc, s, it subtends to the length of the radius vector, r, of the circle in which it
that is, s θ =rFor angle θ , in figure 3-4, view B, which intercepts an arc equal to two times the length of the radius vector, θ equals two radians. For angle θ , in figure 3-4, view C, which intercepts an arc equal to three times the length of the radius vector, θ equal three radians.EXAMPLE 3: Find the radian measure of the central angle in a circlewith a radius of 10 inches if the angle subtends an arc of 5 inches.SOLUTION:θ = =s r 5 10= 0.5 radians Recall from plane geometry that the circumference of a circle is 2? times the radius or C = 2?r Hence, the radius vector can be laid off on the circumference 2? times. (See fig. 3-4, view D). Since the arc length of the circumference is 2? radians and the circumference encompasses one complete revolution of 360 ° , then2? radians = 360 °One-half of a revolution equals 180 °or ? so, ? radians = 180 ° (3.1)3-6By dividing both sides of equation (3.1) by ? , we find that 1 radian =180 ° π= 57.2958 °(rounded) = 57°17' 45& By dividing both sides of equation (3.1) by 180, we find that 1 °=π 180radians= 0.01745 radians (rounded) NOTE: The degree symbol (° ) is customarily used to indicate degrees, and a pure number with no symbol attached is used to indicate radians. For example, sin 3 should be understood to represent &sine of 3 radians,& whereas the &sine of 3 degrees& would be written sin 3 ° . The following list indicates other relationships frequently used in trigonometric problems: Radians ?/6 ?/4 ?/3 ?/2 ? 3? /22?Degrees 30 45 60 90 180 270 360EXAMPLE: Express 160 °in radians, using ? in the answer.SOLUTION: 1°=π 180radians160°= 160 x 1° = 160 x =8π 9 π 180radiansradiansEXAMPLE: Express ?/20 in degrees. SOL UTION: 1 radian =180 ° ππ 20radians==π 20 π 20x 1 radianx180 ° π= 9° Refer to figure 3-5. We can see that if 6 represents the number of radians in a central angle, r the length of the radius of the circle, and s the length of the intercepted arc, then the length of the arc equals the number of radians multiplied by the length of the radius orS =θ rEXAMPLE: In a circle having a radius of 11 inches, an arc subtends a central angle of 3 radians. Find the length of UTION: theSOL arc in inches.S =θ r = 3.11 = 33 inchesrF i g ure 3 -5 . -L e ng th o f ar c .PRACTICE PROBLEMS: 1. Find the number of radians in the central angle subtended by an arc 18 inches long in a circle whose radius is 8 inches. Express the following angles in radians, using ? in the answer: 2. 420 ° 3. 135°Express the following angles in degrees: 4. 20? 5. 5?/6 6. In a circle whose radius, r, is 4 inches, find in inches thelength of arc, s, whose central angle is 1 1 /4 radians. ANSWER: 1. 9/4 radians 2. 7?/3 3. 3?/4 4. 3,600° 5. 150° 6.5 inchesBecause of the relationship of the radian to arc length, the radian has some special applications in measurements of angular velocity and area of a sector. Angular Velocity Another type of problem that radian measurement simplifies is that which relates the rotating motion of the wheels of a vehicle 3-9to its forward motion. Here we will not be dealing with angles alone but also with angular velocity. Let's analyze this type of motion. Consider the circle at the left in figure 3-6 to indicate the original position of a wheel. As the wheel turns, it rolls so that the center moves along the line CC', where C' is the center of the wheel at its final position. The contact point at the bottom of the wheel moves an equal distance PP'; but as the wheel turns through angle 6, arc s is made to coincide with line P P ' ; so,Figure 3-6.-Angular rotation.s=PP' =dor the length of arc is equal to the forward distance, d, the wheel travels. But sinces = rOthen the forward distance that the wheel travels isd=rODividing both sides of the previous equation by t givesd_rO t-tWhen a vehicle moves with a constant velocity, v, in time, t, the distance, d, the vehicle travels is expressed by the formulad=vtSolving this formula for v, we have dv =t The fraction d/t expresses the linear v e l o c i t y of the vehicle, and 6/t is the angular velocity. If we let w (Greek letter omega) stand for the angular velocity, then the equationd_r6 t - tbecomesv=rco where w is measured in radians per unit time.3-10E XAMPL E : A car wheel is rotating at 1,050 revolutions perminute (rpm). Find 1. the angular velocity in radians per second. 2. the linear velocity in meters per second on the tire tread, 2 5 centimeters from the center.S O L U T I ON :1. To find the angular velocity, we need to convert rev/min to rad/sec. To do this, we will apply unit conversions (multiples of one) as follows: c v = 1,050 rev x 2n rad x 1 min min_ (1,050)(2n) radrev 60 sec sec60= 35n radians per second 2. We find the linear velocity as follows: v = rw= 25 cm x 35nsec= 875n sec NOTE: When no unit of angular measure is indicated, the angle is understood to be expressed in radians. We now need to convert cm/sec to m/sec. We will again apply a unit conversion: v = 875n m x sec 1W m 875rr m 100 sec = 8.75n meters per secondE X A M P L E : A car is traveling 40 miles per hour. If the wheelradius is 16 inches, what is the angular velocity of the wheels in 1. radians per minute? 2. revolutions per minute? 3-11SOL UTION: 1. We know that v=rw Thus, w_ 5 __ mi ___ 5,280 ft 12 in 1 hr 2 hr x in x ______ 1 mix 1 ft x 60 min (5)(5,280)(12) rad (2)(60) min = 2,640 radians per minute 2. Since 2n radians = 360°and 360°= 1 revolution, then w = 2,640 rad x rev min 2n rad 2,640 rev 2n min = 420.2 revolutions per minute EXAMPLE: Determine the distance a truck will travel in 1 minute if the wheels are 3 feet in diameter and are turning at the rate of 5 revolutions per second. HINT: Diameter = 2 x radius SOL UTION: v=rw d t = rw d = rtw= 2 ft x_ (3)(l0n)(60) ____ ft 2 = 2,827.43 feetrasec1 minx (5 sec x In rev )= 2 ft x l minx 10n sec x 60 min3-12Area of a Sector From plane geometry we find that the area of the sector of a circle is proportional to the angle enclosed in the sector. Consider sector AOB of the circle shown in figure 3-7. If 6 is increased to 2? radians (360 ° ), it encompass so the area of the circle is proportional to 2? ? radians. Hence,area of sector area of circle=θ 2πFigure 3-7.-Sector of a circle.But the area of a circle can be found by the formulaA = ?r 2By substitution, we find area of sector=θ 2π(?r 2 )Therefore, the area of a sector of a circle can be found by the formula A=1 2r2 θwhere θ is expressed in radians. EXAMPLE: Find the area of a sector of a circle with a radiusof 6 inches having a central angle of 60 ° .SOLUTION: A = r2 θ2 1= =1 2(6 in)2 (60°× in2π180°)36 π? 2 (3)= 6?square inches3-13The area of a sector of a circle can also be found if the radius and arc length are known. Since S = rthen A= = =1 2 1 2 1 2r2 θr(r θ) rsEXAMPLE: What is the diameter of a circle if a sector of thecircle has an arc length of 9 inches and an area of 18 square inches? SOLUTION: IfA= rs 2 thenr= = But2A s 2(18 in 2 )1= 4 in9 ind=2rTherefore,d = 2(4 in) = 8 inches3-14PRACTICE PROBLEMS: 1. A car travels 4,500 feet in 1 minute. The diameter of the wheels is 36 inches. What is the angular velocity of the wheels in radians per minute? 2. How far in feet does a car travel in 1 minute if the radius of the wheels is 18 inches and the angular velocity of the wheels is 1,000 radians per minute?3. Find the area of a sector of a circle whose central angle is?/3 and whose diameter is 24 inches. Leave the answer in terms of it. 4. Find the area of a sector of a circle in inches whose arc length is 14 inches and whose radius is 2/3 feet.ANSWERS: 1. 3,000 radians per minute 2. 1,500 feet 3. 24? square inches4. 56 square inchesMILS The mil is a unit of small angular measurement, which is not widely used but has some military applications in ranging and sighting. The mil is defined in two ways:1. As 1/6,400 of the circumference of a circle.2. As the angle subtended by an object 1 unit long, perpendicular to the line of sight, at a distance of 1,000 units.3-15From definition 1 we can see that since 360 °= 6,400 mils then 1°= = Also, since 6,400 mils = 360 ° then 1 mil = =360 ° 6,400 9° 160 6,400 360 160 9milsmils= 17.78 mils (rounded)= 0.05625°EXAMPLE : Convert 240 mils to degrees. SOLUTION:1 mil =9° 160240 mils = 240 x 1 mil 9° = 240 x160=27° 2= 13.5°EXAMPLE: Convert 27 °to mils. SOLUTION: 1°=160 9mils27°= 27 × 1° 3-16= 27 ×160 9mils= 480 milsSince l mil =and9° 1601 °= thenπ 180radian1 mil = = =9 160 9 160 π× 1° ×π 180radians3,200radians= 0.00098 radians (rounded) We see that 1 mil is approximately 0.001 or 1/1,000 radians. We also see that 1 radian 1,000 mils.EXAMPLE: Convert 25 mils to an approximate radianmeasure.SOLUTION:1 mil ≈1 1,000radian25 mils = 25 × 1 mil 1 ≈ 25 × 1,000 ≈25 1,000radianradian≈ 0.025 radianConvert measurement in mils.EXAMPLE.6 .48radians to an approximateSOLUTION :3-171 radian ≈ 1,000 mils 6.48 radians = 6.48 × 1 radian ≈ 6.48 × 1,000 mils ≈ 6,480 mils Referring to figure 3- 8, when an angle, θ , subtended by an arc, s, is very small and the radius, r, is large, the chord, c, is almost equal to the arc, s. The formula for the length of arc of a circle, as previously stated, iss=rθFigure 3-8.-Relationship of chord and arc.where θ is in radian measurement. If the measurement of the arc is made in mils, we must divide the mil measure by1,000 to obtain the radian measure. Since, 1 mil =1 1,000radians (approximately)then,m mils = So, S = r(= 1,000 m 1,000radians)rm 1,000Now, since the chord, c, in figure 3-8, is approximately equal to the arc, s, then mr c=1,000Now consider definition 2. If 3-18r = 1,000 yds and m = 1 mil then c= =mr 1,000 1,000×1 1,000= 1 yard We also know that the arc, s, is approximately equal to 1 yard since s ≈ c. The military uses the fact that a mil subtends a yard at a distance of 1,000 yards for quick computations in the field. EXAMPLE: Find the length of a target if, at a right angle to the line of sight, it subtends an angle of 15 mils at a range of 4,000 yards. SOLUTION:rm 1,000 4,000×15 1,000c===60 yardsEXAMPLE: A building known to be 80 feet long and perpendicular to the line of sight subtends an angle of 100 mils. What is the approximate range to the building?SOLUTION: Since c =100rm3-19then r= =1,000c m 1,000×80 100= 80 feetPRACTICE PROBLEMS: 1. Convert 3,456 mils to degrees. 2. Convert 12 degrees to mils. 3. Convert 27,183 mils to an approximate radian measure. 4. Convert 431 radians to an approximate measurement in mils. 5. A tower 500 feet away subtends a vertical angle of 250 mils. What is the height of the tower? 6. If points X and Y are 48 yards apart and are 4,000 yards from an observer, how many mils do they subtend?ANSWERS: 1. 194.4° 2. 213.3 mils 3-203. 27.183 radians 4. 431,000 mils 5. 125 feet6. 12 mils PROPERTIES OF RIGHT TRIANGLES Mathematics, Volume 1, contains information on the trigonometric ratios and other properties of triangles. This section restates some of the properties of right triangles for review and reference. PYTHAGOREAN THEOREM The P y t h a g o r e a n t h e o r e m states that in a right triangle, the square of the length of the hypotenuse (longest side) is equal to the sum of the squares of the lengths of the other two sides. In the right triangle shown in figure 3-9, this relationship is expressed asr2 = x2 + y2ywhere r i s the length of the hypotenuse and x and y are the lengths of the other two sides. This relationship is useful in solving many problems and in developing trigonometric concepts. EXAMPLE: In figure 3-10, what is the length of the hypotenuse of the right triangle if the lengths of the other two sides are 3 and 4? SOLUTION:x Figure 3-9.-Pythagorean relationship.343-21Figure 3.10.-Right triangle with hypotenuse unknown.So, r = 25 = 5 NOTE: We will use the positive value of the square root since we are dealing with lengths.3-22EXAMPLE: Figure 3-11 shows a right triangle with a hypotenuse equal to 40 and one of the other sides equal to 10. What is the length of the remaining side?SOL UTION: r2 = x 2 + y2,40y=iox:2 Figure 3-11.-Right triangle with one side unknown.orx2 = r2 - y2= 402 - 102 = 1,600 - 100 = 1,500 So, x = 1,500 = 38.7 (rounded) SIMILAR RIGHT TRIANGLES Another relationship of right triangles that is useful in trigonometry concerns similar triangles. Whenever the angles of one triangle are equal to the corresponding angles in another triangle, the two triangles are said to be similar. For example, right triangle A in figure 3-12 is similar to right triangle B. Since the two triangles are similar by definition, the following proportions involving the lengths of the corresponding sides are true:a _ b _ c a'- b'- c'This relationship can be used to find the lengths of unknown sides in similar triangles. 3-23Figure 3-12.-Similar triangles.EXAMPLE: Assume right triangles A and B in figure 3-13 are similar with lengths as shown. Find the lengths of sides b' and c ' .SOLUTION: Sincea _ b _ c a' b' c'C'7then 10 11.18 _ 5 7 = b' c' Side b' can be solved for using the first two ratios: 10 _ 7 - 11.18b'Figure 3-13.-Similar triangles, solution example.So,b =, 1 1 . 1 8 x 7 10 78.26 10 = 7.826Side c ' can be solved for using the first and third ratios: 10 7 _ 5 x 7 10 = 3.5 Recall from plane geometry that the c sum of the interior angles of any triangle is equal to 180° . Using this fact, we can assume that two triangles are similar if two angles of one are equal to two angles of the other. The remaining angle in any triangle must be equal to 180°minus the sum of the other two angles. So, 5 c'3-24If an acute angle of one right triangle is equal to an acute angle of another right triangle, the triangles are similar because the right angles in the two triangles are also equal to each other. Hence, if 0 is one of the acute angles in a right triangle, then (90 °- 0) is the other acute angle, such that 9 0 °+ 0 + ( 9 0 °- 0 ) = 180° Therefore, two right triangles are similar if an acute angle of onetriangle is equal to an acute angle of the other triangle.Many practical uses of trigonometry are based on the fact that two right triangles are similar if an acute angle of one triangle is equal to an acute angle of the other triangle. In figure 3-14 triangle A is similar to triangle B since an acute angle in triangle A is equal to an acute angle in triangle B. Since triangle A is similar to triangle B, then x x' y y' r r'Interchanging terms in the proportions givesFigure 3-14.-Similar right triangles.andwhich are considered among the main principles of numerical trigonometry.PRACTICE PROBLEMS: Refer to figure 3-15 in solving the following problems: 1. Use the Pythagorean theorem to calculate the unknown length in triangle A.Figure 3-15.-Triangles for practice problems.3-252. Use the Pythagorean theorem to calculate the unknown length in triangle B.3.Triangles C and D are similar triangles. Find the length of sides a and b in triangle D.ANSWERS:1.V2. 2 f3.a = 1 5 \ / 4 , b = 3/2TRIGONOMETRIC RATIOS, FUNCTIONS, AND TABLES The properties of triangles given in the previous section provide a means for solving many practical problems. Certain practical problems, however, require knowledge of right triangle relationships other than the Pythagorean theorem or the relationships of similar triangles before solutions can be found. For example, the following two problems require additional knowledge: 1. Find the values of the unknown sides and angles in a right triangle when the values of one side and one acute angle are given. 2. Find the value of the unknown side and the values of the angles in a right triangle when two sides are known. The additional relationships between the sides and angles of a right triangle are called trigonometric ratios. These ratios were introduced in Mathematics, Volume 1, and are reviewed in the following paragraphs. The basic foundations of trigonometry rest upon these ratios and their associated trigonometric functions.3-26TRIGONOMETRIC RATIOS AND FUNCTIONS The sides of a right triangle form six ratios. In figure 3-16 we will use the acute angle 6 and the three sides x, y, and r two at a time to define the trigonometric ratios. These ratios and the trigonometric functions associated with each ratio are listed as follows: the sine of 6 the cosine of 6 = the tangent of 6= r , written sin 6written cos 8Figure 3-16.-Right triangle for determining ratios.= x, written tan 8 the cotangent of 6= y, written cot 6the secant of 8 = X, written sec 8 the cosecant of 8 = y, written csc 8 The trigonometric functions of a right triangle are remembered easier by the convention of naming the sides. Refer to figure 3-17. The side of length y is called the side opposite angle 6, the side of length x is called the side adjacent to angle e, and the side of length r is called the hypotenuse. Using this terminology causes the six trigonometric functions to be defined as: opposite ? - hypotenuse x adja cent ? hypotenuse _ y _ opposite x - adjacent x _ adjacent y oppo site _ r _ hypotenuse x - adjacent 3-27Figure 3-17.-Names of sides of a right triangle.? _ hypotenuse y opposite3-28Remember that the six trigonometric ratios apply only to the acute angles of a right triangle. EXAMPLE: Give the values of the trigonometric functions of the angle in the right triangle for figure 3-18, view A. SOLUTION: s i n 0 = r = 5 = 0.6 cos = x = 4 = 0.8tan0=X=4=0.75 0 cot sec 0 csc 0 x 4 3 = 1.33333 (rounded) -y r 5 = 1.25 x 4 r 5 = 1.66667 (rounded) y - 3Figure 3 - 1 8 . - P r p ctice triangles.EXAMPLE: Give the values of the trigonometric functions of the angle in the right triangle for figure 3 -18, view B. SOLUTION: Only two sides are given. To find the third side of the right triangle, use the Pythagorean theorem:r2= x2 + y2andy2 =r2 -x23-29Now, using the values of x, y, and r, we find the values of the six trigonometric functions are as follows: sin 8 cos 0 y x cot 0 tan 0 x yr= Y_ = 3r r-= 0.86603 (rounded)xsec 0 csc 0xr6 -2 3 = 0.5 1 6 23 ~ = = 3 _ 3 1 3 6 2 3 6 33 _ 2= 1.73205 (rounded) = 0.57735 (rounded) 3= 2y33= 1.15470 (rounded)TABLES OF TRIGONOMETRIC FUNCTIONSTables of trigonometric functions give the numerical values of the ratios of the sides of a right triangle that correspond to the trigonometric functions. Appendixes II and III are tables oftrigonometric functions. These tables give values rounded to five decimal places of trigonometric functions for each minute from 0 ° to 90 ° . Appendix II consists of tables of natural sines andcosines. Appendix III consists of tables of natural tangents and cotangents.For example, if we wanted to find sin 3 °25 ', we would use appendix II, Natural Sines and Cosines, to first locate 3 ° on the first row of the table. Next, we would locate sin under 3 °on the second row. Then, we would locate 25 along the first column of the table. Now, reading left to right across from 25 and from top to bottom under sin 3 ° , we find sin 3 ° 25 ' = 0.05960. If we wanted to find cos 86 °35 ', we would first locate 86°on the last row of the table. (The degrees on the top row range from 0 ° to 44 ° , and the degrees on the last row range from 45°to 90 ° .) Next, we would locate cos above 86 °on the next to the last row. Then, we would locate 35 along the last column of the table. Now, reading right to left across from 35 and from bottom to top above cos 86 ° , we find cos 86 °35 ' = 0.05960. Note that sin 3 °25 ' = 0.05960 = cos 86° 35 '. The reason for this will be discussed in chapter 4. 3-30The tables in appendix III, Natural Tangents and Cotangents, are arranged in the same format as the tables in appendix II and are used in the same way. NOTE: Scientific calculators will give you the same values rounded to five decimal places as supplied in the tables in appendixes II and III. Most tables list the sine, cosine, tangent, and cotangent of angles from 0 °to 90 ° . Very few give the secant and cosecant since these functions of an angle are seldom used. When needed, they may be found from the values of the sine and cosine as follows: sec e = r 1 x - x - cose r 1 = 1 = sin e 1and csc e = yrr Hence, the reciprocal of the secant function is the cosine function, and the reciprocal of the cosecant function is the sine function. The tangent and cotangent functions may also be expressed in terms of the sine and cosine functions as follows: y tans - x - x csin e ose _r and x cote= y = y = sn8r In additiori, the cotangent function may be determined as the reciprocal of the tangent function as follows: cote =yt a ng x=y=NOTE: These relationships are the fundamental trigonometric identities that will be used extensively in solving more complex identities in chapter 6.3-31USE OF TRIGONOMETRIC RATIOS AND FUNCTIONS The trigonometric ratios and trigonometric functions furnish powerful tools for use in problem solving of right triangles. Finding the remaining parts of a right triangle is possible if, in addition to the right angle, the length of one side and the length of any other side or the value of one of the acute angles is known. EXAMPLE: Find the length of side y in figure 3-19, view A. SOLUTION: We can use tang = x since we know one side and one angle. Thus, tan 35°=20 A B20Figure 3-19.-Practical use of ratios.From appendix III (or calculator), we find that tan 35 °= 0.70021 So, 0.70021 = 20 y = (0.70021)(20) = 14.0042 We could have also used cos 0, cot 0, or sec 0 to find side y. EXAMPLE: Find the value of r in figure 3-19, view B. SOLUTION: sin 0 = r sin 65 °= 5 r 5 __ r _ - sin 65 ° 5 _ r _ - 0.90631 = 5.51688 We could have also used csc 0 to find side y. 3-32PRACTICE PROBLEMS: Refer to figure 3-20 in working problems 1 through 4. 1. Find the values of the trigonometric functions of angle 0 for the right triangle in view A. 2. Find the value of side y in view B using the sine function.i52128AB3. Find the value of side x in view C using the cosine function. 4. Find the value of side y in view D using the tangent function.Figure 3-20.-Triangles for practice problems.ANSWERS: 1. sin 0 = 21/35 = cos 0 = tan 0 = cot 0 = sec 0 = csc 0 = 2. 294.. 12.. = 0.628/35 = 4/5 = 0.8 21/28 = 3/4 = 0.75 28/21 = 4/3 = 1. = 5/4 = 1.25 35/21 = 5/3 = 1.66667SUMMARY The following are the major topics covered in this chapter: 1. Terminology:Radius vector-The line that is rotated to generate an angle. Initial position-The original position of the radius vector. Terminal position-The final position of the radius vector. Positive angle-The angle generated by rotating the radiusvector counterclockwise from the initial position.Negative angle-The angle generated by rotating the radiusvector clockwise from the initial position.2. Degrees: The degree system is the most common system ofangular measurement. In this system a complete revolution is divided into 360 equal parts called degrees. 1 revolution = 360° 1 °= 6 0 ' 1 ' = 60& For convenience, the 360 °are divided into four equal parts of 90°each called quadrants. If 0 °& 0 & 90° , then 0 is in quadrant I. If 90°& 0 & 180° , then 0 is in quadrant II. If 180°& 0& 270° , then 0 is in quadrant III.If 270°& 0 & 360° , then 0 is in quadrant IV. If 0 & 360 ° , then 0 lies in the same quadrant as 0 - n (360 ° ), where n = 1, 2, 3, . . . and n(360° ) & O. 3. Radians: An even more fundamental method of angular measurement involves the radian. A radian is defined as an angle that, if its vertex is placed at the center of a circle, intercepts an arc equal in length to the radius of the circle.2n radians = 360 °itradians = 180 ° 3-341 radian = 180 °It180 °radians The radian measure of an angle, 0, is the ratio of the length of the arc, s, it subtends to the length of the radius vector, r, of the circle in which it is the central angle or4. Other frequently used relationships between radians and degrees:Radiansn/6 n/4 n/3 n/2 n 3n/2 2n5.Degrees30 45 60 90180270 360Length of arc:s = Orwhere 0 represents the number of radians in a central angle, r the length of the radius of the circle, and s the length of the intercepted arc. 6. Angular velocity:where 0 is measured in radians and t is the unit time. 3-357. Linear velocity:where d is the distance and t is the unit time. v=rw where r is the radius and w is the angular velocity. 8. Area of a sector of a circle:A = 2 r 20where 0 is expressed in radians.A = Zrswhere r is the radius and s is the arc length. 9. Mils: The m i l is a unit of small angular measurement that has military applications. The m i l is defined as follows: 1. 1/6,4W of the circumference of a circle. 360 °= 6,400 mils 10_190 mils 901 mil = 1602. The angle subtended by an object 1 unit long, perpendicular to the line of sight, at a distance of 1,000 units. 1 mil - 1 ___ ~~ radians 1 radian 1,000 mils 3-3610. Pythagorean theorem: The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse, r, is equal to the sum of the squares of the lengths of the other two sides, x and y, or r2= x 2+ y211. Similar triangles: Whenever the angles of one triangle are equal to the corresponding angles in another triangle, the two triangles are said to be similar and the following proportions involving the lengths of their corresponding sides are true:a - _ b _ c a'- b'- c'12. Similar right triangles: Two right triangles are similar if an acute angle of one triangle is equal to an acute angle of the other triangle. The following proportions involving the lengths of their corresponding sides are true:X y x' - - y' - r'r13. Trigonometric ratios and functions: _ y - - opposite sin e cos 8 tan 8 cot 0 ?x_hypotenuse _ adjacent hypotenuse opposite ? adjacent ? adjacent opposite sec 0 r csc 8 = y = _ r _ hypotenuse x adjacent hypotenuse opposite?14. Tables of trigonometric functions: Tables of trigonometric functions give the numerical values of the ratios of the sides of a right triangle that correspond to the trigonometric functions. Appendix II consists of tables of natural sines and and cosines. Appendix III consists of tables of natural tangents and cotangents. 3-37ADDITIONAL PRACTICE PROBLEMS1.In which quadrant is the angle5,370 °?2. Find the radian measure of the central angle in a circle with radius it inches if the angle subtends an arc of 3n/5 inches. 3. Express 4,320°in radians, using 4. Express 11n/12 in degrees. 5. If the length of the radius of a circle is 5 meters, find the length of arc subtended by a central angle with measure it radians. 6. Kim and Tom are riding on a Ferris wheel. Kim observes that it takes 30 seconds to make a complete revolution. Their seat is 35 feet from the axle of the wheel. a. What is their angular velocity in radians per second? b. What is their linear velocity in feet per minute? 7. Find the area of a sector of a circle if its central angle is 45 ° and the diameter of the circle is 28 centimeters. 8. Convert 17 7/9 mils to degrees. 9. Convert 3.6 degrees to mils. 10. Convert 9/5 mils to an approximate radian measure. 11. Convert mils.0.00145itin the answer.radians to an approximate measurement in12. An airplane with a wing span of 84 feet is flying toward an observer. What is the distance of the plane from the observer when the plane subtends 7 mils? 13. The length of the hypotenuse of a right triangle is 17, and the length of one of the other sides is 8. What is the length of the remaining side? 14. Assume similar right triangles A and B have sides x, y, r, and x', y', r' , respectively. If x = 6, y = 8, r = 10, and y' = 1/2, what are the values of x' and r'? 15. Find the values of the trigonometric functions 0 of in a right triangle if the hypotenuse is 25 and the side adjacent to 0 is 24. 16. If in a right triangle one of the acute angles is 56 °17 ' and the hypotenuse is 10, what are the lengths of the other two sides? 3-38ANSWERS TO ADDITIONAL PRACTICE PROBLEMS 1. 4th 2. 3/5 3. 24n 4. 165° 5. 5n meters 6. a. n/15 radians per second b. 140n feet per minute 7. 49n/2 square centimeters 8. 1 ° 9. 64 mils 10. 0.0018 radians 11. 1.45 mils 12. 12,000 feet 13. 15 14. x ' = 3/8 z' = 5/8 15. sin 9 = 7/25 = 0.28 cos 0 = 24/25 = 0.96 tan 0 = 7/24 = 0.29167 (rounded) cot 0 = 24/7 = 3.42857 (rounded) sec 0 = 25/24 = 1.04167 (rounded) csc 0 = 25/7 = 3.57143 (rounded) 16. 5.5509 and 8.CHAPTER 4TRIGONOMETRIC ANALYSISLEARNING OBJECTIVES Upon completion of this chapter, you should be able to do the following: 1. Use the rectangular coordinate system to determine the algebraic signs and the values of the trigonometric functions and to locate and define the trigonometric functions. 2. Relate any angle in standard position to its reference angle. 3. Determine the trigonometric functions of an angle in any quadrant, of negative angles, of coterminal angles, of frequently used angles, and of quadrantal angles. 4. Express the trigonometric functions of an angle in terms of their complement. 5. Recognize characteristics of the graphs of the sine, cosine, and t a ngent functions. INTRODUCTION This chapter is a continuation of the broad topic of trigonometry introduced in chapter 3. The topic is expanded in this chapter to allow analysis of angles greater than 90 ° . The chapter is extended as a foundation for analysis of t that is, an angle of any number of degrees. Additionally, the chapter introduces the concept of both positive and negative angles. RECTANGULAR COORDINATE SYSTEM The rectangular, or Cartesian, coordinate system introduced inMathematics, Volume 1, was used
in this4-40chapter it is used to analyze the generalized angle. The following is a brief review of the rectangular coordinate system: 1. The vertical axis (Y axis in fig. 4-1) i s considered positive above the origin and negative below the origin. 2. The horizontal is (X is in fig. 4-1) i s positive to the right of the origin and negative to the left of the origin.3. A point, P(x,y), anywhere in a rectangular5-4 ( ,?) 4+I0I F i F i 1 F3- 2it 2 3 4 5 6 7- - 2 3- 8 -7 -6 -5 - 4 -3 -2' 4 P(4,-5) - 5 - - 6 - - 7 - ). ( * ,coordinate system may be located by two numbers. The value of x is called the abscissa. The value of y is called the ordinate. The abscissa and ordinate of a point are itscoordinates.4._-8Fi gur e 4 - 1 . - R e c t a ngu l a r coor di na te s yst em .In notation used to locate points, the coordinates are conventionally placed in parentheses and separated with a comma, with the abscissa always written first. The general form of this notation is P(x,y). Thus, point P in figure 4-1 would have the notation P ( 4 , - 5 ) .5. The quadrants are numbered in the manner described in chapter 3 of this course (shown as Roman numerals in figure 4-1). 6. The x coordinate is positive in the first (I) and fourth (IV) quadrants and negative in the second (II) and third (III) quadrants. The y coordinate is positive in the first and second quadrants and negative in the third and fourth quadrants. The signs of the coordinates are shown in parentheses in figure 4-1. The algebraic signs of the coordinates of a point are used in this chapter for determining the algebraic signs of trigonometric functions.ANGLES IN STANDARD POSITION To construct an angle in standard position, first lay out a rectangular coordinate system. Then draw the angle, 9, so that its vertex is at the origin of the coordinate system and its initial or original side is lying along the positive X is as shown infigure 4-2. The terminal or final side of the angle will lie in any of the quadrants or on one of the axes separating the quadrants. When the terminal side falls on an is, the angle is called a quadrantal angle, which will be discussed later in this chapter. In figure 4-2 the terminal side lies in quadrant II. The quadrant in which an angle lies is determined by the terminal side. When an angle is placed in standard ____________ 1 ___ x position, the angle is said to lie in the quadrant containing ° the terminal side. For example, the negative angle, 0, shown in standard position in figure 4-3, is said to lie in the second quadrant. When two or more angles in standard position have their terminal sides located at the same position, they are said to be coterminal. If 0 is any general angle, then 0 Figure 4-2.-Angle in standard posiplus or minus an integral multiple of 360 °yields a tion. coterminal angle. For example, the angles 0, +, and a in figure 4-4 are y said to be coterminal angles. If then 0=45°x+ = 0 - 3 6 0 °= 45°- 360 °= -315°Figure 4-3.-Negative angle in quadrant H.anda = 0 + 3 6 0 °= 45 °+ 360 °= 405 ° The relationship of coterminal angles can be stated in a general form. For any angle 0 measured in degrees,any angle + coterminal with 0 can be found by + = 0 + n(360 ° )where n is any integer (positive, negative, or zero); that is,Figure 4-4.-Coterminal angles.n = 0 , ±1 , ±2 , ±3 , . .The principle of coterminal angles is used in developing other trigonometric relationships and other phases of trigonometric analysis. An expansion of this principle, discussed later in this chapter, states that the trigonometric functions of coterminal angles have the same value.PRACTICE PROBLEMS: Determine whether or not the following sets of angles are coterminal: , -3W° , 420° 1. 60 ° , 360 ° , 180 ° 2. 0 ° 3. 45 ° , - 45 ° , 345 °4. 735 ° , -345° ,-705°ANSWERS: 1. Coterminal 2. Not coterminal 3. Not coterminal4. CoterminalDEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS So far, the trigonometric functions have been defined as follows:1. By labeling the sides of a right triangle x, y, and r.2. By naming the sides of a right triangle adjacent, opposite, and hypotenuse.YVxAFigure 4-5.-Functions of general angles.In this chapter we will introduce a third set of definitions using the nomenclature of the coordinate system. Note that each definition defines the same functions using different terminology. To arrive at the third set of definitions, construct an angle in standard position on a coordinate system as shown in figure 4-5, view A. Choose point P(x,y) on the final position of the radius vector. Distance OP is denoted by the positive number r for the length of the radius. By constructing a right triangle using P(x,y) and r, as in figure 4-5, view B, the six trigonometric functions are classified as follows: sing = r cosh= r tano = x cot 8 = x y sec 0 = r xXordinate length of radius abscissa length of radius ordinate abscissa abscissa ordinate length of radius abscissa length of radius ordinateThe value of each function is dependent on angle 6 and not on the selection of point P(x,y). If a different point were chosen, the length of r, as well as the values of the x and y coordinates, would change proportionally, but the ratios would be unchanged. EXAMPLE: Find the sine and cosine of angle 6 in figure 4-5, view A, for the point P(3,4).SOLUTION: To determine the sine and cosine of 6, we must find the value of r. Since the values of the x and y coordinates correspond to the lengths of the sides x and y in figure 4-5, view B, we can determine the length of r by using the Pythagorean theorem or by recalling from Mathematics, Volume 1, the 3-4-5 triangle. In either case, the length of r is 5 units. Hence,sin 6 = ordinate ______ length of radius 4 _ -5 and cos 0 = length of radius abscissa _3 - 5 NOTE: For the remainder of this chapter, all angles areunderstood to be in standard position, unless otherwise stated. PRACTICE PROBLEMS: Find the sine, cosine, and tangent of the angles whose radius vectors pass through the following points: 1. P(5,12) 2. P(1,1) 3. P(1, \ ) 4. P(3,2)ANSWERS: 1. sin 6 = 12/13cos 6 = 5/13 tan 0 = 12/52.sin 6 = 1 / '1 1= /2cos 0 = 1/1 =/ V = /2 tane=3. sin 6 = cos 6 = 1 /2/2tan6='/1 =4.sin6 = 2/ 13 = 2 / 1 3 13/13cosh=3/ 1 3 = 3tan 0 =2/3QUADRANT SYSTEM The quadrants formed in the rectangular coordinate system are used to determine the algebraic signs of the trigonometric functions. The quadrants in figure 4-6 show the algebraic signs of the trigonometric functions in the various quadrants. In the first quadrant the abscissa and ordinate are always positive. The radius vector is always taken as positive. Therefore, all the trigonometric ratios arepositive for angles in the first quadrant. For angles in the second quadrant, only the ratios involving the ordinate and the radius vector are positive. These are the sine and cosecant ratios. For angles in the third quadrant, where the ordinate and abscissa are both negative, only the ratios involving the abscissa and the ordinate are positive.II sin e = +/+ = + cos 0 = -/+ = I sin cos tan cot sec csc 0 =tan sec csce = +/e = +/e = +/+= = = +cot 0 = -/+ = -e = + e = +/+ _6 = 0 = 0 =+/+ _+ +/+ _ + +/+ = + +/+ _ +IQsin 0 = -/+ = - sin cos e = -/+ _ - cos tan 0 = -/- = + tan cot e = -/- = + cot sec e = +/- _ - sec csc e = +/- = - csce e e e e e= -/+ _ = +/+ _ + = -/+ = = +/- _ = +/+ = + = +/-S+/+ + _Figure 4-6.-Signs of functions.These are the tangent and cotangent ratios. For angles in the fourth quadrant, ratios involving the radius vector and the abscissa are positive. These are the cosine and the secant ratios.NOTE: In each quadrant the sine and cosecant have the same sign, the cosine and the secant have the same sign, and the tangent and cotangent have the same sign. The last group of practice problems involved angles in the first quadrant only, where all of the functions were positive. When an angle lies in one of the other quadrants, the trigonometric functions may be positive or negative.EXAMPLE: Find all of the trigonometric functions of 9 if tan 9= 5/12, sin 9 & 0, and r = 13.SOLUTION: Reference to figure 4-6 shows thatan angle with a positive tangent and a negative sine can only occur in the third quadrant. The point in the third quadrant has coordinates (- 12, - 5). (See fig. 4-7) We can now read the trigonometric ratios from the figure: sin B - ____________ ordinate ____ _ - 5 - length of radius = 13 cos 9 = ____ abscissa ______ - 12 length of radius 13 tan - ordinate = - 5 _ 5 abscissa 12 = 12 cot 9 abscissa = - 12 = 12 ordinate -5 5 - 13 12 - 13 5YxFigure 4-7.-Finding the trigonometric functions for a third-quadrant angle.sec - length of radius = _____ 13 abscissa - 12 csc 9 - length of radius = 13 ordinate -5EXAMPLE: Find all of the trigonometric functions of 9 if csc 9 = - 17/15 and cos 9 & 0. SOLUTION: The cosecant is negative in the same quadrants as the that is, quadrants III and IV. The cosine is negativein quadrants II and III. Therefore, the cosecant and cosine are both negative in quadrant III. (Refer to fig. 4-6.) The ordinate in the third quadrant is - 15 and the radius is 17. NOTE: The fraction - 17/15 indicates that either the numerator or denominator is negative, but not both. In this case, we know that the ordinate (denominator) is negative since the radius (numerator) is always positive. From the Pythagorean theorem the abscissa in the third quadrant is x2 = r 2 - y2= (17) 2 - ( -15) 2 = 289 - 225 = 64Therefore, referring to figure 4-8, the six trigonometric functions are as follows:Ysin 0 = -15/17 cos 0 = -8/17 tan 0 = - 1 5 / - 8 = 15/8 cot 0 = - 8 / - 1 5 = 8/15-15 17xsec 9 = 1 7 / - 8 = - 1 7 / 8 csc 9 = 17/ - 15 = - 17/ 15P(-8,-151Figure 4-8.-Construction of a triangle in quadrant 3.tan 8 = - 7/24, find the other four trigonometric ratios of 0.EXAMPLE: If sec 0 = - 25/24 andSOLUTION: The secant and tangent are both negative in the second quadrant. In the second quadrant the abscissa is - 24, the ordinate is 7, and the radius is 25 (refer to fig. 4-9); so, sin 6 = 7/25 cos 6 = - 24/25 cot 6 = -24/7 csc 8 = 25/7YxPRACTICE PROBLEMS: Without using tables, find the six trigonometric functions of 0 under the following conditions: 2. tan 6 = -21/20, r = 29, and cos 6 & 0. 3. cos = - 3 / 5 and cot 8 = 3/4. 4. tan 6 = -8/15 and csc 6 is positive. Indicate the quadrant in which the terminal side of 6 lies for the following conditions: 5. sin 6 & O and cos 6 & 0 6. cos 6 & O and csc 6 & 0 7. sec 6 & O and cot 0 & 0Figure 4-9.-Construction of a triangle in quadrant 2.1. tan 6 = 3/4, r = 5, and 6 is not in the first quadrant.ANSWERS: 1. sin 9 = -3/5 cos 6 = -4/5 tan 8 = - 3/ - 4 = 3/4 cot 0 = - 4/ - 3 = 4/3 sec 6 = 5 / - 4 = - 5 / 4 csc 6 = 5/-3 = -5/32.sin 0 = - 21 /29 cos 0 = 20/29 tan 0 = - 21 /20 cot 0 = 20/ - 21 = -20/21 sec 0 = 29/20 csc 0 = 29/ - 21 = -29/213.sin 0 = -4/5 cos 0 = -3/5 tan 0 = - 4/ - 3 = 4/3 cot 0 = - 3/ - 4 = 3/4 sec 0 = 5/-3 = -5/3 csc 0 = 5/-4 = -5/44.sin 0 = 8/17 cos 0 = -15/17 tan 0 = 8/-15 = -8/15 cot 0 = -15/8 sec 0 = 17/-15 = -17/15 csc 0 = 17/85. 6. 7.2 3 4REFERENCE ANGLE The reference angle, 0', for any angle, 0, in standard position is the smallest positive angle between the radius vector of 0 andthe X axis, such that 0 °& 9 & 90 ° . In general, the reference angle for 0 is 6' = n(180° ) ±e where n is any integer. Expressed in an equivalent form0 ' = n n ±ewhere again n is any integer and 0 & 6 ' & n/2. Refer to figure 4-10. If Pis any point on the radius vector, a perpendicular from P to the point A on the X is forms a right triangle with sides O A , A P , and O P . We call this triangle the reference triangle. The relationship between 0, 6 ', and the reference triangle in each quadrant is shown in figure 4-10. FUNCTIONS OF ANGLES IN ANY QUADRANT In addition to the reference triangle, formulas are used for determining the signs of the functions at any angle. These are called reduction formulas. This section shows the geometrical development of some of the most commonly used reduction formulas. In general, reduction formulas provide a means of reducing the functions of any angle to an equivalent expression for the function in terms of a positive acute angle, O . The reduction formulas can be used in the solution of some trigonometric identities and in other applications requiring analysis of trigonometric functions.Figure 4-10.-Reference triangles in each quadrant.The function of 9 and the reduction formulas of the functions of 180 °- e , 180°+ 0, and 360°- 9 are summarized in the following paragraphs according to their respective quadrants. QUADRANTI Any angle in the first quadrant can be represented by 9; that is, sin 6 = cos e = rxrtan 0 = cot9=xx Y sec 0 = r xcsc 9 = r Y QUADRANT II Any angle in the second quadrant can be represented by 180°9; that is, sin (180°- 9)=r=xsin 0cos (180°- 6) tan (180°- 9) cot (180°- 0)=-= -cos 9= - x = -tan 9= - y = -cot esec (180°- e) = - x = -sec 9 csc (180°- 0) = r = csc 9EXAMPLE: Use a reduction formula and appendix III to find the cotangent of 112° . SOLUTION: Since 112°is in the second quadrant, where cot (180 °- 0) = -cot 0 then cot 112°= cot (180°- 68° )= -cot 68° = -0.40403 QUADRANT III Any angle in the third quadrant can be represented by 180°+ 9; that is, sin (180°+ 0)= - r = -sin 0cos (180°+ 9) = - x = -cos 0 tan (180°+ 0) = x = tan 0 cot ( 1 8 0 °+ 0 ) = - = c o t 0 sec (180°+ 0) = - r = -sec 0 x csc (180°+ 0) = - r = -csc 9 EXAMPLE: Use a reduction formula and appendix II to find the sine of 220° . SOLUTION: Since 220°is in the third quadrant, where sin (180°+ 0) = -sin 0then sin 220 °= sin (180 °+ 40 ° ) = -sin 40° _ - 0.64279 QUADRANT IV Any angle in the fourth quadrant can be represented by 360 °- 6; that is, sin (360 °- 6) = - r= - sin 0cos (360 °- 8) = x = cos 6 - 0) cot tan (360 ° - 6) = - x = - tan 0 = - y = - cot 6sec (360°- 6) = r = sec 6 x csc (360°- 6) = - y = -csc 6EXAMPLE: Find cos 324° .SOLUTION: Sincecos (360°- 0) = cos 0 then cos 324°= cos (360°- 36° ) = cos 36° = 0.80902FUNCTIONS OF NEGATIVE ANGLES The following relationships enable us to change a function with a negative angle into the same function with a positive angle: sin (-0) = -r = - sin 9cos (-0) = x = cos 0 tan (- 0) = -X = - tan 0 cot (-0) = - x = -cot 0 sec (-0) = r = sec 6 x csc (-6) = - y = -csc 6 EXAMPLE: Find tan (-350'). SOLUTION: Since tan L e ) = - tan 0 then tan (-350° ) = - t a n 350 ° and -tan 350 °= -tan (360 °- 10 ° )= - ( - t a n 10° ) = 0.17633 FUNCTIONS OF COTERMINAL ANGLES For a coterminal angle in the form of 0' = n(360 ° )+9where n is any integer 8 and is an integral multiple of 9 ', the trigonometric functions of 9 ' are equal to those of O. In other words, 0 is the remainder obtained by dividing 8 ' by 360, and n is the number of times 360 will divide into 0 ' . Thus, we can find the ratios of a coterminal angle greater than 360 °by dividing 9 ' by 360 and finding the functions of the remainder. EXAMPLE: Find the cosine of -2,080° . (Refer to fig. 4-11.) SOLUTION: Divide 2,080 by 360. 5Y,800 280-280°So, cos (-2,080° ) = cos (-280° ) and cos (-280° ) = cos (280° ) = cos (360°- 80° )Figure 4-11.-Coterminal angles of - 2,080 ° .= cos 80°= 0.17365PRACTICE PROBLEMS: Use reduction formulas and appendixes II and III to find the values of the sine, cosine, and tangent of 8 given the following angles: 1. 137° 2. 214° 3. 325° 4. -70° 5. 1,554°ANSWERS: 1. sin 137 °= sin 43°= 0.68200 cos 137 °= -cos 43°= -0.73135 tan 137 °= -tan 43°= -0.93252 2. sin 214' = -sin 34°= -0.55919 cos 214°= -cos 34°= -0.82904 tan 214 °= tan 34°= 0.67451 3. sin 325 °= -sin 35°= -0.57358 cos 325 °= cos 35°= 0.81915 tan 325 °_ -tan 35°= -0.7?21 4. sin (-70') = -sin 70°_ -0.93969 cos (- 70 ° ) = cos 70°= 0.34202 tan (-70') = -tan 70°= -2.74748 5. sin 1,554°= sin 114°= sin 66°= 0.91355 cos 1,554° = cos 114°= -cos 66°= -0.40674 tan 1,554 °= tan 114 °= -tan 66°= -2.24604COFUNCTIONS AND COMPLEMENTARY ANGLESComplementary angles are angles whose sum is 90'. Twotrigonometric functions that have equal values for complementary angles are called cofunctions.Inspect the triangle in figure 4-12. We will compare the six trigonometric functions of 0 with the six trigonometric functions of 9 0 °- 0 . Functions of 0 sin 0 Functions of 90°- 8 cos (90 °- 0) ==rrcos 0 = x r tan 0 = zsin (90 °- 0) = cot (90 °- 0) =rrFigure 4-12.-Complementary angles.cot 0 = x tan (90 °- 0) = sec 0 = r x csc 8 = Y csc (90°- 9) = r x sec (90°- 0) = yWe see from t he above relationships that sin 0 = cos (90 ° - 0) cos 0 = sin (90 ° - 0) tan 0 = cot (90°- 0) cot 0 = tan (90 ° - 0) sec 0 = csc (90 ° - 0) csc 0 = sec (90°- 0) Hence, a trigonometric function of an angle is equal to the confunction of its complement. NOTE: These relationships may explain to you how the cosine, cotangent, and cosecant functions received their names. The confunction principle accounts for the format of the tables of trigonometric functions in appendixes II and III. For example, in appendix II sin 21 °30 ' = 0.36650and cos 68°30' = 0.36650 Notice that 21° 30' + 68°30' = 90° PRACTICE PROBLEMS: Express the following as a function of the complementary angle: 1. sin 27° 2. tan 38°17' 3. csc 41 ° 4. cos 16°30' 5. sec 79°37 ' 6. cos 56° 7. cot 48° 22& 16&ANSWERS: 1. cos 63 ° 2. cot 51 °43 ' 3. sec 49° 4. sin 73°29' 5. csc 10°22' 6. sin 34° 7. tan 42 ° 38& 44&SPECIAL ANGLES Two groups of angles are discussed in this section. The first group of angles is considered because the angles can be determined geometrically and are used frequently in problem solving. The second group is considered because the radius vectors of the angles fall on one of the coordinate axes, not in one of the quadrants. FREQUENTLY USED ANGLES As stated previously, the approximate values of the trigonometric functions for any angle can be read directly from tables or can be determined from tables by the use of the principles stated in this text. However, certain frequently used simple angles exist for which the exact function values are often used because these exact values can easily be determined geometrically. In the following paragraphs the geometrical determination of these functions is shown. 30 0 -60 °Angles The trigonometric functions of 30 °and 60 °can be determined geometrically. Construct an equilateral triangle with side lengths of 2 units, such as triangle OYA in figure 4-13. (The functions to be determined are not dependent on the lengths of the sides being 2 this size was selected for convenience.) Drop a perpendicular from angle Y to the base of the triangle at point X. The right triangles YXO and YXA are formed by the perpendicular, which also bisects angle Y forming a 30 °angle. Moreover, since side OA is 2 units long, then OX is 1 unit long and YX is v5 units long (using the Pythagorean theorem). Figures 4-14 and 4-15 show a 30 ° and a 60 ° reference triangle, respectively. From these figures we can determine the trigonometric ratios of 30 °and 60 ° , which are summarized in table 4-1. 1Figure 4-13.-Geometrical construction of 30°and 60°right triangles.YFigure 4 - 1 5 . - 6 0 °reference triangle. Table 4-1.-Trigonometric Functions of Special Angles0 30' ?tn 6 I2 60' 45?costan 9ea I? cc9C.C a2a,I 2I 4~ 4 f12 r3~ 31T4a7as f1 r3T fNOTE: The values of the confunctions interchange since 30 ° and 60 ° are complementary angles. An easy way to recall the values of the functions of right triangles with 30° and 60 ° complementary angles is to remember that the ratio of the sides is always 1, 2, and where the largest side value represents the length of the hypotenuse. EXAMPLE: Find the six trigonometric functions of 300 ° . SOLUTION: Referring to figure 4-16, 300 °is in the fourth quadrant and its reference angle is 60 ° . Therefore, sin 300 °= - / 2 cos 300 °= 1 /2 tan 300°= - ~ / 1 = cot 300 °= 1/ - ' = sec 300 °= 2/1 = 2 csc 300 °= 2/ 45 °Angles Refer to figure 4-17. If one of the acute angles of the right triangle OXY is 45 ° , then the other acute angle is also 45 ° . Since triangle OXY is an isosceles triangle, then sides OX and XY are equal. If we let OX and XY be 1 unit long, then bathe Pythagorean theorem, the length of OY is V 2 units. NOTE: This relationship is true of all 45 ° triangles and is not altered by the lengths of the legs. The ratio of the sides of right triangles with 45 °complementary angles will always be 1, 1, and where the largest value represents the length of the hypotenuse. = -2/3 - ~ -v,V/3Figure 4-16.-300°angle in standard position.Figure 4-17.-C~eomctrical construction of a right triangle.as °Figure 4-18 shows a 45 °reference triangle. From this figure we can determine the trigonometric ratios of 45 ° , which are also summarized in table 4-1. Y EXAMPLE: Find the six trigonometric functions of 135 ° . SOLUTION: Referring to figure 4-19, 135 °is in the second quadrant and its reference angle is 45 ° . Therefore, sin 135 °= 1 / v [ = /2cos 135 °= - 1 / v [ = - ' / 2 tan 135°= 1 / - 1 = - 1 cot 135°= - 1 / 1 = - 1 sec 135°= ' / - 1 c s c 1 3 5 °= = QUADRANTAL ANGLES An angle whose terminal side lies on a coordinate is when the angle is in standard position is a quadrantal angle. Angles of 0 ° , f 90 ° , ± 180 ° , and ± 270 °are quadrantal angles. The trigonometric functions of the quadrantal angles are defined in the same manner as before, except for the restriction that a function is undefined when the denominator of the ratio is zero. To derive the functions of the quandrantal angles, we choose points on the terminal sides, where r = 1, as shown in figure 4-20. Then either x or y is zero, and the other coordinate is either positive or negative 1. r Y e=0? U,0) r=I 0=180? X=~Figure 4 -1 8 . - 4 5 °reference triangle.Figure 4 -1 9. -13 5°angle in standard position.YBCDFigure 4-20.-Functions of quadrantal angles.Table 4-2.-Functions of Quadrantal Angles e ~g 0° 90° 180° 270° Had 0 sin 9 0 10cos 9 1 0-1tan 9 0 undefined0cot 0 undefined 0 undefined 0sec 9 1 undefined -1 undefinedcsc 9 undefined 1 undefined -12n 3w-10undefined2Consider view C of figure 4-20 in which 9 = 180 ° . For the point P ( - 1,0) and r = 1, we have sin 180°= 0/1 = 0 cos 180°= - 1/1 = - I tan 180°= 0 / - 1 = 0 cot 180°= - 1 / 0 (undefined) sec 180°= 1 / - 1 = - 1 csc 180° = 1 /0 (undefined) The values of the functions of the other quadrantal angles can be found by a similar procedure and are summarized in table 4-2.EXAMPLE: Determine the six trigonometricfunctions of 990° . SOLUTION.- Referring to figure 4-21, we see that 990°lies on the same quadrantal axes as 270 ° . Therefore, for P(0,-1) and r = 1, we have sin 990°= - 1 / 1 = - 1 cos 990°= 0/1 = 0 tan 990°= - 1 / 0 (undefined) cot 990°= 0 / - 1 = 0Figure 4-21.-990 °angle. Ysec 990°= 1 /0 (undefined) csc 990°= 1 / - 1 = - 1PRACTICE PROBLEMS: Without using the appendixes, determine the trigonometric functions of problems 1 through 5. 1. 0 = 210° 2. 0 = 360 ° 3. 0 = 585° 4. 0 = -180° 5. 0 = -315° Without using the appendixes, evaluate problems 6 through 8. [Note that sin' 0 = (sin 0)'.) 6. sin' 150 °+ cos' 150 ° 7. 2 sin 120°cos 120°8. cos'135 °- sin'135 °ANSWERS: 1. sin cos tan cot sec 210° 210° 210° 210° 210°csc 210° 2. sin 360° cos 360°=tan 360°= cot 360 °= 1 /0 (undefined) sec 360°= 1/1 = 1 csc 360°= 1 /0 (undefined)2/-=-2V/32/-1 = -2 0/1 = 0 1/1 = 1 0/1 = 03. sin 585°= - 1 / V = - / 2 cos 585°= = - v 9 / 2 tan 585°= - 1 / - 1 = 1 cot 585°_ - 1 / - 1 = 1 sec 585 °= csc 585 °= /- 1 = - V /- 1 = -4. sin (-180° ) = 0/1 = 0 cos (-180° ) = -1/1= -1tan (-180° ) = 0/-1= 0 cot (-180° ) = -1/0 (undefined) sec (-180° ) = 1 / - 1 = - 1 csc (-180° ) = 1 /0 (undefined) 5. sin (-315° ) = 1 /V = cos (-315° ) = = tan (-315° ) = 1/1 = 1 cot (-315° ) = 1/1 = 1 sec (-315° )= csc (-315° )= 6. (1/2) 2 +( /= = 2) 2 = 17. 2 ( / 2 ) ( - 1 / 2 ) _ - ' / 2 8. (-1/) 2 - (1 / \ ) 2 = 0Figure 4-22.-Graph of the sine function.PERIODS OF THE TRIGONOMETRIC FUNCTIONS A trigonometric function of an angle is not changed in value when the angle is changed by any multiple of 360 °or 2n radians. For this reason the functions are said to be periodic. In the following paragraphs, the graphs of the sine, cosine, and tangent functions are discussed. GRAPH OF THE SINE FUNCTION Figure 4-22 shows two periods of the sine function. The graph shows that the value of the sine function varies between + 1 and 1 and never goes beyond these limits as the angle varies. The graph also shows that the sine function increases from 0 at 0 °or 0 radians to a maximum value of + 1 at 90° or n/2 radians. It decreases back to 0 at 180 °or it radians and continues to decrease to a minimum value of - 1 at 270 ° or 3n/2 radians. It then increases to a value of 0 at 360 °or 2n radians. If we extend the graph (in either direction), the sine function will continue to repeat itself every 360 ° or 2n radians. Therefore, the period of the sine function is 360 °or2n radians.GRAPH OF THE COSINE FUNCTIONThe cosine function also has a period of 360 °or 2n radians. Figure4-23shows two periods of the cosine function. The range yFigure 4-23.-Graph of the cosine function.of the values the cosine function takes on also lies between + 1 and - 1. However, as seen on the graph, the cosine function decreases from 1 at 0° or 0 radians to 0 at 90° or n/2 radians and continues to decrease to a minimum value of - 1 at 180 °or it radians. It then increases to 0 at 270 °or 3n/2 radians and continues to increase to a maximum value of + 1 at 360 °or 2n radians. This completes one period of the cosine function. GRAPH OF THE TANGENT FUNCTION Figure 4-24 shows the graph of the tangent function from 0 radians to 2n radians. Notice that the tangent function is 0 at 0 °or 0 radians and increases to positive infinity (without bounds) between 0 °and 90 °or 0 radians and n/2 radians. Remember that the tangent function is undefined for 90 ° + n(180 ° ) or n/2 + nn, where n is any integer. The dashed vertical lines in figure 4-24 represent the undefined points. The tangent function increases from negative infinity to 0 between 90 ° and 180 ° or n/2 radians and it radians. At 180 °or it radians, the tangent function is 0. The function continues to increase from 0 to positive infinity between 180 °and 270 °or it radians and 3n/2 radians. Between 270 ° and 360 ° or 3n/2 radians and 2n, it again increases from negative infinity to 0 at 360 ° or 2n radians. If we extend the graph (in either direction), the curve will repeat itself every 180 °or it radians. Therefore,the period of the tangent function is 180°or it radians. EXAMPLE: Using the graphs in figures 4-22 through 4-24,xFigure 4-24.-Graph of the tangent function.determine the values of 0, where sin 0 and tan 0 increase together, if0&0&n. SOLUTION: In figure 4-22 the sine function increases between 0 and n/2 radians for the interval of 0 & 0 & i t. (The sine function does not increase or decrease at points 0 or n/2.) In figure 4-24 the tangent function also increases between 0 and n/2 radians for theinterval of 0 & 0 & i t . (The tangent function does not increase or decrease at 0 and is undefined at n/2.) Therefore, the values of 0, where sin 0 and tan 0 increase together, are 0 & 0 & n/2.PRACTICE PROBLEMS: Use the graphs in figures 4-22 through 4-24 to answer the following problems (use appendixes II and III to verify your answers):1. For what values of 8 does cos 8 increase if 0 & 0 & n? 2. For what values of 0 do sin 8 and cos 0 decrease together if 0 & 8 & 2 n ? 3. For what values of 8 do cos 8 and tan 8 increase together if n/2 & 8 3n/2? 4. For what values of 8 do sin 8, cos 8, and tan 8 increasetogether if 0 & 8 & 2n? ANSWERS: 1. None2. n / 2 & 0 & n 3. n & 8 & 3 n / 2 4.3n/2&0&2nSUMMARY The following are the major t}