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# Trigonometric Equations

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#### Trigonometric Equations - Lesson Summary

Equations:
A mathematical statement that shows equality between two expressions is called an equation.

Trigonometric Equations:
An equation involving trigonometric functions of a variable is known as a trigonometric equation.

Example:

a cos Î¸ + b sin Î¸ = 0,

p tan2 Î¸ + q sec2 Î¸ + r = 0

sin x = 1/2

Solution of a trigonometric equation:
The value of the unknown angle that satisfies a given trigonometric equation.

Example:

sin x = 0 â‡’ x = 0, Ï€, 2Ï€,...â€¦

2. cos x = 0 â‡’ x = Ï€ 2 , 3 Ï€ 2 ,5 Ï€ 2 ...â€¦
The solution of a trigonometric equation, for which 0 â‰¤ x < 2Ï€, are called the principal solutions.

0 and Ï€ are the principal solutions of sin x = 0.

Ï€/2 and 3Ï€/2 are the principal solutions of cos x = 0.

Find the principal solutions of tan x = -âˆš3.

But tan Ï€ 3 = âˆš3.

Also, the tan function is negative in the second and the fourth quadrants.

tan(Ï€ - Ï€ 3 ) = -tan Ï€ 3 = -âˆš3 and tan(2Ï€ - Ï€ 3 ) = - tan  Ï€ 3 = -âˆš3

i.e. tan 2 Ï€ 3 = tan 5 Ï€ 3 = -âˆš3

Hence, 2 Ï€ 3 and 5 Ï€ 3 are the principal solutions

Ex: Find the principal solutions of sin x = 1/2.

But sin Ï€ 6 = ½.

Also, sin(Ï€ - Ï€ 6 ) = sin 5 Ï€ 6 = ½.

Hence, Ï€ 6 and 5 Ï€ 6 are the principal solutions.

Trigonometric functions are periodic functions. The functions sin, cos, cosec and sec repeat after an interval of 2Ï€, while the functions tan and cot repeat after an interval of Ï€.

sin q = 0 â‡’ q = nÏ€, n âˆˆ Z

When the solution set of a trigonometric equation is an expression involving an integer n, the solution is called the general solution of the trigonometric equation.

cos q = 0 â‡’ q = (2n+1) Ï€ 2 , n âˆˆ Z

Theorem

For any real numbers x and y, sin x = sin y implies x = nÏ€ + (-1)ny, where n âˆˆ Z .

Proof

For any real numbers x and y,

sin x = sin y.

â‡’sin x - sin y = 0

â‡’ 2cos ( x + y 2 ). sin ( x - y 2 ) = 0/2 = 0

â‡’ Either cos ( x + y 2 ) = 0 or sin ( x - y 2 ) = 0

sin q = 0 â‡’ q =nÏ€ and, cos q = 0 â‡’ q = (2n+1)Ï€/2,n âˆˆ Z

Therefore, ( x + y 2 ) = (2n+1)Ï€/2 or ( x - y 2 ) = np, where n âˆˆ Z

i.e. x= (2n + 1)Ï€ - y or x = 2nÏ€ + y, where n âˆˆ Z

(-1)2n+1 = -1 and (-1)2n = 1, where n âˆˆ Z

Hence, x = (2n + 1) p + (-1)2n+1y or x = 2nÏ€ + (-1)2ny, where n âˆˆ Z.

Combining these two results, we get

x = nÏ€ + (-1)ny, where n âˆˆ Z.

Theorem

For any real numbers x and y, cos x = cos y, implies x = 2nÏ€±y, where n âˆˆ Z

Proof

For any real numbers x and y,

cos x = cos y.

â‡’ cos x - cos y = 0.

â‡’ -2 sin ((x+y)/2). sin ((x-y)/2) = 0

â‡’ sin ((x+y)/2). sin ((x-y)/2) = 0/2 = 0

â‡’ sin ((x+y)/2) = 0 or sin ((x-y)/2) = 0

sin q = 0 â‡’ q = nÏ€, n âˆˆ Z

Therefore, ((x+y)/2) = np or ((x-y)/2) = n p, where n âˆˆ Z

i.e. x = 2nÏ€-y or x = 2nÏ€+y where n âˆˆ Z

Hence, x = 2nÏ€ ± y where n âˆˆ Z.

Theorem

If x and y are not odd multiples of Ï€/2, then tan x = tan y implies x = np+ y, where n âˆˆ Z.

Proof

Suppose x and y are not odd multiples of Ï€/2 and

tan x = tan y.

â‡’ tan x - tan y = 0

â‡’ sin x / cos x - sin y / cos y = 0

â‡’ (sin x cos y - cos x sin y)/ cos x cos y = 0

â‡’sin(x - y)/cos x cos y = 0

â‡’ sin(x- y) = 0

sin Î¸ = 0 â‡’ Î¸ = nÏ€, n âˆˆ Z

Therefore, x - y = nÏ€

i.e. x = nÏ€ + y, where n âˆˆ Z

Solution set of Trigonometric Functions

The solution set of sin Î¸ = k, k âˆˆ R and -1 â‰¤ k â‰¤ 1, is {nÏ€ + (-1)ny/n âˆˆ Z}, where y is the principal solution.
The solution set of cos Î¸ = k, k âˆˆ R and -1 â‰¤ k â‰¤ 1,  {2nÏ€ + y/n âˆˆ Z} where y is the principal solution.
The solution set of tan Î¸ = k, k âˆˆ R is {nÏ€ + y/n âˆˆ Z}, where y is the principal solution.