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Mathematics

Mathematics. Trigonometric Equations – Session 2. Session Objectives. Session Objectives. Removing Extraneous Roots. Avoiding Root Loss. Equation of the form of a cosx +b sinx = c. Simultaneous equation. Trigonometric Equation – Removing Extraneous Solutions. _J30. Extraneous solutions

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Mathematics

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  1. Mathematics

  2. Trigonometric Equations – Session 2

  3. Session Objectives

  4. Session Objectives • Removing Extraneous Roots • Avoiding Root Loss • Equation of the form of a cosx +b sinx = c • Simultaneous equation

  5. Trigonometric Equation – Removing Extraneous Solutions _J30 Extraneous solutions The solutions, which do not satisfy the trigonometric equation. • Origin of Extraneous solutions ? • Squaringduring solving the trigo. equations. ( ±)2= + • Solutions, which makes the equation undefined. ( denominator being zero for a set of solutions) Esp . equations containing tan or sec, cot or cosec

  6. Trigonometric Equation – Extra root because of squaring _J30 Illustrative problem Solve sec - 1 = ( √ 2 – 1) tan Equation can be rewritten as sec = ( √ 2 – 1) tan + 1 On squaring , we get sec2 = (2+1–2√2)tan2 +1+2(√2–1)tan sec2 - tan2 = (2–2√2)tan2 +1 + 2(√2–1)tan   (2 – 2√2 ) tan2 + 2(√2 – 1) tan = 0 tan  = 0 and tan  = +1

  7. Trigonometric Equation – Extra root because of squaring _J30 Solve sec - 1 = ( √ 2 – 1) tan tan  = 0 or tan  = +1  = n or  = n + /4 Putting =  in given equation , we get L.H.S. = ( -1 –1) = -2 R.H.S. = ( √ 2 – 1).0 = 0 Extraneous solutions :  = odd integer multiple of π WHY ?? Answer : = 2n or  = 2n + /4

  8. Trigonometric Equation – Solutions, which makes the equation undefined _J30 Solve tan5 = tan3 Solution will be 5 = nπ + 3  = nπ/2 , where n  Z solutions  = π/2, 3π/2…..etc. will make tan3 and tan5 undefined Answer :  = mπ, where m  Z

  9. Trigonometric Equation – Avoiding root loss _J31 • Reason for root loss ? • Canceling the terms from both the sides of the equation • Use of trig. Relationship , which restricts the acceptable values of  ,i.e., the domain of  changes.

  10. Trigonometric Equation – Cancelling of terms from both sides _J31 Illustrative problem Solve sin .cos = sin If we cancel sin  from both the sides cos  = 1   = 2n , where n  Z However , missed the solution provided by sin = 0   = n ,where n  Z Answer :  = n ,where n  Z

  11. If we use Equation can be written as : Trigonometric Equation – Changes in domain of  _J31 Illustrative Problem Solve : sin - 2.cos = 2

  12. Trigonometric Equation – Changes in domain of  _J31 Solve : sin - 2.cos = 2  = 2nπ + 2 , where n  Z and  = tan-1 2 Root loss : = (2n+ 1) π where , n Z As  = π , 3 π , 5 π …. satisfy the given trig. equation Answer :  = (2nπ+2) U (2n+1)π, where n  Z and = tan-1 2

  13. Divide both sides of the equation by • The equation now will be as Trigonometric Equation – a sin  + b cos  = c _J32 Reformat the equation in the form of cos(  -  ) = k.

  14. Hence , the given equation can be written as a b For real values of  , Trigonometric Equation – a sin  + b cos  = c _J32 Compare with sin.sin + cos.cos = k

  15. Trigonometric Equation – a sin  + b cos  = c - Algorithm _J32 Step 1: Reformat the equation into cos(-) = k Step 2: Check whether real solution exists Step 3: Solve the equation cos(-) = k

  16. Trigonometric Equation – a sin  + b cos  = c - Problem _J32 Illustrative Problem Solve sin + cos = 1 Step 1: Reformat the equation into cos(-) = k

  17. Trigonometric Equation – a sin  + b cos  = c - Problem _J32 Solve sin + cos = 1 Step 2: Check whether real solution exists Real solution exists

  18. Trigonometric Equation – a sin  + b cos  = c - Problem _J32 Solve sin + cos = 1 Step 3: Solve the equation cos(-) = k

  19. Simultaneous Trigonometric Equations _J33 Case I : Two equations and one variable angle Step 1 – Solve both the equation between 0 and 2π . Step 2 – Find common solutions. Step 3– Generalise the solution by adding 2nπ to common solution as per step 2.

  20. Simultaneous Trigonometric Equations - Problem _J33 Find the general solution of tan = -1, cos = 1/2 Step 1 – Solve both the equation between 0 and 2π

  21. Simultaneous Trigonometric Equations - Problem _J33 Illustrative Problem Find the general solution of tan = -1, cos = 1/2 Step 2 – Find common solutions Step 3– Generalise the solution Answer

  22. Simultaneous Trigonometric Equations _J33 Case II : Two equations and two variable angles(θ,φ) and smallest positive values of the angles satisfying the equations needs to be found out Find the smallest positive values of θ and φ , if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3

  23. Simultaneous Trigonometric Equations _J33 Step 1 – Solve both the equations between 0 and 2π . Step 2 – Find two equations with the conditions such as ( θ+φ) > (θ-φ) for positive θ and φ Step 3 - Solve the two equations to determine θ and φ

  24. Simultaneous Trigonometric Equations - Problem _J33 Illustrative Problem Find the smallest positive values of θ and φ , if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3 Step 1 – Solve both the equations between 0 and 2π .

  25. Step 3 - Solve the two equations to determine θ and φ Simultaneous Trigonometric Equations - Problem _J33 Find the smallest positive values of θ and φ , if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3 Step 2 – Find two equations with the conditions such as ( θ+φ) > (θ-φ) for positive θ and φ

  26. Trigonometric Equation – Misc. Tips : • cosθ = k is simpler to solve compared to sinθ = k • Check for extraneous roots and root loss • In case of ‘algebraic function of angle’ is a part of the equation use of the following properties: • Range of values of sin and cos functions • x2 ≥ 0 • A.M. ≥ G.M.

  27. Trigonometric Equation – Misc. Tips : • Equation with multiple terms of the form (sinθ ± cosθ) and sinθ.cosθ : • Put sinθ+cosθ = t. • Equation gets converted in to a quadratic equation in t sin x + cos x = 1+ sin x. cos x

  28. Trigonometric Equation – Misc. Tips : • Equation with terms of sin2θ , cos2θ and sinθ.cosθ • Try dividing by cos2θ to get a quadratic equation in tan θ. Solve the equation 2 sin2 x – 5 sinx.cos x – 8 cos2 x = -2 • Solution to sin2θ = sin2 , cos2θ = cos2  and tan2θ = tan2 is n 

  29. Class exercise Q1 _J30 Number of solution of the equation tanx+secx = 2cosx lying in the interval [0,2] (a) 2 (b) 3 (c ) 0 (d) 1 Solution: cos x  0

  30. Class exercise Q1 _J30 Number of solution of the equation tanx+secx = 2cosx lying in the interval [0,2]

  31. Class exercise Q1 _J30 Number of solution of the equation tanx+secx = 2cosx lying in the interval [0,2]

  32. Class exercise Q2 _J30 Solution:

  33. Class exercise Q2 _J30 As we have squared both the sides, we should check for extraneous roots Similarly, for n=1, 3, 5 …the values of x do not satisfy the question. Hence, the solution is

  34. Class exercise Q3 _J32 For nz, the general solution of the equation Solution:

  35. Class exercise Q3 _J32 For nz, the general solution of the equation

  36. (b) x=65o , y=15o (a) x=15o , y=25o (d) x=45o , y=15o (c) x=45o , y=45o Class exercise Q4 _J33 the values of x and y lying between 0o and 90o are given by Solution:

  37. Class exercise Q4 _J33 the values of x and y lying between 0o and 90o are given by

  38. Class exercise Q5 _J33 Solution: L.H.S =0 if sinx-cosx=0, sinx-1=0 cosx-1=0  sinx=cosx, sinx=1, cosx=1 Which is not possible for any value of x.  No Solution

  39. Class exercise Q6 _J30 Solution:

  40. Class exercise Q6 _J30

  41. Class exercise Q7 _J32 Solution: the given equation of the form a cos + b sin = c for real solution

  42. Class exercise Q7 _J32

  43. Class exercise Q7 _J32 Taking positive sign

  44. Class exercise Q7 _J32 Taking Negative sign

  45. In such types of problems we divide both sides by cos2x which yield a quadratic equation in tanx. Class exercise Q8 _J30 Solution: In this equation if cosx = 0, the equation becomes 2sin2x=-2 or sin2x=-1 which is not possible hence on dividing the equation by cos2x we get

  46. Class exercise Q8 _J30 2tan2x-5tanx-8 = -2sec2x 2tan2x+2(1+tan2x)-5tanx-8 = 0 or 4tan2x-5tanx-6 = 0 or 4z2-5z-6 = 0 where z = tanx or 4z2-8z+3z-6 = 0  4z(z-2)+3(z-2)=0  z=2,-3/4

  47. Class exercise Q8 _J30

  48. Class exercise Q9 _J33 Determine for which value of ‘a’ the equation a2–2a+sec2(a+x)=0 has solution and find the solution The equation involves two unknown a and x so we must get two condition for determining unknowns since R.H.S is zero. So break the L.H.S of the equation as sum of two square.

  49. Class exercise Q9 _J33 Determine for which value of ‘a’ the equation a2–2a+sec2(a+x)=0 has solution and find the solution  a=1 and tan(a+x)=0 but a=1

  50. Class exercise Q10 _J30 Solve cot – tan = sec 

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