Trigonometric Equations Made Easy

Trigonometric Equations are those equations that involve the trigonometric functions of a variable. In this article, you will learn how to find solutions for the given equations. These equations consist of one or more unknown angles. Let us consider an example, cos m - sin2 m = 0, which is a trigonometric equation that does not satisfy all the values of m. Hence for such equations, you will have to either find the value of m or you will have to find the solution.

Previously, you have learned that the values sin x and cos x are repeated after an interval of 2π and tan x and these values repeat itself after the interval of π. Principal Solutions are those solutions that lie in the interval of [ 0, 2π ] of such given trigonometry equations. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation and it is expressed in a generalized form in terms of ‘n’. The general representation of these equations comprises the formula of the trigonometric equation;

E1 ( sin m, cos m, tan m ) = E2 ( sin m, cos m, tan m )

Here,

E1 - is a rational function.

E2 - is a rational function.

Since sine, cosine, and tangent are the three major trigonometric functions, the solutions will be derived for these equations will comprise of only these three ratios. Although the solutions for the other three ratios such as secant, cosec, and cotangent can be obtained with the help of those solutions.

Let us consider a basic equation to understand this concept. Equation: sin m = 0 and 0, π, and 2π  will be the principal solutions for this case since these values satisfy any given equation which lies in between the [ 0, 2π ]. If the values of the sin m = 0, then the value of m = 0, π, 2π, - π, -2π, -6π, etc of the given equation. Therefore, the general solution for sin m = 0 will be m = nπ, where n belongs to Integers.

Solutions for Trigonometric Equations

Proofs for Solutions of Trigonometric Equations

Theorem 1:  For any real number j and k, sin j = sin k implies that j = nπ + ( - 1 ) . n . k, where n Є Z

Proof: consider the equation, sin j = sin k. Now, let us try and find the general solution of this equation.

sin j = sin k

⇒ sin j – sin k = 0

⇒ 2 cos ( j + k ) / 2 sin ( j – k ) / 2 = 0

⇒ cos ( j + k ) / 2 = 0 or sin ( j – k ) / 2 = 0

Upon taking the common solution from both the conditions, we get:

j = n π + (-1)n k, where n ∈ Z

Theorem 2: For any real numbers j and k, cos j = cos k, implies j = 2nπ ± k, where n Є Z.

Proof: Similarly, the general solution of cos j = cos k will be:

cos j – cos k = 0

2 sin ( j + k ) / 2 sin ( k – j ) / 2 = 0

sin ( j + k ) / 2 = 0 or sin ( j – k ) / 2 = 0

( j + k ) / 2 = ( n * π ) or ( j – k ) / 2 = ( n * π )

On taking the common solution from both the conditions, we get:

j = 2 * n * π ± k, where n ∈ Z

Theorem 3: Prove that if  and k are not odd multiple of π / 2, then tan j = tan y implies that j = nπ + k, wheren Є Z.

Solution:

Similarly to find the solution of equations involving tan x or other functions, we can use the conversion of trigonometric equations.

In other words, if tan x = tan y then;

\[\frac{{sin j }}{cos k}\] = \[\frac{{cos j }}{cos k}\]

sin j * cos k = sin k * cos j

sin j cos k – sin k cos j = 0

sin  ( j – k ) = 0     [By trigonometric identity]

Hence, j – k = ( n * π ) or j = ( n * π + k ), where n ∈ Z.