Trigonometric Functions | Class 11 Maths

Equations involving one or more trigonometric functions of a variable are called trigonometric equations.  It is expressed as ratios of sine(sin), cosine(cos), tangent(tan), cotangent(cot), secant(sec), cosecant(cosec) angles. For instance, sin3x = 1/2, tan2x = √3, 2sinx + 1 = 0 are all examples of trigonometric equations.

The Solution of a Trigonometric Equation

The values of an unknown angle that satisfy any given trigonometric equation are known as the solution of a trigonometric equation and the process of finding the solution set is called solving the trigonometric equation. Let’s consider a simple equation, say tan x = 1. Here, as you can see x = pi/4, 5pi/4, 9pi/4, etc. are some of the solutions which can satisfy the above equation.

Note: It is not necessary that every trigonometric equation has a solution. For example, tan x = 5 as no solution at all!

General Solution

The set of all solutions is called the solution set or general solution of a trigonometric equation. In other words, a general solution is an expression involving integer ‘n’ which gives all the solutions. Consider the equation, 2cos x + 1 = 0 or cos x = -1/2. This equation is clearly satisfied with x = 2pi/3, 4π/3, etc. Here, x = 2pi/3, 2pi ± 2pi/3, 4pi ± 2pi/3…. are all solutions of the above trigonometric equation, 2cosx + 1 = 0. Now, as you can see these solutions can be put together in the compact form as 2n.pi ± 2pi/3, where ‘n’ can take the value of integers.

Note: All trigonometric functions are periodic in nature. So, if a trigonometric equation has a solution, it will have infinitely many solutions.

General solution of some standard trigonometric equations:

Equation	General Solution
sin x = 0	x = n.pi
cos x = 0	x = n.pi + pi/2
tan x = 0	x = n.pi
sin x = sin y	x = n.pi + (-1)n, where y ∈ [-pi/2, pi/2]
cos x = cos y	x = 2n.pi ± y, where y ∈ (0, pi]
tan x = tan y	x = n.pi + y, where y ∈ (0, pi]
sin 2x = sin 2y	x = n.pi ± y
cos 2x = cos 2y	x = n.pi ± y
tan 2x = tan 2y	x = n.pi ± y

Examples: Find the general solution of the following equations?

(i) sin mx + sin nx = 0

Solution:

We have, 

sin mx + sin nx = 0

⇒ 2 sin((m + n) / 2)x cos ((m – n) / 2)x = 0

⇒ sin((m + n) / 2)x or cos ((m – n) / 2)x = 0

Now, sin((m + n) / 2)x = 0 

⇒ ((m + n) / 2)x = k.pi, k ∈ Z

⇒ x = 2k.pi / (m + n), k ∈ Z

And, cos ((m – n) / 2)x = 0

⇒ ((m – n) / 2)x = (2r + 1) pi/2, r ∈ Z

⇒ x = (2r + 1)pi / (m – n), r ∈ Z

Hence, the general solution of the given equation is

x = 2k.pi / (m + n) or, x = (2r + 1)pi / (m – n) where k, r ∈ Z

(ii) sin x + sin 3x + sin 5x = 0

Solution:

We have,

sin x + sin 3x + sin 5x = 0

⇒ (sin x + sin 5x) + sin 3x = 0

⇒ 2 sin3x cos2x + sin 3x = 0

⇒ sin 3x (2cos 2x + 1) = 0

⇒ sin 3x = 0 or 2cos 2x + 1 = 0

⇒ sin 3x = 0 or cos 2x = -1/2

Now, sin 3x = 0 ⇒ 3x = n.pi, n ∈ Z ⇒ x = n . pi/3, n ∈ Z.

And, cos 2x = -1/2

⇒ cos 2x = cos 2pi/3

⇒ 2x = 2m.pi ± 2pi/3, m ∈ Z

⇒ x = m.pi ± pi/3, m ∈ Z.

Hence, the general solution of the given equation is

x = n.pi/3 or, x = m.pi ± pi/3 where n, m ∈ Z.

(iii) 2 cos²x + 3 sin x = 0

Solution:

We have, 2cos²x + 3sin x = 0

⇒ 2 (1 – sin²x) + 3sin x = 0

⇒ 2sin²x – 3sin x – 2 = 0

⇒ 2sin²x – 4sin x + sin x – 2 = 0

⇒ 2sin x (sin x – 2) + 1 (sin x – 2) = 0

⇒ (sin x – 2) (2sinx + 1) = 0

⇒ 2sin x + 1 = 0                                       [ Note: sin x ≠ 2 ]

⇒ sin x = -1/2

⇒ sin x = sin (-pi/6)

⇒ x = n.pi + (-1)ⁿ(-pi/6), n ∈ Z

Hence, x = n.pi + (-1)ⁿ⁺¹(pi/6), n ∈ Z.

Principal Solution

The solution in which the absolute value of the angle is the least is called the principal solution. In other words, the solutions of a trigonometric equation for which ‘x’ lies in the interval 0 to 2pi, i.e., 0 ≤ x ≤ 2pi, are called the principal solutions.  

Note: If a given trigonometric equation has a solution, then it will always have two principal solutions. 

Let us tell you why it is that so. Consider a simple equation, say sin x = 1/2. In a quadrant system, whatever trigonometric equation we are given, the absolute value on the right-hand side will either be positive or negative, or zero. Now, as we know that sine function can be positive in two quadrants (I and II) and negative in two quadrants (III and IV). And this is why when we talk of principal solutions, we will always have two solutions.  

We know that, sin pi/6 = ½

Also, sin 5pi/6 = sin (pi – pi/6)

Now, as sin (pi – x) = sin x  

Hence, sin 5pi/6 = sin pi/6 = ½  

So, here x = pi/3 and 5pi/6 are the two principal solutions of sin x = ½.

Note:

Conversion techniques for finding principal solution.  

In quadrant I, θ

In quadrant II, pi – θ  

In quadrant III, pi + θ

In quadrant IV, 2pi – θ

Example: Find the principal solution of the following equations?

(i) sin θ = √3/2

Solution:

We have sin θ = √3/2

               sin θ = sin pi/3 or sin (pi – pi/3)

               θ = pi/3 or pi – pi/3

Hence, θ = pi/3 or 2pi/3

(ii) cos 3θ = -1/2

Solution:

We have cos 3θ = -½

               cos 3θ = cos (pi – pi/3) or cos (pi + pi/3)

               3θ = 2pi/3 or 4pi/3

Hence, θ = 2pi/9 or 4pi/9

(iii) tan 5θ = 1/√3

Solution:

We have tan 5θ = 1/√3

               tan 5θ = tan pi/6 or tan (pi + pi/6)

               5θ = pi/6 or 7pi/6

Hence, θ = pi/30 or 7pi/30

The General Solution of sin θ = sin ∝

We learned that the general solution is solutions expressing all the values which would satisfy the given trigonometric equation and can be expressed in a generalized form in terms of ‘n’. Let us discuss general solutions of equations of the form sin θ = sin α.  

The general solution of sin θ = sin ∝ is given by: θ = n.pi + (-1)ⁿ, n ∈  Z  

Now let’s prove it.

We have,     sin θ = sin ∝

               ⇒ sin θ − sin ∝ = 0

               ⇒ 2 sin((θ − ∝) / 2) cos((θ + ∝) / 2)) = 0

               ⇒ sin((θ − ∝) / 2) cos((θ + ∝) / 2)) = 0

               ⇒ (θ − ∝) / 2 = m.pi and (θ + ∝) / 2 = (2m + 1) pi/2  

               ⇒ θ = 2m.pi + ∝ and θ = 2m.pi + pi – ∝

So, the general solution of sin θ = sin ∝ is given by: θ = n.pi + (-1)ⁿ, n ∈ Z.

Let’s go through an example here to get a better understanding of it.

Example: Solve the equation: sin θ = -√3/2.

Solution:

We have, sin θ = -√3/2

                sin θ = sin (-pi/3)  

                     θ = n.pi + (-1)ⁿ(-pi/3), n ∈ Z

         Hence, θ = n.pi + (-1)ⁿ⁺¹(pi/3), n ∈ Z.