Transcription
version: 1.1CHAPTER9Fundamentals ofTrigonometry
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.PunjabeLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjab9.2.1. Sexagesimal System: (Degree, Minute and Second).9.1 IntroductionTrigonometry is an important branch of Mathematics. The wordTrigonometry has been derived from three Greek words: Trei (three), Goni (angles) andMetron (measurement). Literally it means measurement of triangle.For study of calculus it is essential to have a sound knowledge of trigonometry.It is extensively used in Business, Engineering, Surveying, Navigation, Astronomy,Physical and Social Sciences. If the initial ray OA rotates in anti-clockwise direction in such a way that it coincideswith itself, the angle then formed is said to be of 360 degrees (360 ).One rotation (anti-clockwise) 360 1rotation (anti-clockwise) 180 is called a straight angle21rotation (anti-clockwise) 90 is called a right angle.49.2 Units of Measures of AnglesConcept of an AngleTwo rays with a common starting point form an angle. One of the rays of angle is calledinitial side and the other as terminal side. The angle is identiied by showing the direction ofrotation from the initial side to the terminal side.An angle is said to be positive/negative if the rotation is anti-clockwise/clockwise. Anglesare usually denoted by Greek letters such as a (alpha), b (beta), g (gamma), q (theta) etc.In igure 9.1 AOB is positive and COD is negative.1rotation 180 21 rotation 360 1rotation 90 41 degree (1 ) is divided into 60 minutes ( 60′ ) and 1 minute ( 1’) is divided into 60 seconds( 60′′ ). As this system of measurement of angle owes its origin to the English and because 90,60 are multiples of 6 and 10, so it is known as English system or Sexagesimal system.Thusigure 9.11 rotation (anti-clockwise)One degree (1 )One minute (1′ ) 360 . 60’ 60”9.2.2. Conversion from D M’ S” to a decimal form and vice versa.There are two commonly used measurements for angles: Degrees and Radians. whichare explained as below:(i)16 30′ ′ 30 16.5 (As1 20.5 )version: 1.1version: 1.123
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry(ii)45.25 eLearn.Punjab45 15′ (0.25 25 1 1004Example 1: Convert 18 6′ 21′′ to decimal form. 1 1 ′′′Solution:1′ and1 60 60 1 60 60 60 415′)Consider a circle of radius r. Construct an angle AOB at thecentre of the circle whose rays cut of an arc AB on the circle whoselength is equal to the radius r.1 radian.Thus m AOB 1 1 18 6′ 21′′ 18621 60 60 60 (18 0.1 0.005833) 18.105833 9.3 Relation between the length of an arc of a circle and thecircular measure of its central angle.Example 2: Convert 21.256 to the D M ′ S ′′ form andProve that q 0.256 (0.256)(1 ) (0.256) ( 60′ )0.36′ (0.36)(1′)eLearn.Punjabmost useful for the study of higher mathematics. Specially in Calculus, angles are measuredin radians.Deinition: Radian is the measure of the angle subtended at the center of the circle by anarc, whose length is equal to the radius of the circle. Solution:eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry15.36′ (0.36)(60′′)lrwhere r is the radius of the circle l, is the length of the arc and q is the circular measureof the central angle.Proof:21.6′′Therefore,21.256 21 0.256 21 15.36′ 21 15′ 0.36′By deinition of radian;An angle of 1 radian subtends an arc AB on the circle of length 1.r 21 15′ 21.6′′ 21 15’ 22”An angle ofrounded of to nearest second11radian subtends an arc AB on the circle of length .r22An angle of 2 radians subtends an arc AB on the circle of length 2.r An angle of q radian subtends an arc AB on the circle of length q.r9.2.3. Circular System (Radians)There is another system of angular measurement, called the Circular System. It is AB q .rversion: 1.1version: 1.145
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjab l q q .r l rAlternate ProofLet there be a circle with centre O and radius r. Suppose that length of arc AB l andthe central angle m AOB q radian. Take an arc AC of length r.1 radian.By deinition m AOC We know from elementary geometry that measures of central angles of the arcs of acircle are proportional to the lengths of their arcs. m AOBm AB m AOCm ACq radianrq 2p radianThus we have the relationship2p radian 360 p radian 180 1 radian 1 q lrThus the central angle q (in radian) subtended by a circular arc of length l is given bylq , where r is the radius of the circle.r ql r6 23pp180radian.3.1416 0.0175 radain180Example 3: Convert the following angles in degree:2pradain(ii)32p2 radains(p radain)Solution:(i)333 radians.3 radains 3(1 radain)(ii)2(180 )3 120 3(57.296 ) 171.888 Example 4: Convert 54 45 ‘ into radians. 45 3 ′ 54 45 54 54 60 4 9.3.1 Conversion of Radian into Degree and Vice VersaSolution:We know that circumference of a circle of radius r is 2p r (l ), and angle formed by onecomplete revolution is q radian, therefore,q 180 57.296 3.1416180 (i)Remember that r and l are measured in terms of the same unit and the radian measureis unit-less, i.e., it is a real number.For example, if r 3 cm and l 6 cmthen eLearn.Punjab2p rq Furtherl 1 radianreLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry 219 4219 (1 )4lrversion: 1.1version: 1.167
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry eLearn.Punjab219(0.0175) radinas4 0.958 radains.Most calculators automatically would convert degrees into radians and radians intodegrees.eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.PunjabExample 7: Find correct to the nearest centimeter, the distance at which a coin of diameter‘1’ cm should be held so as to conceal the full moon whose diameter subtends an angle of31’ at the eye of the observer on the earth.Example 5: An arc subtends an angle of 70 at the center of a circle and its length is132 m.m. Find the radius of the circle.Solution:3.1416701170 70radains (3.1416)radain radains. (p180180911 qradain9 q and PS can be taken as the arc .of the circle with centre O andradius OP.l 132m.m.l9r 132 108 m.m.11qOP r , l 1 cmNow Example 6: Find the length of the equatorial arc subtending an angle of 1 at the centre ofthe earth, taking the radius of the earth as 6400 km. q 1 Solution: r Now q q p180radains3.1416180and3.1416radain180lrql ,1 60 18031 pq 31′ 31 pradains60 18060 180 110.89 cm.31 3.1416Hence the coin should be held at an approximate distance of 111 cm. from the observer’sr 6400 km.eye.Note: If the value of p is not given, we shall take p 3.1416.lrl rqm ACO ( m POC ) is very very small.Now since lr3.1416)Solution: Let O be the eye of the observer. ABCD be the moon and PQSR be the coin, so thatAPO and CSO are straight line segments. We know that mPS 1 cm, m AOC 31′ 6400 314161800000 111.7 kmversion: 1.1version: 1.189
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjab9.Exercise 9.11.Express the following sexagesimal measures of angles in radians:i)30 ii)45 iii)60 iv)75 v)90 vi)105 vii) 120 viii) 135 ix)xiii)2.vi)xi)5.6.7.8.120′ 40′′10 15′x)xiv)154 20′′ xi)35 20′xii)xv)0 xvi)75 6′ 30′′3′′Convert the following radian measures of angles into the measures of sexagesimalsystem:i)3.4.150 17p24iv)ix)xiv)p37p1225p36v)x)xv)p29p519p32What is the circular measure of the angle between the hands of a watch at 4 O’clock?Find q , when:i)l 1.5 cm,r 2.5cmii)l 3.2m,r 2mFind l, when:q p randains,r 6cmi)ii)q 65 20′Find r, when:r 18mmi)l 5 cm,q ii)l 56 cm,q 45 eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry10.11.12.13.14.15.16.17.eLearn.PunjabA railway train is running on a circular track of radius 500 meters at the rate of 30 kmper hour. Through what angle will it turn in 10 sec.?A horse is tethered to a peg by a rope of 9 meters length and it can move in a circlewith the peg as centre. If the horse moves along the circumference of the circle, keepingthe rope tight, how far will it have gone when the rope has turned through an angle of70 ?The pendulum of a clock is 20 cm long and it swings through an angle of 20 eachsecond. How far does the tip of the pendulum move in 1 second?Assuming the average distance of the earth from the sun to be 148 x 106 km and theangle subtended by the sun at the eye of a person on the earth of measure 9.3 x 10-3radian. Find the diameter of the sun.A circular wire of radius 6 cm is cut straightened and then bent so as to lie along thecircumference of a hoop of radius 24 cm. Find the measure of the angle which itsubtends at the centre of the hoop.1Show that the area of a sector of a circular region of radius r is r 2 q , where q is the2circular measure of the central angle of the sector.Two cities A and B lie on the equator such that their longitudes are 45 E and 25 Wrespectively. Find the distance between the two cities, taking radius of the earth as6400 kms.The moon subtends an angle of 0.5 at the eye of an observer on earth. The distanceof the moon from the earth is 3.844 x 105 km approx. What is the length of the diameterof the moon?The angle subtended by the earth at the eye of a spaceman, landed on the moon, is1 54’. The radius of the earth is 6400 km. Find the approximate distance between themoon and the earth.1radian29.4 General Angle (Coterminal Angles)What is the length of the arc intercepted on a circle of radius 14 cms by the arms of acentral angle of 45 ?Find the radius of the circle, in which the arms of a central angle of measure 1 radiancut of an arc of length 35 cm.version: 1.110There can be many angles with the same initial and terminal sides. These are called coterminal angles. Consider an angle POQ with initial side OP and terminal side OQ withversion: 1.111
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometryvertex O.Let m POQ q radian, where 0 qeLearn.PunjabeLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjab General angle is q 2kp , k Z ,2p9.5 Angle In The Standard Position Now, if the side OQ comes to its present position after one or more complete rotationsin the anti-clockwise direction, then m POQwill beq 2p , after one revolutionq 4p , after two revolutions,ii)i)An angle is said to be in standard position if its vertex lies at the origin of a rectangularcoordinate system and its initial side along the positive x-axis.The following igures show four angles in standard position:An angle in standard position is said to lie in a quadrant if its terminal side lies in thatquadrant. In the above igure:Angle a lies in I Quadrant as its terminal side lies is I QuadrantAngle b lies in II Quadrant as its terminal side lies is II QuadrantAngle g lies in III Quadrant as its terminal side lies is III Quadrantand Angle q lies in IV Quadrant as its terminal side lies is IV QuadrantIf the terminal side of an angle falls on x-axis or y-axis, it is called aHowever, if the rotations are made in the clock-wise direction as shown in the igure,m POQ will be:i)q - 2p , after one revolution,ii)q - 4p , after two revolution, It means that OQ comes to its original position after every revolution of 2p radians inthe postive or negative directions.In general, if angle q is in degrees, then q 360k where k Z , is an angle coterminalwith q . If angle q is in radians, then q 2kp where k Z , is an angle coterminal with q .quadrantal angle.i.e., 90 , 180 , 270 and 360 are quadrantal angles.9.6 Trigonometric FunctionsConsider a right triangle ABC with C 90 and sides a, b, c, as shown in the igure. Letm A q radian.version: 1.1version: 1.11213
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.PunjabeLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjabthe angle be (x, y).If rx 2 y 2 denote the distance from O (0, 0) to P (x, y), then six trigonometricfunctions of q are deined as the ratiosThe side AB opposite to 90 is called the hypotenuse (hyp),The side BC opposite to q is called the opposite (opp) andthe side AC related to angle q is called the adjacent (adj)yrsin q ; cscq ( y 0) ; tan qryxrcosq ; secq ( x 0); cot qrxWe can form six ratios as follows:a b a c cb, , , ,andc c b a baIn fact these ratios depend only on the size of the angle and not on the triangle formed.Therefore, these ratios are called trigonometric functions of angle q and are deined asbelow:Sin q:Sin q a copp; Cosecant q : csc qhypcahyp;oppCosine q : Cos q b cadj; Secant qhypcbhyp;adjTangent q : tan q aopp ; Cotangent q :badj: sec qcot q ba( x 0)(y0)Note: These deinitions are independent of the position of the point P on the terminal sidei.e., q is taken as any angle.9.8 Fundamental IdentitiesFor any real number q , we shall derive the following three fundamental identities:sin 2 q cos 2 q 1i)adj.opp1 tan 2 qii)We observe useful relationships between these six trigonometric functions asfollows:11sin qcscq;sec q; tan q; sin qcosqcosqcosq1cot q;cot q; sin qtan q9.7 Trigonometric Functions of any angle sec 2 qiii) 1 cot 2 qcsc 2 q .Proof:(i) Refer to right triangle ABC in ig. ( I) by Pythagoras theorem, we have, Dividinga2 b2 c2 both sides by c2, we geta2b2c2 2 c2ca2 Now we shall deine the trigonometric functions of any angle.q radian inConsider an angle XOP standard position.Let coordinates of P (other than origin) on the terminal side of a b 1 c c (sin q ) 2 (cosq ) 2 1221 sin 2 q cos 2 q (1)version: 1.1version: 1.114yxxy15
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometryii)eLearn.Punjab Again as a2 b2 c2Dividing both sides by b2, we get(ii)a2b2c2 2 b2bb2 (iii) a c 1 b b 2(iii)1 ta n 2 q sec 2 qAgain as a2 b2 c2Dividing both sides by a2, we get(iv)a2b2c2 a2a2a2 ve 0x0, -cosqryve 0, tanq ve x0If q lies in Quadrant IV, then a point P(x, y) on its terminal side has positive x-coordinate.and negative y-coordinate.ysin q- ver0x cosq rve0- tanq ve0These results are summarized in the following igure. Trigonometric functionsmentioned are positive in these quardrants.21 cot 2 q csc 2 qNote: (sinq ) 2sin 2 q , (cosq ) 2xy- ve 0, tanq rxIf q lies in Quadrant III, then a point P(x, y) on its terminal side has negative x-coordinate.and negative y-coordinate. b c 1 a a (cscq ) 2 1 (cot q ) 2 2If q lies in Quadrant II, then a point P(x, y) on its terminal side has negative x-coordinate.and positive y-coordinate.y sin q ver(2)eLearn.PunjabAll trigonometric functions are ve in Quadrant I.y sin q -ve 0, cosqr2(tan q ) 2 1 (secq ) 2eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry(3)cos 2 q and (tanq ) 2tan 2 q etc.9.9 Signs of the Trigonometric functionsIf q is not a quadrantal angle, then it will lie in a particular quadrant. Becauser x 2 y 2 is always positive, it follows that the signs of the trigonometric functions canbe found if the quadrant of 0 is known. For example,(i)If q lies in Quadrant I, then a point P(x, y) on its terminal side has both x, y co-ordinates veIt is clear from the above igure thatsin (-q ) - sin q ;csc(-q ) - cscqcos(-q ) cosq ;tan(-q ) - tan q ;sec(-q ) secqcot(-q ) - cot qversion: 1.1version: 1.11617
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryExample 1: If tan q eLearn.Punjab8and the terminal arm of the angle is in the III quadrant, ind the15tan q cot q 815 The terminal side of the angle is in the IV quadrantsec q Now115 tan q82a2289secq 225 1715cos q sin q and1 sec q 115 1717158 15 tan q . cos q - 15 17 25169sin-q5 131113cosec q 5sin q513cot q 1117 sin q - 88171441 1695 13tan q 8sin q 17csc q sin q2As the terminal side of the angle is in the IV quadrant where sin q is negative.The terminal arm of the angle is in the III quadrant where sec q is negative17 sec-q 15Now113 12 1213 12 sin q - 1 cos -q 1 - 13 2 1 cosq264 289 8 sec q 1 tan q 1 1 225 225 15 2eLearn.PunjabSolution: The terminal side of the angle is not in the I quadrant but cos q is positive,values of the other trigonometric functions of q .Solution:eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometrysin q cosq1 tan q-513 - 5121213112 5512Exercise 9.212Example 2: Find the value of other ive trigonometric functions of q , if cos q and the13terminal side of the angle is not in the I quadrant.1.Find the signs of the following:i)sin 160 ii)cos 190 iii)tan115 iv)sec 245 v)cot 80 vi)cosec 297 version: 1.1version: 1.11819
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry2.Fill in the blanks:iii)tan(-182 ) . tan 182 iv)cot(-173 ) . cot137 v)sec (-216 ) . sec 216 vi)cosec(-15 ) . cosec 15 v)cot q 0 and sin q 0,iv)vi)8.sec q 0 and sin q 0,cos q 0 and tan q 0?cos q iii)cos q -iv)tanv)1and the terminal arm of the angle is not in quad. III.sin q 2Find cot q 3and the terminal arm of the angle is in quad. III.2If cosec q If cot q 5and the terminal arm of the angle is in the I quad., ind the value of23 sin q 4cos q.cos q - sin q(a)Case 1 when m A 45 p4randian45 then m B and the terminal arm of the angle is in quad. II.15and the terminal arm of the angle is not is quad. I, ind the values of8cos q and cosec q .6.1and the terminal arm of the angle is not in the III quad., ind the values of7Consider a right triangle ABC with m C 90 and sidesa, b, c as shown in the igure on right hand side.9and the terminal arm of the angle is in quad. IV.41 -If tan q 9.10 The values of Trigonometric Functions of acute angles 45 ,30 and 60 12sin q and the terminal arm of the angle is in quad. I.13ii)eLearn.Punjabcsc 2 q - sec 2 q.csc 2 q sec 2 qFind the values of the remaining trigonometric functions:i)5.7.In which quadrant are the terminal arms of the angle lie whensin q 0 and cos q 0,ii)cot q 0 and cosec q 0,i)tan q 0 and cos q 0,eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometryratios. . cos 75 ii)iii)4.cos(-75 )sin(-310 ) . sin 310 i)3.eLearn.Punjabm 1p and m 0 0 q , ind the values of the remaining trigonometric2m2 2 ABC is right isosceles.As values of trigonometric functions depend only on the angle and not on thesize of the triangle, we can take a b 1By Pythagoras theorem, c2 a 2 b2 c2 1 1 2c 2\ Using triangle of ig. 1, with a b 1 and c 2version: 1.1version: 1.12021
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry sin 45 cos 45 tan 45 (b)a cb ca 1bCase 2: when m A 30 then m B 60 eLearn.Punjab csc 45 1;21;21sin 45 1 sec 45 cos 45 cot 45 ;p61tan 45 2;2;\randiansec 30 cot 30 3 4 -1 3 sin 60 a3 ;c2b 1 cos60 ;c 2a tan 60 3;b1 csc 60 sin 60 1sec60 cos60 1 cot 60 tan 60 2;32;1.3Example 3: Find the values of all the trigonometric functions of csc30 pa2 c2 - b2\ Using triangle of fig.3, with a 3, , b 1 and c 2(i)\ Using triangle of fig.2, with a 1, b 3 and c 2when m A 60 a 2 b2 c2 a 3 b 3Case 3: 1. b 2 c 2 - a 2 4 -1 3(c)eLearn.Punjabhalf the hypotenuse.Let c 2 then b 1\By Pythagoras theorem ,By elementary geometry, in a right triangle the measure of the side opposite to 30 ishalf of the hypotenuse.Let c 2 then a 1\By Pythagoras theorem , a2 b2 c4a 1 sin 30 ;c 23bcos30 ;2ca1 tan 30;b3eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry1tan 30 1 sin 30 1 cos30 Solution:(i)2;2;33.radian, then m B 30 420 (ii)-7p4(iii)19p3We know that q 2kp q , where k Z420 60 1(360 )(k 1) 60 32 sin 420 sin 60 ; csc 420 231 cos 420 cos60 ;sec 420 221tan 420 tan 60 3;cot 420 3By elementary geometry, in a right triangle the measure of the side opposite to 30 isversion: 1.1version: 1.12223
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry(ii)-7pp (-1)2p44 eLearn.Punjab(k -1) so41 -7p p -7p sin sin ;csc 2 4 4 4 1 -7p p -7p cos sec cos ; 2 4 4 4 -7p p -7p tan 1 ; cot tan 4 4 4 (iii) (k p csc 4 p sec 4 p cot 4 2;1; 3)(b) 19p p 2csc ; csc 3 3 3 1 19p p ;sec sec 2 ;2 3 3 19p p 19p p 1 tan 3 ;cot . tan cot 3 3 3 3 3 9.11r and y 0x2 y 2 1oWhen q 90The point (0, 1) lies on the terminal side of angle 90 .332x 1y011 sin 0 0 csc 0 (undefined)r1sin 0 0x 11 1 sec 0 cos0 1r1cos 0 y011 0 tan 0 cot 0 (undefined) x1tan 0 0 2;pp 19p sin ( ) sin 3 3 p 19p cos ( ) cos3 3 eLearn.PunjabThe point (1,0) lies on the terminal side of angle 0 p19p p 3 ( 2p )33eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry; x 0sor x2 y 2 1y 1 sin 90 1;r 1x 0cos90 0;r 1y 1tan 90 (undefined);x 0The values of the Trigonometric Functions of angles0 , 90 , 180 , 270 , 360 .When terminal line lies on the x- axis or the y- axis, the angle q is called a quadrantalangle.Now we shall ind the values of trigonometric functions of quadrantal angles 0 , 90 ,180 , 270 , 360 and so on.(a) When q 0 and y 11csc90 1; sin 90 11 sec90 (undefined); cos90 0x 0 90 cot0.y 1(c)oWhen q 180 The point (-1, 0) lies on the terminal side of angle 180 .x -1 and y 0version: 1.1version: 1.12425
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometryr soeLearn.Punjabx2 y 2 1(d)oWhen q 270 The point (0, -1) lies on the terminal side of angle 270 .x 0 and y -1r csc 360 is undefined;1 cos 360 cos0 1; sec 360 1cos0 tan 360 tan 0 0;cot 360 is undefined. (ii)Now -(iii)Solution: We know that q 2 kp q ,(i) Now 360 0 1(360 ), 0 (iii)2 3p (-1)2p25p3p2 csc sec cot p - - 1; 2 p is undefined;2 p 02 5p p 2(2 p ) (k 2)Now pcsc 5p is undefined;psec5 1;cot 5p is undefined; Exercise 9.3where k Z(k 1)1.Verify the following:(i)sin60 cos 30 - cos 60 sin 30 sin 30 (ii)sin 2p6 sin 2p3 tan 2p42 version: 1.1version: 1.126 -(k -1) sin 5p sin p 0;cos 5p cos p1;p 0;tan 5ptan Example 4: Find the values of all trigonometric functions of-p2pwhere k Z p 3p sin - sin - 1; 2 2 p 3p - cos cos 0 ; 2 2 p 3p tan - tan - is undefined; 2 2 y -1r 1 - sin 1; 1;270csc270r 1y -1x 0r 1 0; sec 270 cos 270 (undefined);r 1x 0y -1x0 270 tan 270 (undefined);cot0.x 0y-1(ii)We know that q 2 kp q , x2 y 2 1(i) 360 eLearn.Punjab sin 360 sin 0 0;y 0r 1 sin 180 0;csc 180 (undefined);r 1y 0x -1r 1cos 180 1; sec 180 1; r 1x -1y 0x 1 tan 180 0; cot 180 (undefined).x -1y 0soeLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry27
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjab(iii)132 sin 45 cosec 45 22(iv)sin 22.Evaluate the following:pptan - tan36i)pp1 tan tan363.iii)v)4: sin 2p3: sin 2p2 1: 2 : 3: 4.ii)sin 2q 2sin q cos q cos 2qii)2cos 2 q - 1iv)2 tan qtan 2q 1 - tan 2 q1 - tan 2We list the trigonometric functions and fundamental identities, learnt so far mentioningtheir domains as follows:sin q,for all q R(i)p1 tan 2pcos 2q(ii)3iv)vii)-p9- p2- 2430 ii)-3pcos 2 q - sin 2 q(iv) sec q cos 2q 1 - 2sin 2 qv)- 15pviii)iii)235p2tan q (v)5p2vi)1530 ix)407p2iv)vii)-675 25p6v)viii)vi)ix) (vi) cot q(vii)for all q R,1sin q1cos qsin q cosq ,for all qRbut qnp , ,for all qRbut q 2n 1 p , 2 for all qR but q( 2nfor all qRbut qnp ,but q( 2n,cosqsin q ,2sin q cos 2 q 1 , (viii) 1 tan 2 q sec 2 q, (ix) 1 cot 2 q csc 2 q ,for all qR for all qR for all qRbut q13p3Example 1:Prove that cos 4 q - sin 4 q cos 2 q - sin 2 q ,Solution: L.H.S p1) ,2n Zn Zn Zp1) ,2n Zn Z-1035 for all q Rcos 4 q - sin 4 qversion: 1.1version: 1.128np ,n ZNow we shall prove quite a few more identities with the help of the above mentionedidentities.Find the values of the trigonometric functions of the following angles:i)390 ii)- 330 iii)765 -17p3-71p6cos q (iii) csc q3Find x, if tan2 45 - cos2 60 x sin 45 cos 45 tan 60 .Find the values of the trigonometric functions of the following quadrantal angles:i)6.pVerify the following when q 30 , 45 i)4.5.6: sin 2eLearn.Punjab9.12 Domains of Trigonometric functions and of FundamentalIdentities peLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry29
eLearn.Punjab1. Quadratic Equations9. Fundamentals of TrigonometryeLearn.Punjab ( cos q ) - ( sin q ) q sin q )( cos( cos (1) ( cos q - sin q ) 22222 222 q sin q ) 2 q ( sincos 2 q 1)cos q - sin q R.H.S.2HenceExample 2: 2 cos 4 q - sin 4 q cos 2 q - sin 2 qProve that sec 2 A L.H.S Solution: 22np , n Z cosec 2 A sec 2 A cosec 2 A Where A 2 sec A cosec A211sin A cos A cos 2 A sin 2 A cos 2 A sin 2 A1 sin 2 A cos 2 A 1 22 cos A sin A2of2Solution: R.H.S 1 - sin q1 sin q(1 - sin q )(-q(1 - sin q )21 - sin 2 qcos 2 q1 - sin qcos q1sin q secq - tan q R.H.Scosq cosqcot 4 q cot 2 q(cosec 2 q - 1)cosec 2 qcosec 4q - cosec 2q R.H.S.Hence cot 4 q cot 2 q cosec4q - cosec2q .Exercise 9.4Prove the following identities, state the domain of q in each case:1 - sin q(rationalizing.)1 - sin q1.2q)qq2 1 - sin q1 sin qL.H.S. (1 - sin q ) cot 2 q (cot 2 q 1)p1 - sin q sec q - tan q , where q is not an odd multiple of .21 sin qProve that- q1 - sin q.L.H.S. Hence sec 2 A cosec 2 A sec 2 A. cosec 2 A.Example 3:2- q1 sin qqqShow that cot 4 q cot 2 q cosec 4 q - cosec2 q , where q is not an integral multipleSolution: 11. 22cos A sin A sec 2 A. cosec 2 Ap eLearn.Punjab1 - sin q sec q - tan q .1 sin qHenceExample 4:2eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry-qqtan q cot q cosec q secq2.sec q cosec q sin q cos q 1version: 1.1version: 1.13031
eLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry3.5.cos q tan q sin q sec qcot 2 q - cos 2 q cot 2 q cos 2 q8.2 cos q - 1 1 - 2sin q12.13.14.15.16.17.18.19.4.sec 2 q - cosec 2 q tan 2 q - cot 2 q6.10.eLearn.Punjab22cosq - sin q cot q - 1 cosq sin q cot q 17.9.11.cot 2 q - 1 2cos 2 q - 121 cot qcosec q tan q secq cosecq sec 2 q20.(sec q tan q ) (sec q - tan q ) 11 - tan 2 qcos q - sin q 21 tan q22eLearn.Punjabsin 3 q - cos3 q (sin q - cos q )(1 sin q cos q )21.sin 6 q - cos 6 q (sin 2 q - cos 2 q ) (1 - sin 2 q cos 2 q )22.sin 6 q cos 6 -q 1 3sin 2 q cos 2 q23.sin q cot q cosec q1 cos qeLearn.Punjab1. Quadratic Equations9. Fundamentals of Trigonometry24.11 2sec 2 q1 sin q 1 - sin qcos q sin qcos q - sin q2 cos q - sin qcos q sin q 1 - 2sin 2 q1 cos q (cosec q cot q ) 21 - cos q1 - sin q(sec q - tan q ) 2 1 sin q2 tan q 2sin q cos q1 tan 2 q1 - sion qcos q cos q1 sin q(tan q cot q ) 2 sec 2 q cosec2 qtan q sec q - 1 tan q - sec q 1tan q sec q1111 cosec q - cot q sin qsin qcosec q cot qversion: 1.1version: 1.13233
9. Fundamentals of Trigonometry eLearn.Punjab 9. Fundamentals of Trigonometry eLearn.Punjab 8 version: 1.1 version: 1.1 9 219 (0.0175) radinas 4 0.958 radains. Most calculators automatically would convert degrees into radians and radians into degrees. Example 5: An arc subtends an angle of 70 at the center of a circle and its length is
