Prove that sin &#x2061;<!-- ⁡ --> x + sin &#x2061;<!-- ⁡ --> y = 1 doe

oleifere45

oleifere45

Answered question

2022-06-24

Prove that sin x + sin y = 1 does not have integer solutions
Suppose x and y are angles measured in radians. Then how to show that the equation
sin x + sin y = 1
does not have a solution ( x , y ) N × N ?
This question is prompted by curiosity. I don't have any ideas how it can be approached.

Answer & Explanation

Angelo Murray

Angelo Murray

Beginner2022-06-25Added 23 answers

No, and there is not even a solution for ( x , y ) Q × Q .
We can quickly exclude x=y, which would require that sin x = 1 2 , but that is only true for x = n π 6 for certain nonzero integers n, and none of these produce a rational. Similarly we can easily exclude x=0, y=0, or x=−y.
Now, using Euler's formula, rewrite the equation to
(*) e i x + e i y e i x e i y = 2 i e 0
and apply the Lindemann–Weierstrass theorem which in one formulation says that the exponentials of distinct algebraic numbers are linearly independent over the algebraic numbers. But { ± i x , ± i y , 0 } are all algebraic and (by our assumptions so far) different, so (*) would be one of the linear relations that can't exist.
This argument generalizes to show that the only algebraic number that can be written as a rational combination of sines of algebraic (radian) angles is 0.

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