Possible?
Is there a value of θ satisfying
sin θ = a b + b c + c a a 2 + b 2 + c 2
for positive integers a , b and c ?
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2 solutions
I don't think the answer is completely correct. Because if you let a = 1 , b = 1 , c = 1, then there is a θ such that s i n θ = 3 / 3 = 1 , and that θ is 2 π . To discuss the general case, if we define u = ( a , b , c ) , and w = ( b , c , a ) then from Schwarz Inequality, u ⋅ w ≤ ∣ u ∣ ∣ w ∣ , with equality only if u is a positive multiple of w .
This means that a b + b c + c a a 2 + b 2 + c 2 ≥ 1 . Equality will occur if a b = b c = c a . If we call this ratio x , then a b ⋅ b c = x 2 = a c = x 1 . Thus x must be 1. And this means, the only way that the given expression is a sine of an angle is if and only if a = b = c . And this case that angle is 2 π .
Sin(90)=1......
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There would be no natural solution even for sin 9 0 = 1
Take for example, a = b = 0 , c = 1 \
a b + b c + c a a 2 + b 2 + c 2 = 0 1 = undefined.
Because as nobody ever said: The most valid proof for something is to show it won't work with 2 random numbers.
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The thing is there would be no natural solution. Look at the condition in the problem.
For example a = b = c = 1 ⟹ s i n θ = 3 3 = 1 , which is true for θ = 2 π .
Take for example, sin ( − 3 0 ) = − sin ( 3 0 ) = − 2 1 Since, sin θ = a b + b c + c a a 2 + b 2 + c 2 , ( a , b , c ) ∈ N There is no possibility that sum of natural numbers is a negative number. Take another for example sin ( 6 0 ) = 2 3 ⟹ a b + b c + c a a 2 + b 2 + c 2 = 2 3 LHS is rational where as RHS is irrational which can never happen Even for sin ( 3 0 ) = 2 1 , there will be no natural solutions for a, b, c satisfying the condition.