2018-G11-Math-E

91.. FQuunaddarmaetinctaElqs uoaf tTiroingosnometry eLearn.Punjab eLearn.Punjab = -q +q = -q -q 1+ sinq 1- sinq = (1- sinq )2 1 - sin2 q== (1- sinq )2 1- sin q = cos2 q cos q 1 - sinq =secq - tan q = R.H.S cosq cosqHence 1 - ss=iinnqq sec q - tan q. 1 +Example 4: Show that cot4 q + cot2 q= cosec4 q - cosec2 q , where q is not an integral multipleof p . 2Solution: L=.H.S. cot4 q + cot2 q = + cot2 q (cot2 q 1)= (cosec2 q -1)cosec2 q = cosec4q - cosec2q = R.H.S. Hence cot4 q + cot2 q = cosec4q - cosec2q. Exercise 9.4 version: 1.1 Prove the following identities, state the domain of q in each case:1. tanq + cotq =cosec q secq 2. sec q cosec q sin q cos q = 1 31

19.. FQuunaddarmaetinctaElqsuoaftTioringosnometry eLearn.Punjab eLearn.Punjab3. cos q + tan q sinq =sec q 4. cosec q + tanq secq =cosecq sec2 q5. sec2 q - cosec2 q =tan2 q - cot2 q6. cot2 q - cos2 q =cot2 q cos2 q 7. (sec q + tan q ) (sec q - tan q ) =18. 2 cos2 q -1 = 1 - 2sin2 q 9. cos2 q - sin2 q =11+- ttaann 2 q 2 q10. cosq - sinq = cotq -1 11. sin q + cot q =cosec q cosq + sinq cotq +1 1+ cos q12. 1co+t=c2 oqt2-q1 2cos2 q -113. 1 =+- ccooss qq (cosec q + cot q )2 114. (sec q - tan q )2 =11+- ssiinn qq15. 2 tan q = 2sin q cos q 1 + tan2 q16. 1 - sion q = cos q cos q 1+ sin q17. (tan q + cot q )2 =sec2 q cosec2 q18. tan q + sec qq=+- 11 tan q + sec q tan q - sec19. 1 - 1 = 1 - 1 cosec q - cot q sin q sin q cosec q + cot q version: 1.1 32

91.. FQuunaddarmaetinctaElqs uoaf tTiroingosnometry eLearn.Punjab eLearn.Punjab20. sin3q - cos3 q = (sin q - cos q )(1 + sin q cos q )21. sin6 q - cos6 q = (sin2 q - cos2 q ) (1 - sin2 q cos2 q )22. sin6 q + cos6 -q =1 3sin2 q cos2 q23. 1 + 1 - 1 q =2sec2 q 1+ sin q sin24. cos q + sin q + cos q - sin q =1 - 2s2in2 q cos q - sin q cos q + sin q version: 1.1 33

CHAPTER version: 1.110 Trigonometric Identities

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjab10.1 Introduction In this section, we shall irst establish the fundamental law of trigonometry beforediscussing the Trigonometric Identities. For this we should know the formula to ind thedistance between two points in a plane.10.1.1 The Law of CosineLet P (x1 , y1) and Q (x2 , y2) be two points. If “ d “ denotes the distance between them, then, d = PQ = (x1 - x2 )2 + ( y1 - y2 )2 = (x2 - x1)2 + ( y2 - y1)2i.e., square root o f the sum of square the diference of x-coordinates and square thediference o f y-coordinates.Example 1: Find distance between the following points: i) A(3,8) , B(5,6) ii) P(cosx , cosy), Q(sinx , siny)Solution:i) Distance = AB = (3 - 5)2 + (8 - 6)2 = 4 + 4 = 8 = 2 2 = (5 - 3)2 + (6 - 8)2 = 4 + 4 = 8 = 2 2ii) Distance = (cos x - sin x)2 + (cos y - sin y)2 = cos2 x + sin2 x - 2cos x sin x + cos2 y + sin2 y - 2cos y sin y - =2 2cos-xsin x 2cos y sin y version: 1.1 - =2 2(co+s xsin x cos y sin y 2

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab10.1.2 Fundamental Law of trigonometry eLearn.Punjab version: 1.1Let a and b any two angles (real numbers), then cos(a=- b ) cosa cos b + sina sin bwhich is called the Fundamental Law of Trigonometry.Proof: For our convenience, let us assume that a > b > 0.Consider a unit circle withcentre at origin O. Let terminal sides o f angles a and b cut the unit circle atA and B respectively.Evidently ∠AOB =a - b Take a point C on the unit circle so that ∠XOC =∠m AO=B -a b . Join A,B and C,D.Now angles a b and a - b are in standard position.∴ The coordinates o f A are (cos a, sin a) the coordinates o f B are (cos b, sin b)the coordinates o f C are (cosa - b , sin a - b )and the coordinates o f D are (1, 0).Now ∆AOB and ∆COD are congruent. [(SAS) theorem]∴ AB =CD⇒ AB 2 =CD 2Using the distance formula, we have:(cosa - cos b )2 + (sina - sin b=)2 [(cos(a - b ) -1]2 + [sin(a - b ) - 0]2⇒ cos2 a + cos2 b - 2cosa cos b + sin2 a + sin2 b - 2sina sin b = cos2 (a - b ) +1 - 2cos(a - b ) + sin2 (a - b )⇒ 2 - 2 (cosa cos b + sina sin b ) = 2 - 2cos(a - b ) Hence cos(a=- b ) cosa cos b + sina sin b . 3

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.PunjabNote: Although we have proved this law for a > b > 0. it is true for all values of a and b Suppose we know the values of sin and cos of two angles a and b. we can indcos (a - b) using this law as explained in the following example:Example 1: Find the value of cos p . 12Solution: As p =15 =45 - 30 =p - p 12 4 6∴ cos =p cos  p - p6= cos p cos p + sin p sin p 12 4 4 6 46 = 1 . 3 + 1 .1 = 3 +1. 2 2 22 2 210.2 Deductions from Fundamental Law1) We know that: cos(a -=b ) cosa cos b + sina sin bPutting a = p in it, we get 2 cos  p =- b  cos p cos b + sin p sin b 2 22⇒ cos  p - b  = 0. cos b + 1. sin b  co=s p 0, si=n p 1 2 2 2∴ cos  p - b  =sin b 22) We know that: cosa cos b + sina sin b cos (a -=b )Putting b = - p in it, we get 2 cos a - =- p2  cosa . cos  - p  + sina sin  - p  2 2 version: 1.1 4

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab⇒ cos a + p = cosa. 0 + sin a.(-1) scions--p2p2-==cos-isn=p2p=2 0 1 eLearn.Punjab 2 (ii) version: 1.1 ∴ cos  p +a  = - sin a 23) We known that cos  p - b  =sin b [(i) above] 2 Putting b= p + a in it, we get 2 cos p2 -  p + a = sin  p +a  2 2 ⇒ cos(-a=) sin  p +a  2 ⇒ cos=a sin  p + a  {-=cos( a ) cos a} 2 ∴ sin  p + a  =cos a (iii) 24) We known that cos(a=- b ) cosa cos b + sina sin b replacing b by -b we get cos[a -=(-b )] cosa cos(-b ) + sina sin(-b ) { cos(-b ) =cos b ,sin (-b ) =-sin b}7 ⇒ cos(a=+ b ) cosa cos b - sina sin b5) We known that cos(a=+ b ) cosa cos b - sina sin b replacing a by p + a we get 2 5

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjabcos  p + a  +=b  cos  p + a  cos b - sin  p + a  sin b eLearn.Punjab 2 2 2 version: 1.1⇒ cos p2 + (a + b ) = - sina cos b - cosa sin b⇒ - sin(a + b ) = - [sina cos b + cosa sin b ]∴ sin(a=+ b ) sina cos b + cosa sin b6) We known thatsin(a =+ b ) sina cos b + cosa sin b [from (v) above]replacing b by -b we getsin(a -=b ) sina cos(-b ) + cosa sin(-b )  sin(-b ) =- sin b cos(-b ) =cos b∴ sin(a =- b ) sina cos b - cosa sin b (vi)7) We known that cos(a=- b ) cosa cos b + sina . sin b=Let a 2=p and b q∴ co=s(2p -q ) c+os 2p . cos q sin 2p sin q= 1 . cos q + 0 . sin q  scions22pp = 1 = cos q = 08) We known that sin(=a - b ) sina . cos b - cosa . sin b ∴ sin(=2p -q ) sin 2p . cosq - cos 2p sin q 6

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab eLearn.Punjab = 0 . cosq -1 . sin q  scions22pp = 0 = - sin q = 1 (viii)9) tan(a + b ) = sin(a + bb=)) sina cos b + cosa sin b cos(a + cosa cos b - sina sin b = sina cos b + cosa sin b -ndeceuoDnmsuivameirdiacnitnooagstrobarnd cosa cos b - cosa cos b cosa cos b sina sin b cosa cos b cosa cos b ∴ tan(a + b ) =1t-antaan+a tan b (ix) tan b10) tan(a - b )= sin(a --=bb)) sina cos b - cosa sin b cos(a cosa cos b + sina sin b = sina cos b - cosa sin b ndeeDunmiuvmeidriainntoagtroarnd cosa cos b cosa cos b cosa cos b sina sin b cosa cos b + cosa cos b ∴ tan(a - b ) =1t+antaan-attaannbb (x)10.3 Trigonometric Ratios of Allied Angles The angles associated with basic angles of measure q to a right angle or its multiplesare called allied angles. So, the angles of measure 90° ± q , 180° ± q , 270° ± q , 360° ± q , areknown as allied angles. Using fundamental law, cos(a - b ) = cos a cos b + sin a sin b and its deductions, wederive the following identities: version: 1.1 7

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjabss=iinn  pp22 +-qq= =ccooss-qq =,, ccooss-pp22 qq+ sin q , tan  p q  cotq =si-nq , 2 q+ =co-tq tan  p 2ssiinn((pp - q ) =sin q , cos(p q-) =co-sq , tan(p q ) - tan=q- + q ) =-sinq , cos(p q+) =co-sq , tan(p q ) +tanq=ssiinn  3p -q -=cosq , cos  3p -q -=sinq , tan  3p -q  =cotq  2 + q -=cosq , 2 2 3p cos  3p +q  =sinq , tan  3p q+- =cotq 2 2 2=ssiinn((22pp -+qq=)) =s-isni+qnq=, ,cocso(+s2(p2p q-) c=osq , , tan(2p q) - tan=q- q) cosq tan(2p q) tanqNote: The above results also apply to the reciprocals of sine, cosine and tangent. Theseresults are to be applied frequently in the study of trigonometry, and they can beremembered by using the following device:1) If q is added to or subtracted from odd multiple of right angle, the trigonometric ratios change into co-ratios and vice versa. i.e, sin ←→cos, tan ←→cot, sec←→cosese.=g. sin =p2 -q  cos+q and cos  3p q  sinq 22) If q is added to or subtracted from an even multiple of p the trigonometric ratios shall 2remain the same.3) So far as the sign of the results is concerned, it is determined by the quadrant in which the terminal arm of the angle lies. e.g. s=in(p -q ) sinq , t=an(p + q ) tan q , co=s(2p -q ) cos q version: 1.1 8

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab Measure of the Quad. eLearn.Punjab angle I version: 1.1 p -q II 2 III Iv p + q orp -q 2 p + q or 3p -q 23p + q or 2p -q2a) In sin  p -q  , sin  p +q ,sin  3p -q  and sin  3p +q  2 2 2 2 odd multiplies of p are involved. 2 ∴ sin will change into cos. Moreover, the angle of measure i)  p -q  will have terminal side in Quad.I, 2 So sin  p - q  =cos q ; 2 ii)  p +q  will have terminal side in Quad.II, 2 So sin  p + q  =cos q ; 2 iii)  3p -q  will have terminal side in Quad.III, 2 9

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjab So sin  3p -q -=cosq ; 2iv)  3p +q  will have terminal side in Quad.Iv, 2 So sin  3p + q -=cos q. 2b) In cos(p -q ), cos(p + q ), cos(2p -q ) and cos(2p + q ) even multiples of p are involved. 2 ∴ cos will remain as cos. Moreover, the angle of measure i) (p -q ) will have terminal side in Quad. II, ∴ cos(p -q ) =-cos q ; ii) (p + q ) will have terminal side in Quad. III, ∴ cos(p + q ) =-cos q ; iii) (2p -q ) will have terminal side in Quad. IV: ∴ cos(2p -q ) =cos q ; iv) (2p + q ) will have terminal side in Quad. I ∴ cos(2p + q ) =cos q.Example 2: Without using the tables, write down the values of: i) cos 315° ii) sin 540° iii) tan (-135°) iv) sec(-300°)Solution:i) cos315 =cos(270 + 45) =cos(3× 90 + 45) =+sin 45 =1ii) sin 540 2 =sin(540 + 0) =sin(6 × 90 + 0) =+sin 0 =0iii) tan(-135-) =t-an=135 - ta-n(1=80 ×45)- -ta-n=(2 90 =45) ( tan 45) 1iv) sec(-300) = sec300= sec(360 - 60)= sec(4 × 90 - 60)= sec60= 2 version: 1.1 10

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab eLearn.PunjabExample 3: Simplify: sin(360 - q )cos(180 - q ) tan(180 + q ) sin(90 + q )cos(90 - q ) tan(360 + q )Solution:  =tsainn((138600= +-qq)) =t-ans+iqnq,, cos(180 q-) =co-sq = cos(9=0 -q ) sin+q , sin(90 q ) cosq tan(360 q ) tanq∴ sin(360 - q )cos(180 - q ) tan(180 + q ) =(-csoinsqq ).(s-incoqsq.ta)ntaqnq 1 = cosq. sinq . tanq Exercise 10.11. Without using the tables, ind the values of: i) sin (-780) ii) cot (-855) iii) csc(2040) iv) sec(-960) v) tan (1110) vi) sin (-300)2. Express each of the following as a trigonometric function of an angle of positive degree measure of less than 45°. i) sin 196 ii) cos 147 iii) sin 319 iv) cos 254 v) tan 294 vi) cos 728 vii) sin (-625 ) viii) cos (-435 ) ix) sin 1503. Prove the following: i) sin (180 + a ) sin (90 - a ) -=sina cosa ii) sin 780 sin 480 + cos120 sin 30 =1 2 iii) cos306 + cos 234 + cos162 + cos18 =0 iv) cos 330 sin 600 + cos120 sin150 =1 - version: 1.1 11

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjab4. Prove that:  3p  2 sin2 (p + q ) tan +qi)  3p cos2 (p = cos q 2 cot 2 -q -q )coses(2p -q)ii) cos(90 + q )sec(-q ) tan(180 - q ) = -1 sec(360 - q )sin(180 + q )cot(90 - q )5. If a, b, g are the angles of a triangle ABC, then prove thati) sin (a + b ) =sing ii) cos  a + b  = sin giii) cos(a + b ) -=cosg 2 2 iv) tan (a + b ) + tang =0.10.4 Further Application of Basic IdentitiesExample 1: Prove that sin(a + b ) sin(a - b ) = sin2 a - sin2 b (i) = cos2 b - cos2 a (ii)Solution: L.H.S. =sin(a + b ) sin(a - b ) cosa sin b ) =(s+ina cos b cosa sin b )(s-ina cos b= -sin2 a cos2 b cos2 a sin2 b = sin2 a (1 - sin2 b ) - (1 - sin2 a )sin2 b = sin2 a - sin2 a sin2 b - sin2 b + sin2 a sin2 b = sin2 a - sin2 b (i) = (1 - cos2 a ) - (1 - cos2 b ) = 1 - cos2 a -1 + cos2 b = cos2 b - cos2 a (ii)Example 2: Without using tables, ind the values of all trigonometric functions of 75°.Solution: As =75 45 + 30 sin 7=5 sin (45 + 30= ) sin 45 cos30 + cos 45 sin 30 version: 1.1 12

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab=  1   3  +  1   1 = 3 +1 eLearn.Punjab 2 2 2 2 22 version: 1.1cos 7=5 cos (45 + 30= ) cos 45 cos30 - sin 45 sin 30=  1   3  -  1   1 = 3 -1 2 2 2 2 22tan 7=5 tan (45 + 30= ) tan 45 + tan 30 1 - tan 45 tan 30= 1+ 1 3 +1 =1 -1. 31 3-1 3co=t 75 =tan175 3 -1 3 +1cos=es 75 =sin175 22 and se=c 75 =cos175 22 3 +1 3 -1Example 3: Prove that: cos11 + sin11 = tan 56. cos11 - sin11Solution: ConsiderR.H.S=. tan 56= tan(45 + 11 )= tan 45 + tan11 1 - tan 45 tan11= =11+- ttaann1111 1=1+- sccsioionnss11111111 cos11 + sin11 = L.H.S. cos11 - sin11 13

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.PunjabHence cos11 + sin11 = tan 56. cos11 - sin11Example 4: I-f cosa =2=4 , tan b 9 the terminal side of the angle of measure a is in the II 25 40quadrant and that of b is in the III quadrant, ind the values of:i) sin (a + b ) ii) cos (a + b )In which quadrant does the terminal side of the angle of measure (a + b ) lie?Solution: We know that sin2 a + cos2 a =1∴ sina±=-1 cos2=a ± -1± 57=6 ±= 49 7 625 625 25As the terminal side o f the angle of measure of a is in the II quadrant, where sin a ispositive.∴ sin a =7 25Now sec b =± 1 + tan±2 b =+ 1 =81± 41 1600 40As the terminal side of the angle of measure of b in the III quadrant, so sec b isnegative∴ sec-b =41 and cos b = - 40 40 41 sin b = ± 1 - cos2 b = ± 1 - 1600 =± 9 1681 41As the terminal arm of the angle of measure b is in the III quadrant, so sin b is negative∴ sin-b =9 41∴ sin (a +=b ) sina cos b + cosa sin b version: 1.1 14

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab eLearn.Punjab =  7   - 40  +  - 24   - 9  25 41 25 41 = -280 + 216= - 64 1025 1025and cos(a + b =) cosa cos b - sina sin b = - 24   - 40  -  7   - 9  25 41 25 41 = 960 + 63 1025 = 1023 1025 sin (a + b ) is - ve and cos(a + b ) is + ve\ The terminal arm of the angle of measure (a + b ) is in the IV quadrant.Example 5: If a , b ,g are the angles of ∆ABC prove that: i) tana + tan b + tang =tan a tan b tan g ii) tan a tan b + tan b tan g + tan g tan a =1 22 22 22Solution: As a , b , g are theangles of ∆ ABC∴ a + b + g =180i) a + b = 180 - g∴ tan (a + b ) = tan (180 - g )⇒ tana + tan b =tan g - 1- tana tan b version: 1.1 15

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjab⇒ tana + tan b = - tang + tana tan b tang∴ tana + tan b + tang =tana tan b tangii) As a =+ b + g 180 ⇒ a +=b + g 90 222So a + b = 90 - g 22 2∴ tan  a + =b2  tan  90 - g  2 2⇒ tan a + tan b ==cot g 1 g 2 a 2 2 tan 1 - tan tan b 2 22⇒ tan a tan g + tan b t-an g =1 tan a tan b 22 22 22∴ tan a tan b + tan b tan g + tan g tan a =1 22 22 22Example 6: Express 3 sin q + 4 cos q in the form r sin(q + φ), where the terminal side of theangle of measure φ is in the I quadrant.S=o=lution: Let 3 r cosφ and 4 r sinφ ∴ 32 +=42 r2 cos2 φ + r2 sin2 φ⇒ 9=+ 16 r2 (cos2 φ + sin2 φ) ⇒ 4= r sinφ⇒ 25 =r2 3 r cosφ⇒ 5 =r 4 =tanφ 3⇒ r =5 ∴ tanφ =4 3∴ 3sinq + 4=cosq r cos φ sinq + r sinφ cos q = + r(sinq cosφ cosq sinφ) version: 1.1 16

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab = r sin (q + φ) eLearn.Punjab==Where r 5 and φ tan-1 4 version: 1.1 3 Exercise 10.21. Prove that ii) cos (180 + q ) =- cosq iv) cos (q -180-) =cosq i) sin (180 + q ) =- sinq vi) sin (q + 27-0) =cosq iii) tan (270 -q ) =cotq viii) cos (360 -q ) =cosq v) cos (270 + q ) =sinq vii) tan (180 + q ) =tanq2. Find the values o f the following:i) sin 15 ii) cos 15 iii) tan 15iv) sin 105 v) cos 105 vi) tan 105 (Hint: 15 =(45 30 ) and -105 (60 45 )=. +3. Prove that:i) sin(45= + a ) 1 (sina + cosa ) 2ii) cos(a +=45) 1 (cosa - sina ) 24. Prove that: i) tan (45 + A) tan (45 - A) =1ii) tan  p -q  + tan  3p +q  =0 4 4iii) sin q + p  + cosq + p  =cosq 6 3iv) sin q - cos q tan q = tan q 2 2 cos q + sin q tan q 2 17

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjab v) 1- tan q tan φ = cos(q +φ) 1+ tan q tan φ cos(q -φ)5. Show that: cos(a + b )cos(a - b=) cos2 a - sin2 =b cos2 b - sin2 a6. Show that: sin(a + b ) + sin(a - b ) = tana cos(a + b ) + cos(a - b )7. Show that: i) cot (a + b ) =ccoottaac+otcbot-b1 ii) cot (a - b ) =ccoottabc-otcbot+a1 tana + tan b sin(a + b ) iii) tana - tan b = sin(a - b )8.==If sina 4 and cos b 40 , where 0<a < p and 0<b < p . 5 41 22 Show that sin (a - b ) =133 . 2059. If sina 4 < and< sinb = 1<2 <where p a p and p b p . Find 5 13 2 2 i) sin (a + b ) ii) cos (a + b ) iii) tan (a + b ) iv) sin (a - b ) v) cos (a - b ) vi) tan (a - b ). In which quadrants do the terminal sides of the angles of measures (a + b ) and (a - b ) lie?10. Find sin (a + b ) and cos(a + b ), given that i) =tan a 3=, cos b 5 and neither the terminal side o f the angle of measure 4 13 a nor that of b is in the I quadrant. version: 1.1 18

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab eLearn.Punjab ii) tan a =15 and sin -b 7 and nei=ther the termina-l side of the angle of 8 25 measure a nor that of b is in the Iv quadrant.11. Prove that cos 8 sin 8 tan 37 cos 8 sin 812. If a , b ,g are the angles of a triangle ABC, show that cot a + cot b + cot g =cot a cot b cot g 222 22213. If a + b + g =180 , show that cota cot b + cot b cotg + cotg cota =114. Express the following in the form r sin (q +φ) or r sin (q - φ) , where terminal sides of the angles of measures q and φ are in the irst quadrant: i) 12 sin q +5 cos q ii) 3 sin q - 4 cos q iii) sinq - cos q iv) 5 sin q - 4 cosq v) sin q + cos q. vi) 3 sin q -5 cos q10.5 Double angle Identities We have discovered the following results: version: 1.1 sin (a=+ b ) sin a cos b + cos a sin b cos (a=+ b ) cos a cos b - sin a sin b and tan (a + b ) =1t-antaan+attaannbb We can use them to obtain the double angle identities as follows:i) P=ut b a in sin (a=+ b ) sin a cos b + cos a sin b ∴ sin (a =+ a ) sin a cos a + cos a sin a 19

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjab Hence sin 2a = 2sin a cos aii) =Put b a in cos (a=+ b ) cos a cos b - sin a sin b cos(a=+ a ) cos a cos a - sin a sin aHence c=os 2a cos2 a - sin2 a =cos 2a cos2 a - sin2 a ( sin2 a =1 -cos2 a )∴ cos 2a = cos2 a - (1 - cos2 a ) = cos2 a -1 + cos2 a\ =cos 2a 2cos2 a -1 co=s 2a cos2 a - sin2 a ( cos2-a =1 sin2 a )∴ cos 2a -=(1 si-n2 a ) sin2 a∴ cos 2a = 1 - 2 sin2 aiii) =Put b a in tan(=a + b ) tan a + tan b 1- tan a tan b tan(a + a ) =1t-antaan + tan a a tan a ∴ tan 2a = 2 tana 1 - tan2 a10.6 Half angle Identities The formulas proved above can also be written in the form of half angle identities, inthe following way:i)  co=s a 2cos2 a -1 ⇒ cos=2 a 1 + cos a version: 1.1 2 22 20

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab ⇒ c±os a =1+ cos a eLearn.Punjab 22 version: 1.1ii)  cosa -=1 2 sin2 a⇒ sin=2 a 1 - cos a 2 22 ⇒ sin±a =1- cos a 22 tan a = sin a 1- cos a 2 2iii) = ± 2 cos a 1 + cos a 22 ⇒ tan a± =11-+ cos a 2 cosa10.7 Triple angle Identities i) s=in 3a 3 sina - 4 sin3 a ii) =cos 3a 4 cos3 a - 3 cos a iii) tan 3a = 3tan a - tan3 a 1 - 3tan2 aProof:i) =sin 3a sin (2a + a ) = + sin 2a cos a cos 2a sin a= 2+sin-a cos a cos a (1 2 sin2 a ) sin a = 2sina cos2 a + sin a - 2 sin3 a = 2sin a (1 - sin2 a ) + sina - 2sin3 a 21

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab = 2sin a - 2sin3 a + sina - 2sin3 a eLearn.Punjab version: 1.1\ =sin 3a 3sin a - 4sin3 aii) =cos 3a cos (2a + a ) = - cos 2a cos a sin 2a sina = (2 cos2 a -1) cos a - 2 sin a cos a sin a = 2 cos3 a - cos a - 2 sin2 a cosa = 2 cos3 a - cos a - 2(1 - cos2 a ) cos a = 2 cos3 a - cos a - 2 cos a + 2 cos3 a\ =cos3a 4 cos3 a - 3 cos aiii) =tan 3a tan (2a + a ) = tan 2a + tana 1- tan 2a tana 2 tan a + tan a 2 tana + tana - tan3 a 1 - tan2 a 1 - tan2 a - 2 tan2 a== 2 tan a - - tan2 a a 1 1 . tan\ tan 3a = 3tana - tan3 a 1 - 3tan2 aExample 1: Prove that sin A + sin 2A = tan A 1 + cos A + cos 2A==Solution: L.H.S sin A + 2sin Acos A sin A(1+ 2cos A) 1 + cos A + 2 cos2 A -1 cos A(1 + 2cos A) = =sin A tan A = R.H.S cos A 22

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab eLearn.Punjab Hence sin A + sin 2A = tan A. 1 + cos A + cos 2AExample 2: Show that i) sin 2q = 2 tanq ii) cos 2q = 1 - tan2 q 1 + tan2 q 1 + tan2 qSolution: i)===sin 2q 2sin q cos q 2sin q cosq 2 sinq cosq 1 cos2 q + sin2 q 2sinq cos q 2 sin q cos2 q cos q cos2 q + sin2 q cos2 q sin2 q cos2 q cos2 q + cos2 q\ sin 2q = 2 tanq 1 + tan2 q cos2 q - sin2 q = cos2 q -sin2 qii) cos 2q = cos2 q -sin2 q = 1 cos2 q + sin2 q cos2 q -sin2 q cos2 q - sin2 q cos2 q cos2 q cos2 q== cos2 q + sin2 q cos2 q sin2 q cos2 q + cos2 q cos2 q\ cos 2q = 1 - tan2 q 1 + tan2 qExample 3: Reduce cos4 q to an expression involving only function of multiples of q , raisedto the irst power.Solution: We know that: co=s2 q 1 + cos 2q 2 cos2 q + =1 cos⇒2q 2 ∴ cos4 q = (cos2 q )2 = 1 + cos 2q 2  2  version: 1.1 23

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab = 1 + 2 cos 2q + cos2 2q eLearn.Punjab 4 version: 1.1 =1 [1 + 2cos 2q + cos2 2q ] 4 =1 1 + 2 cos 2q + 1+ cos 4q  4 2 = 4 1 [2 + 4 cos 2q + 1 + cos 4q ] × 2 =1 [3 + 4cos 2q + cos 4q ] 8 Exercise 10.31. Find the values of sin 2a , cos 2a and tan 2a , when:i) sina = 12 ii) cosa = 3 , where 0<a < p 13 52 Prove the following identities:2. cota - tana =2cot 2a 3. sin 2a = tana 1+ cos 2a4. 1- cosa = tan a 5. cosa +-=ssiinnaa sec 2a - tan 2a sina 2 cosa 1+ sina sin a + cos a coses q + coses 2q = cot q6. 1- sina = 2 2 7. sec q 2 sin a - cos a 228. 1+ tana tan 2a =sec 2a 9. 2sin q sin 2q = tan 2q tanq cosq + cos3q10. sin 3q - cos 3q =2 11. cos 3q + sin 3q =4 cos 2q sinq cosq cosq sinq 24

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab eLearn.Punjab tan q + cot q sin 3q cos 3q12. 2 2 = sec q 13. cosq + sinq =2cot 2q cot q tan q - 2214. Reduce sin4 q to an expression involving only function of multiples of q , raised to the irst power.15. Find the values of sinq and cos q without using table or calculator, when q is i) 18 ii) 36 iii) 54 iv) 72 Hence prove that: cos36 cos 72 cos108 cos 144 = 1 16==Hint : Let q 18 Let q 36==5q 90 5q 180 =(3q + 2q ) =90 + 3q 2q 180 3q =90 2q - -3q 180 2q = sin 3q =sin-(90 2q ) sin=3q - sin(180 2q ) etc. etc.10.8. Sum, Difference and Product of Sines and Cosines We know that: (i) sin (a +=b ) sin a cos b + cosa sin b (ii) sin (a -=b ) sina cos b - cosa sin b (iii) cos(a =+ b ) cosa cos b - sina sin b (iv) cos(a=- b ) cosa cos b + sina sin b (v) Adding (i) and (ii) we get sin (a + b ) + sin (a - b ) =2sina cos b version: 1.1 Subtracting (ii) from (i) we get 25

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjabsin (a + b ) - sin (a - b ) =2cosa sin b (vi) eLearn.Punjab version: 1.1Adding (iii) and (iv) we get (vii) cos(a + b ) + cos(a - b ) =2cosa cos bSubtracting (iv) from (iii) we get (viii) cos(a + b ) - cos(a - b ) =- 2sina sin bSo we get four identities as: 2sina cos b= sin(a + b ) +sin(a - b )2cosa sin b= sin(a + b ) -sin(a - b )2cosa cos b= cos(a + b ) + cos(a - b )-2sina sin b= cos(a + b ) -cos(a - b )Now putting a=+ b P and a=- b Q, we get==a P + Q and b P-Q 2 2sin P + sin Q =2sin P + Q cos P - Q 22sin P - sin Q =2cos P + Q sin P - Q 22cos P + cosQ =2cos P + Q cos P - Q 22cos P - co-sQ =2cos P + Q sin P - Q 22Example 1: Express 2 sin 7q cos3q as a sum or diference.Solution: 2sin 7q cos3=q sin(7q + 3q ) + sin(7q - 3q ) = sin 10q + sin 4qExample 2: Prove without using tables / calculator, that 26

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab sin 19 cos 11 + sin 71 sin 11 =1 eLearn.Punjab 2 version: 1.1 Solutio=n: L.H.S. + sin 19 cos 11 sin 71 sin 11= + 1 [2sin 19 cos 11 2sin 71 sin 11] 2= 1 [{sin(19 + 11 ) + sin(19 -11 )} - {cos(71 + 11 ) - cos(71 -11 )}] 2= 1 [sin 30 + sin 8 - cos82 + cos 60 ] 2= 1  1 + sin 8 - cos(90 - 8 ) + 1  2 2 2= 1  1 + sin 8 - sin 8 + 1  ( cos82= cos(90 - 8=) sin 8 ) 2 2 2= 1  1 + 1  2 2 2=1 2= R.H.S. Hence sin19 cos11 + sin 71 + sin11 =1 2Example 3: Express sin 5x + sin 7x as a product.Solution: sin 5x + sin 7x= 2 sin 5x + 7x cos 5=x - 7x 2sin 6x -cos( x) 22 ==2- sin 6x cos x ( cos( q ) cos q )Example 4: Express cos A + cos3A + cos5A + cos7 A as a product.Solution: cos A + cos3A + cos5A + cos7 A 27

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab= (cos3A + cos A) + (cos7 A + cos5A) eLearn.Punjab version: 1.1= +2cos 3A + A cos 3A - A 2cos 7 A + A cos 7 A - 5A 22 22= +2cos 2Acos A 2cos6Acos A= +2cos A(cos6A cos 2A)= 2 cos A 2 cos 6 A + 2 A cos 6 A - 2 A  2 2 2 cos A(2cos 4A cos 2A) 4cos Acos 2Acos 4A.Example 4: Show that cos 20 cos 40 cos80 = 1 8Solution: L.H.S. = cos 20 cos 40 cos80 = 1 (4cos 20 cos 40 cos80 ) 4 = 1 [(2cos 40 cos 20 ) . 2 cos80] 4= 1 [(cos 60 + cos 20).2 cos80] 4 = 1  1 + cos 20 .2 cos 80 = 4 2 1 (cos80 + 2cos80 cos 20 ) 4 = 1 (cos80 + cos100 + cos 60) 4 = 1 [cos80 + cos(180 -80) + cos 60] 4 = 1  cos 80 - cos 80 + 1  [ cos(180 -q ) =-cosq ] = 4 2 =14  12  1 R.H.S. 8 28

110. .QTuriagdornaotmiceEtrqicuIadteionntisties eLearn.Punjab Hence cos 20 cos 40 cos 80 = 1 . eLearn.Punjab 8 version: 1.1 Exercise 10.41. Express the following products as sums or diferences:i) 2 sin 3q cos q ii) 2 cos 5q sin 3qiii) sin 5q cos 2q iv) 2 sin 7q sin 2qv) cos(x + y) sin (x - y) vi) cos(2x + 30) cos(2x - 30)vii) sin 12 sin 46 viii) sin(x + 45) sin(x - 45)2. Express the following sums or diferences as products:i) sin 5q + sin 3q ii) sin 8q - sin 4qiii) cos 6q + cos 3q iv) cos 7q - cos qv) cos 12 + cos 48 vi) sin (x + 30) + sin(x - 30)3. Prove the following identities:i) sin 3x - sin x = cot 2x ii) sin8x + sin 2x = tan 5x cos x - cos3x cos8x + cos 2xiii) sina - sin b = tan a - b cot a + b sina + sin b 224. Prove that: i) cos 20 + cos100 + cos140 =0ii) sin  p -q sin  p +q  =1 cos 2q 4 4 2 29

110. .QTuriagdornaotmiceEtrqicuaIdteionntisties eLearn.Punjab eLearn.Punjabiii) sinq + sin 3q + sin 5q + sin 7q = tan 4q cosq + cos3q + cos5q + cos7q5. Prove that:i) cos 20 cos 40 cos 60 cos80 = 1 16ii) sin p sin 2p sin p sin 4p = 3 9 9 3 9 16iii) sin10 sin 30 sin 50 sin 70 = 1 16 version: 1.1 30

version: 1.1CHAPTER11 Trigonometric Functions and their Graphs

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjab11.1 Introduction Let us irst ind domains and ranges of trigonometric functions before drawing theirgraphs.11.1.1 Domains and Ranges of Sine and Cosine Functions We have already deined trigonometric functionssin q , cos q , tan q , csc q , sec q and cot q . We know thatif P(x, y) is any point on unit circle with center at the originO such that ∠XOP = q is standard position, then cos q = x and sin q = y⇒ for any real number q there is one and only one value of each x and y .i.e., of each cos q and sin q .Hence sin q and cos q are the functions of q and their domain is R a set of real numbers.Since P(x, y) is a point on the unit circle with center at the origin O.∴ -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1⇒ -1 ≤ cos q ≤ 1 and -1 ≤ sin q ≤ 1Thus the range of both the sine and cosine functions is [-1, 1].11.1.2 Domains and Ranges of Tangent and Cotangent Functions From igure 11.1i=) tan q ≠ y , x0 x⇒  should not coincide with OY or OY’ (i.e., Y-axis) terminal side OP version: 1.1 2

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.Punjab ⇒ q ≠ ± p ,± 3p ,± 5p ,... 22 2 ⇒ q ≠ (2n +1) p , where n ∈ Z 2aa Domain of tangent function =R -{x | x =(2n +1) p , n ∈ Z} 2 and Range of tangent function = R = set of real numbers.ii) From igure 11.1=cot q ≠ x , y0 y ⇒ terminal side  should not coincide with OX or OX’ (i.e., X - axis) OP ⇒ q ≠ 0,± p ,± 2p ,... ⇒ q ≠ np , where n ∈ Z Domain of cotangent function = R -{x | x =np , n ∈ Z} and Range of cotangent function = R = set of real numbers.11.1.3 Domain and Range of Secant Function From igure 11.1=sec q ≠ 1 , x0 x ⇒ terminal side  should not coincide with OY or OY’ (i.e., Y - axis) OP ⇒ q ≠ ± p ,± 3p ,± 5p ,... 22 2 version: 1.1 3

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjabaa ⇒ q ≠ (2n +1) p , where n ∈ Z 2 Domain of secant function = R -{x | x =(2n +1) p , n ∈ Z} 2 As sec q attains all real values except those between -1 and 1 Range of secant function= R - {x | -1 < x < 1}11.1.4 Domain and Range of Cosecant Function From igure 11.1=csc q ≠ 1 , y0 y ⇒ terminal side  should not coincide with OX or OX’ (i.e., X - axis) OP ⇒ q ≠ 0,±p ,± 2p ,... ⇒ q ≠ np , where n ∈ Zaa Domain of cosecant function = R -{x | x =np , n ∈ Z} As csc q attains all values except those between -1 and 1 Range of cosecant function = R - {x | -1 < x < 1} The following table summarizes the domains and ranges of the trigonometric func-tions: Function Domain Range y = sin x -∞ < x < +∞ -1 ≤ y ≤ 1 y = cos x -∞ < x < +∞ -1 ≤ y ≤ 1 version: 1.1 4

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.Punjab y = tan x -∞ < x < +∞, x ≠ (2n +1)p ,n ∈ Z - ∞ < y < +∞ y = cot x 2 y = sec x -∞ < x < + ∞, x ≠ np , n ∈ Z -∞ < y < +∞ y = coses x -∞ < x < + ∞, x ≠ (2n +1)p ,n ∈ Z y ≥ 1 or y ≤ -1 2 n ∈ Z y ≥1 or y ≤ -1 -∞ < x < + ∞, x ≠ np ,11.2 Period of Trigonometric Functions All the six trigonometric functions repeat their values for each increase or decrease of2p in q i.e., the values of trigonometric functions for q and q ± 2np , where q ∈ R , and n ∈ Z,are the same. This behaviour of trigonometric functions is called periodicity. Period of a trigonometric function is the smallest +ve number which, when added tothe original circular measure of the angle, gives the same value of the function. Let us now discover the periods of the trigonometric functions.Theorem 11.1: Sine is a periodic function and its period is 2p .Proof: Suppose p is the period of sine function such that=sin (q + p) sin∈q for all q R (i)Now put q = 0, we havesin (0 + p) =sin 0⇒ sin p =0⇒ p = 0,±p , ± 2p , ±3p ,...i) if p = p , then from (i) (not true) sin (q + p ) =sin q  sin (q + p ) =- sin qa p is not the period of sin q .ii) if p = 2p , then from (i) sin (q + 2p ) =sin q , Which is true version: 1.1 5

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjab As is the smallest +ve real number for whicha sin (q + 2p ) =sin qa 2p is not the period of sin q .Theorem 11.2: Tangent is a periodic function and its period is p .Proof: Suppose p is the period of tangent function such that =tan (q + p) ta∈n q for all q R (ii)Now put q = 0 , we have tan (0 + p) =tan 0 ⇒ tan p =0 ∴ p =0, p , 2p ,3p ,....i) if p = p , then from (i) tan (q + p ) =tan q , which is true. As p is the smallest +ve number for which tan (q + p ) =tan q p is not the period of tan q .Note: By adopting the procedure used in inding the periods of sine and tangent, we can prove thati) 2p is the period of cos q . ii) 2p is the period of cscq .iii) 2p is the period of secq . iv) p is the period of cotq . 6 version: 1.1

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.PunjabExample 1: Find the periods of: i) sin 2x ii) tan x 3Solution: i) We know that the period of sine is 2p∴ sin (2x=+ 2p ) sin 2x ⇒ sin 2=(x + p ) sin 2x It means that the value of sin 2x repeats when x is increased by p . Hence n is the period of sin 2x.ii) We know that the period of tangent is p∴ tan  x +=p  tan x ⇒ tan 1 (x=+ 3p ) tan x 3 3 33It means that the value of tan x repeats when x is increased by 3p . 3Hence the period of tan x is 3p . 3 Exercise 11.1Find the periods of the following functions:1. sin 3x 2. cos 2x 3. tan 4x 4. cot x 25. sin x 6. coses x 7. sin x 3 4 5 8. cos x 69. tan x 10. cot 8x 11. sec 9x 7 14. 2 cos x 15. 3 cos x 12. cosec 10x.13. 3 sin x 5 version: 1.1 7

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjab11.3 Values of Trigonometric Functions We know the values of trigonometric functions for angles of measure 0°, 30°, 45°, 60°,and 90°. We have also established the following identities:sin (-q ) - =sin q -cos ( q=) cos q- - tan ( q=) tan qsin (p -q ) =sin q cos (p q )- - co=s q ta-n (-p =q ) tan qsin (p + q )- =sin q co+s (p- =q ) cos q+ tan=(p q ) tan qsin (2p -q )- =sin q cos (2p -q ) =cos q tan (2p q-)- =tan q By using the above identities, we can easily find the values of trigonometric functionsof the angles of the following measures: -30, -45, -60, -90 ±120, ±135, ±150, ±180 ±210, ±225, ±240, ±270 ±300, ±315, ±330, ±360.11.4 Graphs of Trigonometric FunctionsWe shall now learn the method of drawing the graphs of all the six trigonometricfunctions. These graphs are used very often in calculus and social sciences. For graphing thelinear equations of the form:a1x + b1y + c1 =0 (i)a2x + b2 y + c2 =0 (ii) We have been using the following procedure. i) tables of the ordered pairs are constructed from the given equations, ii) the points corresponding to these ordered pairs are plotted/located,and iii) the points, representing them are joined by line segments. Exactly the same procedure is adopted to draw the graphs of the trigonometricfunctions except for joining the points by the line segments. version: 1.1 8

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.PunjabFor this purpose,i) table of ordered pairs (x, y) is constructed, when x is the measure of the angle and y is the value of the trigonometric ratio for the angle of measure x;ii) The measures of the angles are taken along the X- axis;iii) The values of the trigonometric functions are taken along the Y-axis;iv) The points corresponding to the ordered pairs are plotted on the graph paper,v) These points are joined with the help of smooth ciurves.Note: As we shall see that the graphs of trigonometric functions will be smooth curves and none of them will be line segments or will have sharp corners or breaks within their domains. This behaviour of the curve is called continuity. It means that the trigonometric functions are continuous, wherever they are deined. Moreover, as the trigonometric functions are periodic so their curves repeat after a ixed interval.11.5 Graph of y = sin x from -2p to 2p We know that the period of sine function is 2p so, we will irst draw the graph for theinterval from 0° to 360° i.e., from 0 to 2p .To graph the sine function, irst, recall that -1 ≤ sin x ≤ 1 for all x ∈ R i.e., the range of the sine function is [-1, 1], so the graph will be between the horizon-tal lines y = +1 and y = - 1 The table of the ordered pairs satisfying y = sin x is as follows: version: 1.1 9

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjab To draw the graphi) Take a convenient scale 11ssiiddee of small square on the x - axis =10 of big square on the y - axis =1unitii) Draw the coordinate axes.iii) Plot the points corresponding to the ordered pairs in the table above i.e., (0, 0), (30°, 0.5), (60°, 0.87) and so on,(iv) Join the points with the help of a smooth curve as shown so we get the graph of y = sin x from 0 to 360° i.e., from 0 to 2p . In a similar way, we can draw the graph for the interval from 0° to -360°. This willcomplete the graph of y = sin x from -360° to 360° i.e. from - 2p to 2p , which is given below: The graph in the interval [0, 2p ] is called a cycle. Since the period of sine function is 2p ,so the sine graph can be extended on both sides of x-axis through every interval of 2p (360°)as shown below: version: 1.1 10

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.Punjab11.6 Graph of y = cos x from -2p to 2p We know that the period of cosine function is 2p so, we will irst draw the graph for theinterval from 0° to 360° i.e., from 0 to 2p To graph the cosine function, irst, recall that -1 ≤ sin x ≤ 1 for all x ∈ R i.e., the range of the cosine function is [-1, 1], so the graph will be between the horizontallines y = +1 and y = -1 The table of the ordered pairs satisfying y = cos x is as follows: The graph of y = cos x from 0° to 360° is given below: version: 1.1 11

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjab In a similar way, we can draw the graph for the interval from 0° to -360°. This willcomplete the graph of y = cos x from -360° to 360° i.e. from - 2p to 2p , which is given below: As in the case of sine graph, the cosine graph is also extendedon both sides of x-axis through an interval of 2p as shown above:11.7 Graph of y = tan x from -p To p We know that tan (-x) = - tan x and tan (p - x) = - tan x, so the values of tan x forx = 0°, 30°, 45°, 60° can help us in making the table. Also we know that tan x is undeined at x = ± 90°, wheni) x approaches p from left i.e., x → p - 0, tan x increases indeinitely in I Quard. 22ii) x approaches p from right i.e., x → p + 0 , tan x increases indeinitely in IV Quard. 22iii) x approaches - p from left i.e., x → - p - 0 tan x increases indeinitely in II Quard. 22 version: 1.1 12

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.Punjabiv) x approaches - p from right i.e., x → - p + 0 , tan x increases indeinitely in III Quard. 22 We know that the period of tangent is p , so we shall irst draw the graph for theinterval from -p to p i.e., from -180° to 180° The table of ordered pairs satisfying y = tan x is given below:a version: 1.1 13

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjab We know that the period of the tangent function is p . The graph is extended on bothsides of x-axis through an interval of p in the same pattern and so we obtain the graph ofy = tan x from -360° to 360° as shown below:11.8 Graph of y = cot x From -2p to p We know that cot (-x) = - cot x and cot (p - x) = - cot x, so the values of cot x forx = 0°, 30°, 45°, 60°, 90° can help us in making the table. The period of the cotangent function is also p . So its graph is drawn in a similar way oftangent graph using the table given below for the interval from -180° to 180°. version: 1.1 14

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.PunjabWe know that the period of the cotangent function is p . The graph is extended on both version: 1.1 15

111. .QTuriagdornaotmiceEtrqicuFautinocntsions and their Graphs eLearn.Punjab eLearn.Punjabsides of x - axis through an interval of p in the same pattern and so we obtain the graph ofy = cotx from from -360° to 360° as shown below:11.9 Graph of y = sec x from -2p to 2p We know that sec (-x) =sec x and sec (p x-)- =sec x, So the values of sec x for x = 0°, 30°, 45°, 60°, can help us in making the following tableof the ordered pairs for drawing the graph of y = sec x for the interval 0° to 360°: version: 1.1 16

111. .QTuriagdornaotmiceEtrqicuFautinocntisons and their Graphs eLearn.Punjab eLearn.Punjab Since the period of sec x is also 2p , so we have the following graph of y = sec x from- 360° to 360° i.e., from - 2p to 2p : Graph of y = sec x from - 360° to 360°11.10 Graph of y = csc x from -2p to 2p We know that: csc ( -x) = - csc x and csc (p - x) = csc x So the values of csc x for x = 0°, 30°, 45°, 60°, can help us in making the followingtable of the ordered pairs for drawing the graph of y = csc x for the interval 0° to 360°: version: 1.1 17