When does equality occur in the inequality (a^2+b^2)cos(α-β)<=2ab?

[tiny-tim]

time to tidy-up …

hi harry654!

ok, when we get stuck, it's good idea to go back and check whether we missed anything on the way

and so, looking back to page 3 (!), I've noticed that we got to …

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?

and going from the first inequality to the second, we multiplied by cos(α-β), which can be negative,

and of course if it is negative, then multiplying by it turns the ≤ into a ≥

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! )​

 

[harry654]

Yes THANK YOU tiny-tim! I understand this and I thank you for your patience and assistance.

 

[harry654]

Anyway I have question. Does it apply when sin(α-β) isn't 0 so cos(α-β) is positive or no?

 

[tiny-tim]

sorry, i don't understand

 

[harry654]

Assume that apply sin(α-β) is not 0 then question is : Is cos(α-β) positive?

 

[tiny-tim]

Assume that apply sin(α-β) is not 0 then question is : Is cos(α-β) positive?

sin(α-β) wil be zero only if α = β …

if α ≠ β then cos(α-β) can be either positive or negative

for example if α = 150° and β = 20° then cos(α-β) = cos130° is negative

why are you asking? ​

 

[harry654]

because I thought that I can say cos(α-β) is positive because sin(α-β) is not 0 but I found out that I can not :D OK thank you again:)

 

[harry654]

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]

hi harry654!

Hi tiny-tim!
If α=135 and β=30 then sin²α + sin²β < 1 and apply α+β>90 so what isn't correct?

nothing, that's fine because cos(α-β) = cos105° < 0, so it agrees with …

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! )​