We use some results and general solutions of the basic trigonometric equations to solve other trigonometric equations. These results are as follows:
- For any real numbers x and y, sin x = sin y implies x = nπ + (-1)ny, where n ∈ Z.
- For any real numbers x and y, cos x = cos y implies x = 2nπ ± y, where n ∈ Z.
- If x and y are not odd multiples of π/2, then tan x = tan y implies x = nπ + y, where n ∈ Z.
Now, we can prove these results using trigonometric formulas.
Prove that for any real numbers x and y, sin x = sin y implies x = nπ + (-1)ny, where n ∈ Z
Proof: If sin x = sin y, then sin x – sin y = 0
⇒ 2 cos (x + y)/2 sin (x − y)/2 = 0 — [Using formula Sin A – Sin B = 2 cos ½ (A + B) sin ½ (A – B)]
⇒ cos (x + y)/2 = 0 or sin (x − y)/2 = 0
⇒ (x + y)/2 = (2n + 1)π /2 or (x − y)/2 = nπ, where n ∈ Z —- [Because sin A = 0 implies A = nπ and cos A = 0 implies A = (2n + 1)π/2, where n ∈ Z]
i.e. x = (2n + 1) π – y or x = 2nπ + y, where n ∈ Z.
Hence x = (2n + 1)π + (–1)2n + 1y or x = 2nπ + (–1)2n y, where n ∈ Z.
Combining these two results, we get x = nπ + (–1)ny, where n ∈ Z.
Prove that for any real numbers x and y, cos x = cos y implies x = 2nπ ± y, where n ∈ Z.
Proof: If cos x = cos y, then cos x – cos y = 0
⇒ -2 sin (x + y)/2 sin (x − y)/2 = 0 — [Using formula Cos A – Cos B = – 2 sin ½ (A + B) sin ½ (A – B)]
⇒ sin (x + y)/2 = 0 or sin (x − y)/2 = 0
⇒ (x + y)/2 = nπ or (x − y)/2 = nπ, where n ∈ Z —- [Because sin A = 0 implies A = nπ, where n ∈ Z]
i.e. x = 2nπ – y or x = 2nπ + y, where n ∈ Z.
Hence x = 2nπ ± y, where n ∈ Z.
Prove that if x and y are not odd multiples of π/2, then tan x = tan y implies x = nπ + y, where n ∈ Z.
Proof: If tan x = tan y, then tan x – tan y = 0
⇒ sin x / cos x – sin y / cos y = 0
⇒ (sin x cos y – cos x sin y) / (cos x cos y) = 0
⇒ sin (x – y) / (cos x cos y) = 0 —-
[Using trigonometric formula sin (A – B) = sinA cosB – sinB cosA]
⇒ sin (x – y) = 0
⇒ x – y = nπ, where n ∈ Z — [Because sin A = 0 implies A = nπ, where n ∈ Z]
⇒ x = nπ + y, where n ∈ Z
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