- #71
harry654
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soo I see that it apply but I don't know mathematically explain :(
harry654 said:sin²α + sin²β = 1 when α+β=90 so inequality doesn't apply I think
harry654 said:apply sin²α + sin²β = 1 when α+β = 90° and from that α+β > 90 so sin²α + sin²β >1
so when I prove sin²α + sin²β >1, I proved that (a²+b²)cos(α-β) ≤ 2abcos(α-β)?
harry654 said:I have (a²+b²)cos(α-β) ≤ (a²+b²-h²)/cos(α-β)
and from it I get
(a²+b²)cos²(α-β) ≤ a²+b²-h²
but how can I tidy up later?
harry654 said:Assume that apply sin(α-β) is not 0 then question is : Is cos(α-β) positive?
harry654 said:Hi tiny-tim!
If α=135 and β=30 then sin²α + sin²β < 1 and apply α+β>90 so what isn't correct?
tiny-tim said:so (tidying-up time!) actually we need to prove that, if α + β > 90°, then:
sin²α + sin²β > 1 if cos(α-β) > 0, ie if |α-β| < 90°, and
sin²α + sin²β < 1 if cos(α-β) < 0, ie if |α-β| > 90°;
(alternatively, we could simply point out that we needn't bother with the cos(α-β) < 0 case, since the originally given inequality, (a² + b²)cos(α-β) ≤ 2ab, is obviously true in that case, since the LHS is negative and the RHS is positive!)