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

[harry654]

soo I see that it apply but I don't know mathematically explain :(

 

[tiny-tim]

ok, then explain it in ordinary English first …

what makes you think that it applies?

 

[harry654]

I see from definition, better said I believe that apply :(

 

[harry654]

or c²≤ a²+b²
c²/c² ≤ (a²+b²)/c²
sin²α + sin²β ≥ 1

 

[harry654]

But it doesn't apply! when I have (a²+b²)cos(α-β) ≤ 2ab when a=b and α=β then occurs equality.
But when I have sin²α + sin²β ≥ 1 ,α=β equality doesn't occur ! where is mistake ?

 

[tiny-tim]

i don't understand … what is c ?

EDIT: oh, i didn't see your last post

why are you going back?

you have to prove (sin²α + sin²β) ≥ 1 using only α + β > 90°

 

[harry654]

I don't know how :(

 

[harry654]

oh, I have feeling that I never finish this prove :(
sin²α + sin²β = 1 when α+β=90 so inequality doesn't apply I think

 

[tiny-tim]

sin²α + sin²β = 1 when α+β=90 so inequality doesn't apply I think

yes, but the question specifies an acute angle (BCA), so α+β > 90°

ok, as you say, sin²α + sin²β = 1 when α+β = 90°,

so how can you show that sin²α + sin²β > 1 when α+β > 90 ?

 

[harry654]

certainly β or α >45° so sin²α or sin²β > 0,5 so sin²α+sin²β >1 but how explain it mathematically

 

[tiny-tim]

no that argument doesn't work unless both β and α are > 45°, does it?

we're still looking for a proof of sin²α + sin²β > 1 when α+β > 90°,

using sin²α + sin²β = 1 when α+β = 90°

 

[harry654]

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]

Could someone help me? I am desperate:(

 

[tiny-tim]

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)?

(been out all day )

yes, if you prove that sin²α + sin²β > 1,

then the previous arguments show that (a²+b²)cos(α-β) ≤ 2ab …

but first you have to prove that sin²α + sin²β > 1

 

[harry654]

how?
into sin²(α-β)b²/sin²β ≤ (a²+b²)sin²(α-β)
and so b²/sin²β ≤ (a²+b²)
thats isn't true because when α=β we divide 0

 

[tiny-tim]

yes, we can't divide both sides by sin²(α-β) when sin²(α-β) = 0

we have to deal with the case of α-β = h = 0 separately

(this is one of the things i was referring to when i mentioned tidying up earlier )

 

[harry654]

How can I prove sin²α + sin²β > 1 when α+β > 90°?

 

[tiny-tim]

as you said before, sin²α + sin²β = 1 when α+β=90 …

using that, it's actually very easy to prove it …

just draw a few triangles, some with = 90°, and some with > 90°, and you'll see what i mean

(btw, i'll be out soon, for the rest of the evening)

 

[harry654]

Yes I know what do you mean, but when I use picture so it isn't correct mathematical proof so I don't know ...

 

[tiny-tim]

try drawing a three-dimensional graph …

put α and β along the usual x and y directions, and sin²α + sin²β along the z direction …

that will be a surface …

do it for a "box" with 0 < α < 180° +and 0 < β < 180° …

what does it look like?

draw the line z = 1 on it

 

[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! )​