2018-G11-Math-E

113. .QInuvaedrrsaetTicrigEnqoumateitornicsFunctions eLearn.Punjab eLearn.Punjabwhich is one-to-one and has an inverse. The Inverse Cosecant Function is deined by: y = csc-1 x, if and only if x = csc y where - p ≤ y ≤ p , y ≠ 0 and x ≥ 1 22 The students should draw the graph of y = csc-1 by taking the relection of y = cscx inthe line y = x. This is left an exercise for them.Note. While discusspjpihe Inverse Trigonometric Functions, we have seen that there are in general, no inverses of Trigonometric Functions, but restricting their domain to principal Functions, we have made them as functions.13.6 Domains and Ranges of Principal Trigonometric Functionand Inverse Trigonometric Functions. From the discussion on the previous pages we get the following table showing domainsand ranges of the Principal Trigonometric and Inverse Trigonometric Functions.Functions Domain Range y = sin x -p ≤ x ≤ p -1 ≤ x ≤ 1 22 y = sin-1 x -1 ≤ x ≤ 1 -p ≤ x ≤ p y = cos x 0≤ x≤p 22 y = cos-1 x -1 ≤ x ≤ 1 -1 ≤ x ≤ 1 y = tan x -p < x < p 0≤ x≤p 22 (-∞,∞) or ℜ 9 version: 1.1

113. .QInuvaedrrsaetiTcriEgnqoumateitornicsFunctions eLearn.Punjab eLearn.Punjab y = tan-1 x (-∞,∞) or ℜ -p < x < p y = cot x 0< x<p 22 y = cot-1 x (-∞,∞) or ℜ y = sec x (-∞,∞) or ℜ 0< x<p [0,p ], x ≠ p y = sec-1 x y ≤ -1 or y ≥ 1 y = csc x 2 x ≥ -1 or x ≤ 1 [0,p ], y ≠ p y = csc-1 x 2 [- p , p ], x ≠ 0 22 y ≤ -1 or y ≥ 1 x ≤ -1 or x ≥ 1 [- p , p ], y ≠ 0 22Example 4: Show that cos-1 12 sin-1 5 13 13Solution: Let cos-1 12 =⇒a co=sa 12 13 13∴ sina = ± 1 - cos2 a = ± 1 -  12 2 13 = ± 1- 144 169 ±= 169 -±1=44 ± =25 5 169 169 13a cos a is +ve and domain of a is [0, p], in which sine is +ve. version: 1.1 10

113. .QInuvaedrrsaetTicrigEnqoumateitornicsFunctions eLearn.Punjab eLearn.PunjabThus sina = 5 ⇒ a = sin-1 5Hence 13 13 cos-1 12 = sin-1 5 13 13Example 5: Find the value ofi) sin (cos-1 3 ) ii) cos (tan-1 0) iii) sec [sin-1(- 1)] Solution: 2 2i) we irst ind the value of y, whose cosine is 3 2 =cos y 3 , 0≤ y≤p 2⇒ y =p 6⇒ (cos-1 3 ) =p 26∴ sin(cos-1 =3 ) =sin p 1 2 62ii) we irst ind the value of y, whose tangent is 0tan=y 0, -p < y<p 22⇒ y =0⇒ (tan-1 0) =0∴ cos(tan-1 0=) =cos 0 1iii) we irst ind the value of y, whose sine is - 1 2sin y =- 1 , -p ≤ y≤p 2 22⇒ -y =p 6⇒ sin-1(- 1 ) =- p 26 version: 1.1 11

113. .QInuvaedrrsaetiTcriEgnqoumateitornicsFunctions eLearn.Punjab eLearn.Punjab ∴ sec[sin-1(- 1)] =2 23Example: 6 Prove that the inverse trigonometric functions satisfy the following identities:i) sin-1 x =p cos-1 x and cos--1 x p sin-1 x =- 22 =- =-ii) tan-1 x =p cot-1 x and cot--1 x p tan-1 x 22iii) sec-1 x =p csc-1 x and csc--1 x p sec-1 x 22Proof:Consider the right triangle given in the igure Angles a and b are acute and complementary. ⇒ a + b =p 2 ⇒ a =p - b and b =p - a ...(i) 22Now sina= sin(p - b =) cos b= x (say) 2=∴ a sin-1 x =and b cos-1 x =- Thus from (i) we have: sin-1 x =p cos-1 x and cos--1 x p sin-1 x 22 In a similar way, we can derive the identities (ii) and (iii). Exercise 13.11. Evaluate without using tables / calculator:i) sin-1(1) ii) sin-1(-1) iii) cos-1  3  2iv) tan-1  - 1  v) cos-1  1  vi) tan-1  1  3 2 3 version: 1.1 12

113. .QInuvaedrrsaetTicrigEnqoumateitornicsFunctions eLearn.Punjab eLearn.Punjabvii) cot-1(-1) viii) cos ec-1  -2  ix) sin-1  - 1  3 22. Without using table/ Calculator show that:i) tan-1 5 = sin-1 5 ii) 2cos-1 4 = sin-1 24 12 13 5 25 iii) cos-1 4 = cot-1 4 53- 3. Find the value of each expression:i) cos sin-1 1  ii) sec  cos-1 1  iii) tan  cos-1 3  2 2  2( )iv) csc tan-1(-1) v) sec  sin -1 (- 1 )  ( )vi) tan tan-1(-1) 2vii) sin  sin -1 ( 1 )  viii) tan  sin -1 (- 1 )  ( )ix) sin tan-1(-1) 2 213.7 Addition and Subtraction Formulas1) Prove that: sin-1 A + si=n-1 B sin-1( A 1 - B2 + B 1 - A2 )Proof: Let sin-1 A = x ⇒ sin x = A and sin-1 B = y ⇒ sin y = BNow cos x = ± 1 - sin2 x = ± 1 - A2 In sinx = A, domain = - p , p  , in whichCosine is +ve, 2 2∴ cos x =1 A2 -Similarly, cos y= 1 - B2Now sin(x +=y) sin x cos y + cos xsin y version: 1.1 13

113. .QInuvaedrrsaetiTcriEgnqoumateitornicsFunctions eLearn.Punjab = A 1 - B2 + B 1 - A2 eLearn.Punjab ⇒ x=+ y sin-1( A 1 - B2 - B 1 - A2 ) ∴ version: 1.1 sin-1 A + si=n-1 B sin-1( A 1 - B2 + B 1 - A2 ) In a similar way, we can prove that2) sin-1 A - si=n-1 B sin-1( A 1 - B2 - B 1 - A2 )3) cos-1 A + cos=-1 B cos-1( AB - (1 - A2 )(1 - B2 ))4) cos-1 A - cos=-1 B cos-1( AB + (1 - A2 )(1 - B2 ))5) Prove that: tan-1 A + tan-1 B =tan-1 1A-+ABB =A =BProof: Let tan-1 A = x ⇒ tan x and tan-1 B = y ⇒ tan yNow tan=(x + y) 1t-anta=xn+xttaannyy A+ B 1- AB⇒ x+ y =tan -1 A+ B 1 - AB∴ tan-1 A + tan B =tan-1 1A-+ABBIn a similar way, we can prove that6) tan-1 A - tan-1 B =tan-1 1A+-ABB we getCor. Putting A - B in tan-1 A + tan-1B =tan-1 1A-+ABB , 14

113. .QInuvaedrrsaetTicrigEnqoumateitornicsFunctions eLearn.Punjab eLearn.Punjabtan-1 A + tan-1 A =tan-1 1A-+AA2⇒ 2 tan-1 A =tan-1 1 -2 AA2 Exercise 13.2 Prove the following: 2. tan-1 1 + tan-1 1 =tan-1 91. sin-1 5 + sin-1 7 =cos-1 253 4 5 19 13 25 3253. 2 tan-1 2 = sin-1 12 =H=int : Let tan-1 32 x.and shown sin 2x 12  3 13 134. tan-1 120 = 2cos-1 12 5. sin-1 1 + cot-1 3 =p 119 13 546. sin-1 3 + sin-1 8 =sin-1 77 7. sin-1 77 - sin-1 3 =cos-1 155 17 85 85 5 178. cos-1 63 + 2 tan-1 1 =sin-1 3 65 5 59. tan-1 3 + tan-1 3 - tan-1 8 =p 4 5 19 4Hint : First add tan -1 3 + tan-1 3 and then proceed 4 510. sin-1 4 + sin-1 5 + sin-1 16 =p 5 13 65 2 version: 1.1 15

113. .QInuvaedrrsaetiTcriEgnqoumateitornicsFunctions eLearn.Punjab eLearn.Punjab11. tan-1 1 + tan-1 5 = tan-1 1 + tan-1 1 11 6 3 212. 2 tan-1 1 + tan-1 1 =p 3 7413. Show that cos(sin-1 x) = 1 - x214. Show that sin(2cos-1 x=) 2x 1 - x215. Show that16. Show that cos(2sin-1 x) = 1 - 2x217. Show that18. Show that tan-1(-x) - =tan-1 x sin -1 (- x)- =sin-1 x cos-1(-x) - =p cos-1 x19. Show that tan(sin-1 x) =x 1- x220. Given that x = sin-1 1 , ind the values of following trigonometric functions: sinx, cosx, 2 tanx, cotx, secx and cscx. version: 1.1 16

version: 1.1CHAPTER14 Solutions of Trignometric Equation

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations eLearn.Punjab eLearn.Punjab14.1 Introduction The Equations, containing at least one trigonometric function, are called TrigonometricEquations, e.g., each of the following is a trigonometric equation: =sin x 2 ,=Sec x ta-n x =an+d sin2 x sec x 1 3 54Trigonometric equations have an ininite number of solutions due to the periodicity of thetrigonometric functions. For example If sinq =q then q = 0, ± , ± 2 ,... which can be wr=itten as q ∈n , where n Z. In solving trigonometric equations, irst ind the solution over the interval whoselength is equal to its period and then ind the general solution as explained in the followingexamples:Example 1: Solve the equation sin x = 1 2Solution: sin x = 1 2a sin x is positive in I and II Quadrants with the reference angle x = . 6=∴ x and =x= - 5 , where x ∈[0, 2 ] 6 66∴ General values of x are + 2n and 5 + 2n , n ∈ Z 66Hence solution set =6 + 2n  ∪  5 + 2n  ,n∈Z 6 version: 1.1 2

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations eLearn.PunjabExample 2: Solve the equation: 1 + cos x = 0 eLearn.Punjab version: 1.1Solution: 1 + cos x = 0⇒ cos x = -1Since cos x is -ve, there is only one solution x = p in [0, 2p]Since 2p is the period of cos x∴ General value of x is p + 2np, n∈ZHence solution set = {p + 2np}, n∈ZExample 3: Solve the equation: 4 cos2x - 3 = 0Solution: 4 cos2 x - 3 = 0 ⇒ cos2 x =3 ⇒ ± cos x =3 42i. If cos x = 3 2 Since cos x is +ve in I and IV Quadrants with the reference anglex= 6=∴=x -and x = 2 11 where x ∈[0, 2 ] 6 66As 2p is the period of cos x.∴ General value of x are + 2n and 11 + 2n , n ∈ Z 66ii. if cos x = - 3 2Since cos x is -ve in II and III Quadrants with reference angle x = 6∴ x = - =5 and x = x + = 7 where x ∈[0, 2 ] 66 66As 2p is the period of cos x. 3

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations eLearn.Punjab eLearn.Punjab ∴ General values of x are 5 + 2n and 7 + 2n , n ∈ Z 66 Hence solution set =6 + 2n  ∪ 116 + 2n  ∪ 56 + 2n  ∪ 76 + 2n 14.2 Solution of General Trigonometric Equations When a trigonometric equation contains more than one trigonometric functions,trigonometric identities and algebraic formulae are used to transform such trigonometricequation to an equivalent equation that contains only one trigonometric function. The method is illustrated in the following solved examples:Example 1: Solve: sin x + cos x = 0.Solution: sin x + cos x = 0 ⇒ sin x + cos x =0 (Dividing by cos x ≠ 0) cos x cos x ⇒ tan x +1 =0 ⇒ t-an x =1a tan x is -ve in II and IV Quadrants with the reference angle x= 4∴ x = - =3 , where x ∈[0, ] 44As p is the period of tan x,∴ General value of x is 3 + n , nU Z 4∴ Solution set =  3 + n  ,n U Z. 4 version: 1.1 4

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations eLearn.Punjab eLearn.PunjabExample 2: Find the solution set of: sin x cos x = 3 . 4Solution: sin x cos x = 3 . 4 ⇒ 1 (2 sin x cos x) =3 2 4 ⇒ sin 2x =3 2a sin 2x is +ve in I and II Quadrants with the reference angle 2x = 3∴ 2x =and 2x = - =2 are two solutions in [0,2 ] 3 33As 2p is the period of sin 2x .∴ General values of 2x are + 2n and 2 + 2n , , n U Z 33⇒ General values of x are + n and + n , nU Z 63 , nU Z Hence solution set = = 6 + n  ∪ 3 + n Note: In solving the equations of the form sin kx = c, we irst ind the solution pf sin u = c (where kx = w) and then required solution is obtained by dividing each term of this solution set by k.Example 3: Solve the equation: sin 2x = cos 2xSolution: sin2x = cos2x ⇒ 2sinx cos x = cosx ⇒ 2sinx cos x - cosx = 0 ⇒ cosx(2sinx - 1) = 0 version: 1.1 5

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations eLearn.Punjab ∴ cosx = 0 or 2sinx - 1 = 0 eLearn.Punjabi. If cosx = 0 version: 1.1 ⇒ x= and x = 3 where x U [0,2p] 2 2 As 2p is the period of cos x .∴ General values of x are p + 2np and 3p + 2np, nUZ,2 2ii. If 2 sin x - 1 = 0⇒ sin x = 1 2Since sin x is +ve in I and II Quadrants with the reference angle x = p 6∴ x = p and x =p - p = 5p where x U [0, 2p] 6 66As 2p is the period of sin x.∴ General values of x are and p + 2np and 5 p + 2np, nUZ, 66Hence solution set = p2 + 2np  ∪  3p + 2np  ∪ p6 + 2np  ∪ 5 p + 2np , 2 6n∈zExample 4: Solve the equation: sin2 x + cos x = 1.Solution: sin2 x + cos x = 1⇒ 1 - cos2 x + cos x = 1⇒ - cos x (cos x - 1) = 0⇒ cos x = 0 or cos x - 1 = 0i. If cos x = 0 , where x U [0, 2p]⇒ x = p and x = 3p 22 6

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations eLearn.Punjab eLearn.PunjabAs 2p is the period of cos x∴ General values of x are p + 2np and 3p + 2np , nUZ 22ii. If cos x = 1⇒ x = 0 and x = 2p , where x U [0, 2p]As 2p is the period of cos x∴ General values of x are 0 + 2np and 2p + 2np, nUZ.∴Solution Set =p2 + 2np  ∪ 32p + 2np  ∪{2np} ∪{2p + 2np},n ∈ z {2(n +1)p} ⊂ {2np}, n ∈ zHence the solution set =  p + 2np  ∪ 32p + 2np  ∪{2np},n ∈ z 2 Sometimes it is necessary to square both sides of a trigonometric equation. In sucha case, extaneous roots can occur which are to be discarded. So each value of x must bechecked by substituting it in the given equation. For example, x = 2 is an equation having a root 2. On squaring we get x2 - 4 which givestwo roots 2 and -2. But the root -2 does not satisfy the equation x = 2. Therefore, -2 is anextaneous root.Example 5: Solve the equation: csc=x 3 + cot x.Solution: csc=x 3 + cot x .......(i) ⇒ 1 = 3 + cos x sin x sin x ⇒=1 3 sin x + cos x⇒ 1 - cos x =3 sin x⇒ (1 - cos x)2 =( 3 sin x)2 version: 1.1 7

114. .QSuoaludtriaontiscoEfqTuriagtoiononms etric Equations eLearn.Punjab eLearn.Punjab⇒ 1 - 2cos x + cos2 x =3sin2 x⇒ 1 - 2cos x + cos2 x = 3(1 - cos2 x)⇒ 4cos2 x - 2cos x - 2 =0⇒ 2cos2 x - cos x -1 =0⇒ (2cos x +1)(cos x -1) =0⇒ co-s x =1 or =cos x 1 2i. If cos x = - 1 2Since cos x is -v e in II and III Quadrants with the reference angle x = p 3⇒ x =p - p = 2p and x =p + p = 4p , where x U [0, 2p] 33 33Now x = 4p does not satisfy the given equation (i). 3∴ x =4p is not admissible and so x = 2p is the only solution. 33Since 2p is the period of cos x∴ General value of x is 2p + 2np , nUZ 3ii. If cos x = 1⇒ x = 0 and x = 2p where x U [0, 2p]Now both csc x and cot x are not defined for x = 0 and x = 2∴ x = 0 and x = 2 are not admissible.Hence solution set =  2p + 2np  , nUZ 3 version: 1.1 8

114. .QSuoaludtrioantiscoEfqTuriagtoinoonms etric Equations eLearn.Punjab eLearn.Punjab Exercise 141. Find the solutions of the following equations which lie in [0, 2p]i) sin x = - 3 ii) cosecq = 2 iii) sec x = -2 iv) cotq = 1 2 32. Solve the following trigonometric equations: cot2 q = 1 3i) tan2 q = 1 ii) cos ec2q = 4 iii) sec2 q = 4 iv) 3 3 3Find the values of q satisfying the following equations:3. 3tan2 q + 2 3 tanq +1 =04. tan2 q - secq -1 =05. 2sinq + cos2 q -1 =06. 2sin2 q - sinq =07. 3cos2 q - 2 3 sinq cosq - 3sin2 q =0 [Hint: Divide by sin2q]Find the solution sets of the following equations:8. 4 sin2q - 8cosq + 1 = 09. 3 tan x - sec x -1 =0 [Hint: sin3x = 3sinx - 4sin3x]10. cos 2x = sin 3x11. sec 3q= secq12. tan 2q + cotq = 013. sin 2x + sinx = 014. sin 4x - sin 2x = cos 3x15. sin x + cos 3x = cos 5x16. sin 3 x + sin 2x + sin x = 017. sin 7x - sin x = sin 3 x18. sin x + sin 3x + sin 5x = 019. sin q + sin 3q + sin 5q + sin 7q = 020. cos q + cos 3q + cos 5 q + cos 7q = 0 version: 1.1 9