Mathematics

Trigonometric Equations – Session 2

Session Objectives

Removing Extraneous Roots Avoiding Root Loss Equation of the form of a cosx +b sinx = c Simultaneous equation

Extraneous solutions The solutions, which do not satisfy the trigonometric equation. Trigonometric Equation – Removing Extraneous Solutions Origin of Extraneous solutions ? Squaring during 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 _J30

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

Trigonometric Equation – Extra root because of squaring 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 _J30

Trigonometric Equation – Solutions, which makes the equation undefined 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 _J30

Trigonometric Equation – Avoiding root loss 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. _J31

Trigonometric Equation – Cancelling of terms from both sides 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 _J31

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

Trigonometric Equation – Changes in domain of  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 _J31

Trigonometric Equation – a sin  + b cos  = c Reformat the equation in the form of cos(  -  ) = k. 1.Divide both sides of the equation by 2.The equation now will be as _J32

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

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

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

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

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

Simultaneous Trigonometric Equations 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. _J33

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

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

Simultaneous Trigonometric Equations 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 _J33

Simultaneous Trigonometric Equations 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 _J33

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

Find the smallest positive values of θ and φ, if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3 Simultaneous Trigonometric Equations - Problem 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 _J33

Trigonometric Equation – Misc. Tips : 1.cosθ = k is simpler to solve compared to sinθ = k 2.Check for extraneous roots and root loss 3.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 x 2 ≥ 0 A.M. ≥ G.M.

Trigonometric Equation – Misc. Tips : 4.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

Trigonometric Equation – Misc. Tips : 5.Equation with terms of sin 2 θ, cos 2 θ and sinθ.cosθ Try dividing by cos 2 θ to get a quadratic equation in tan θ. 6.Solution to sin 2 θ = sin 2 , cos 2 θ = cos 2  and tan 2 θ = tan 2  is n   Solve the equation 2 sin 2 x – 5 sinx.cos x – 8 cos 2 x = -2

Class exercise Q1 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 _J30

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

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

Class exercise Q2 Solution: _J30

Class exercise Q2 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 _J30

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

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

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

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

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

Class exercise Q6 Solution: _J30

Class exercise Q6 _J30

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

Class exercise Q7 _J32

Class exercise Q7 Taking positive sign _J32

Class exercise Q7 Taking Negative sign _J32

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

Class exercise Q8 2tan 2 x-5tanx-8 = -2sec 2 x 2tan 2 x+2(1+tan 2 x)-5tanx-8 = 0 or 4tan 2 x-5tanx-6 = 0 or 4z 2 -5z-6 = 0 where z = tanx or 4z 2 -8z+3z-6 = 0  4z(z-2)+3(z-2)=0  z=2,-3/4 _J30

Class exercise Q8 _J30

Class exercise Q9 Determine for which value of ‘a’ the equation a 2 – 2a+sec 2 (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. _J33

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

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

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