IIT JEE Blogs » Trigonometric Equations

Trigonometric Equations and Inequations

Prathap — Wed, 18 Nov 2009 03:30:45 +0000

As the name suggests an equation (or) Inequation In valuing one (or) more trigonometric ratios of unknown angles is called a trigonometric equation (or) inequation.

In this lesson we will learn how to affirm the general solution of trigonometric equations and inequations in valuing one (or) more of the six trigonometric ratios. We shall highlight the areas where a student must be careful while solving such equations and inequations we will also discuss the methods to solve equations / Inequations invaluing both trigonometric and non trigonometric functions.

(i) Basic Trigonometric equations:

Consider the following

cos q – sin q = 1 (1)

sin² q + cos² q = 1 (3)

Equation (1) is satisfied if we put q = 0, 2Π, 4Π …etc.

In it. Equation (2) is satisfied if we put q = 7Π/12 , 11Π/12, -Π/12

etc in it. But equation (3) is satisfied for any value of q. Equation (1) and (2) are called trigonometric equations while (3) is a trigonometric Identify. A trigonometric equation has three kinds of solutions.

(1) Principal solution: Numerically smallest value of the equation unknown angle satisfying the given equation

(2) Particular solution : Any value of angle satisfying the given equation.

(3) General solution: Collection of all particular solution. For example the equation of (q – Π/6) = √3 is satisfied if q = -Π/6 (or) 7Π/6 (or) -5Π/6, etc

→q = Π/3 , 4Π/3 , -2Π/3 etc. Out of these the numerically smallest is q = Out of these the numerically smallest is q =/3.

This is the principal solution. It is a particular solution also.

Also the above values of q taken individually are particular solutions. For general solution we observe that since cotangent function is periodic with period p and it takes the value only once in the Interval
[0, Π], (q , -Π/6) can be equal to any particular solution plus a multiple of Π. If we take the particular solution as the principal solution then q – Π/6 = nΠ + Π/6 , n ε l

q = nΠ+Π/3 , n ε l is the general solution. If in any equation we get two values of the unknown angle which are numerically equal but opposite in sign, the principal solution is assumed as the positive angle (e.g.) principal solution of sec q = 2 is q = Π/3, although q = –Π/3 also satisfies it.

Method to find principal value (numerically least angle)

(i) First draw a trigonometric – circle and mark the quadrant in which the angle may be.

(ii) Select anticlockwise direction for 1^st and 2^nd quadrants and select clockwise for 3^rd and 4^th quadrants

(iii) Find the angle in the first rotation.

(iv) Select the numerically least angle from these two values the angle thus found will be the principal value.

(v) In case two angles (one with positive sign and the other with negative sign) qualify for the numerically least angle then we select the angle with positive sign as the principal value

Some simple equations

Prathap — Sat, 31 Oct 2009 03:30:51 +0000

GENERAL SOLUTION OF SOME SIMPLE EQUATION

sin Θ = 0 → Θ = nΠ

cos Θ = 0 → Θ = (2n + 1) Π/2

tan Θ = 0 → Θ = nΠ

cot Θ = 0 → Θ = (2n + 1)Π/2

sin Θ = 1 → Θ = (4n + 1) Π/2

sin Θ = -1 → Θ = (4n + 3) Π/2

cos Θ = 1 → Θ = 2n Π

cos Θ = -1 → Θ = (2n + 1) Π

tan Θ = not defined → Θ = (2n + 1) Π/2

cot Θ = Not defined → Θ = nΠ

cosec Θ = Not defined → Θ = nΠ

sec Θ = not defined → Θ= (2n + 1) Π/2

ü There are some cautions to be taken while solving some Trigonometric Equations. They are listed down here.

Þ Check the validity of the given equation

e.g., 2sin Θ – cos Θ = 4 can never be true for any Θ as the value (2 sin Θ – cos Θ) can never exceed √(2²+(-1)² = √5. So there is no solution to this equation.

Þ Equation involving sec Θ and / or tan Θ can never have a solution of the form Θ = (2n + 1) Π/2. similarly, equation involving cosec Θ and / or cot Θ can never have solution of the form Θ = nΠ. The corresponding function are undefined at these values of Θ.

Þ Avoid squaring the equations as far as possible because it leads to extraneous solution. If it has to be squared, check whether the solution (s) arrived at satisfy the original unsquared equation or not. e.g., given that x = 4 → x² = 16 → x = ± 4. But originally x = 4 only.

Þ Do not cancel common factor involving the unknown angle on L.H.S. and R.H.S because it may delete some solution. e.g. In the equation sin Θ (2cos Θ – 1) = sin Θ cos²Θ if we cancel sin Θ on both sides we get cos²Θ – 2cos Θ + 1 = 0 → (cos Θ – 1)² = 0 → cos Θ = 1 → Θ = 2n Π.

Þ But Θ = nΠ also satisfies the equation because it makes sin Θ = 0. So, the complete solution is Θ = nΠ, n Î Z.

Þ Denominator terms of the equation if present should never become zero at any stage while solving for any value of Θ contained in the answer.

Þ Something the equation has some limitation also. e.g., cot² Θ cosec² Θ = 1 can be true only if cot² Θ = 0 and cosec² Θ = 1 simultaneously as cosec² Θ ≥ 1. Hence the solution is Θ= (2n + 1) Π/2.

Illustrations

1. Obtain the general solution of secx + tanx =√3

Sol. Since secx and tanx. Both are undefined for x = (2n + 1)Π/2, the final solution should not include any odd multiple of Π/2.

Now,√3 cosx – sin x = 1 => cos(Θ + Π/6) = ½ = cosΠ/3

Þ Θ-+ Π/6 = 2nΠ + Π/3.

With ‘+’ sign, it given Θ = 2nΠ + Π/6

With ‘-‘ sign, it given Θ = 2nΠ – Π/2 = (4n – 1)Π/2

Since (4n -1) is always odd therefore we should delete this solution.

Thus, Θ= (12n + 1)Π/6, n Î Z.

Introduction to Trigonometric Equations

Prathap — Sat, 24 Oct 2009 04:30:45 +0000

Trigonometric equations are natural sequel to the trigonometric ratios and identities which constitute basis of many problems in Mathematics.

The trigonometric equations have a large number of concepts associated with relevant applications.

OBJECTIVE

This chapter focuses on the solutions of different trigonometric equations. After studying this chapter we will learn how to find different possible solutions of a given trigonometric equation or how to find the general solution of a trigonometric equation. We will also be able to distinguish a trigonometric identity from a trigonometric equation.

PRE-REQUISITE

=> sin (A + B) = sin A cos B + cos A sin B

=> sin (A – B) = sin A cos B – cos A sin B

=> cos (A + B) = cos A cos B – sin A sin B

=> cos (A – B) = cos A cos B + sin A sin B

=> tan (A + B) = (tanA+tanB)/ 1-tanA tanB

where A ≠ nΠ +Π/2 , B ≠ nΠ +Π/2

=> tan (A – B) =(tanA-tanB)/1+ tanA tanB and A ± B ≠ mΠ +Π/2

=> cot (A + B) =(cotA cot B – 1)/(cotA + cot B),

where A ≠ nΠ, B ≠n Π

=> cot (A – B) = and A ± B≠ np

=> sin (A + B) sin (A – B) = sin² A – sin² B

= cos² B – cos² A

=> cos (A + B) cos (A – B) = cos² A – sin² B

= cos² B – sin² A

=> sin2 Θ = 2 sinΘ cos Θ =2tanΘ/(1+tan 2Θ)

=> cos 2 Θ =cos²Θ – sin² Θ =2 cos²Θ -1 = 1 – 2sin² Θ =(1-tan²Θ)/(1+tan²Θ)

=> 1 + cos 2 Θ = 2 cos²Θ, 1 – cos 2 Θ = 2 sin² Θ

or (1+cosΘ)/2= cos²Θ, (1-cosΘ)/2 = sin² Θ

=> tan2 Θ = 2tanΘ/(1-tan²Θ), where Θ≠ (2n + 1)Π/4

=> (1-cosΘ)/sinΘ = tan Θ/2, where Θ ≠ (2n + 1)Π

=> (1+cosΘ)/ sinΘ= cot Θ/2, where Θ≠ 2n Π

=> (1-cosΘ)/(1+cosΘ)= tan², where Θ ≠ (2n + 1)Π

=> (1+cosΘ)/(1-cosΘ) = cot^{2 Θ/2}, where Θ ≠ 2n Π

=> sin 3Θ= 3sinΘ -sin³Θ

=> cos3 Θ = 4cos³ Θ – 3cos Θ

=> cos A cos2 A cos2² A … cos2^n-1 A =sin2″A/2″sinA

CORE CONCEPTS

An equation involving one or more trigonometric ratios of unknown angle is called a trigonometric equation. A trigonometric equation can be written as

Q₁ (sinΘ, cosΘ, tanΘ, cotΘ, secΘ, cosecΘ)

= Q₂ (sinΘ, cosΘ, tanΘ, cotΘ, secΘ, cosecΘ)

where Q₁ and Q₂ are rational functions.

cos² x – 4 sin x = 1.

All possible values of unknown which satisfy the given equation are called solution of the given equation.

For complete solution “all possible values” satisfying the equation must be obtained.

This is trigonometric equation as it is not satisfied for all values of x e.g., does not satisfy the given equation.

Identify whether the following are trigonometric equations or trigonometric identities.

1. sin³ A = 3sin A – 4 sin³ A

Sol. Trigonometric Identity

2. cos⁷ x + sin⁴ x = 1

Sol. Trigonometric Equation