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

[harry654]

OK
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?

 

[tiny-tim]

oh, forget about later!

ok, now simplify (a²+b²)cos²(α-β) ≤ a²+b²-h²

 

[harry654]

OK
(a²+b²)cos²(α-β) ≤ a²+b²-h²
(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β)
(a²+b²)cos(α-β) ≤ 2ab

 

[tiny-tim]

OK
(a²+b²)cos²(α-β) ≤ a²+b²-h²
(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β)
(a²+b²)cos(α-β) ≤ 2ab

uhh? … now you're going backwards

simplify (a²+b²)cos²(α-β) ≤ a²+b²-h²

 

[harry654]

(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β) OK?

 

[tiny-tim]

(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β) OK?

you're still going backwards, that's exactly the same as (a²+b²)cos(α-β) ≤ 2ab

 

[harry654]

and how can I simplify (a²+b²)cos²(α-β) ≤ a²+b²-h² ? I can only as (a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β) :(

 

[tiny-tim]

well, for a start, you could try putting all similar terms on the same side

 

[harry654]

similar terms on the same side? uff

 

[harry654]

I don't know how carry on again :(

 

[harry654]

could you write me an inequality at which should I arrive?

 

[harry654]

I tried this
(a²+b²)cos²(α-β) ≤ a²+b²-h²
(a²+b²)cos²(α-β)-(a²+b²)≤ -h²
but how carry on... I am desperate:(

 

[tiny-tim]

good morning!

(a²+b²)cos²(α-β)-(a²+b²)≤ -h²

yes, that's a standard way of simplifying an equation …

putting all the similar things on one side!

ok, now simplify (a²+b²)cos²(α-β)-(a²+b²) ​

 

[harry654]

good morning
(a²+b²)cos²(α-β)-(a²+b²)=
=(a²+b²)(cos²(α-β)-1)=
=(a²+b²)(cos(α-β)-1)(cos(α-β)+1)
OK?

 

[tiny-tim]

good morning
(a²+b²)cos²(α-β)-(a²+b²)=
=(a²+b²)(cos²(α-β)-1)=
=(a²+b²)(cos(α-β)-1)(cos(α-β)+1)
OK?

oh good grief!

no wonder you've been having difficulty with this question

take the morning off and become familiar with using your https://www.physicsforums.com/library.php?do=view_item&itemid=18" …

in particular cos² + sin² = 1 ​

 

[harry654]

should I substitute 1=cos²(α-β)+sin²(α-β)

 

[tiny-tim]

yes, of course

get on with it!​

 

[harry654]

so
I get (a²+b²)(cos(α-β)-cos²(α-β)-sin²(α-β))(cos(α-β)+cos²(α-β)+sin²(α-β))
And now?

 

[tiny-tim]

how is that simpler??

 

[harry654]

(a²+b²)2cos(α-β)
and then?

 

[tiny-tim]

(a²+b²)2cos(α-β)

(i'm fascinated to know where you got that from )

harry, I'm more or less saying things at random while i wait for you to come up with the next step

i get the impression you're saying things at random too

i've given you several big hints, but I'm not going to give you the actual answer

even the next step is not the end of the problem

have you been having similar difficulty with other problems on this course?

if so, this obviously isn't "your thing", and you should seriously consider changing course

 

[harry654]

(i'm fascinated to know where you got that from)
I am sorry I made a mistake...
harry, I'm more or less saying things at random while i wait for you to come up with the next step

i get the impression you're saying things at random too

i've given you several big hints, but I'm not going to give you the actual answer

even the next step is not the end of the problem

have you been having similar difficulty with other problems on this course?
Yes, but I am trying to learn this kind of problems and I need help. I'm sorry to bother you.
(a²+b²)(cos(α-β)-cos²(α-β)-sin²(α-β))(cos(α-β)+cos²(α-β)+sin²(α-β))
(a²+b²)(-sin²(α-β)) is that right?

 

[tiny-tim]

(a²+b²)(-sin²(α-β)) is that right?

at last!

ok, let's recap where we've got to …

we have a triangle BCD, its sides have lengths a b and h, and their opposite angles are α 180°-β and α-β; and we know that α+β > 90°

and we now have (or rather, we need to prove) the formula (a²+b²)sin²(α-β) ≥ h²

we don't particularly want h (it's not in the final answer), so the next step will be to eliminate it …

how? ​

 

[harry654]

substitute h²=a²+b²-2abcos(α-β)

 

[tiny-tim]

the cosine rule … no, beacuse we've already used that … you're going backwards again …

you have a triangle, you can't use the cosine rule now because it takes you the wrong way …

so … ? ​

 

[harry654]

I think that
h²=(sin(α-β)b/sinβ)²
h²=sin²(α-β)b²/sin²β

 

[tiny-tim]

(the sine rule) yes …

that would convert h² ≤ (a²+b²)sin²(α-β)

into sin²(α-β)b²/sin²β ≤ (a²+b²)sin²(α-β)

and so b²/sin²β ≤ (a²+b²)

but that doesn't involve a, and the result you want is symmetric in a and b …

can you think of a slightly different way that involves both a and b, preferably equally?

 

[harry654]

I don't know what do you think but apply sinβa=sinαb

 

[tiny-tim]

worth a try …

what does that give you? ​

 

[harry654]

a²/sin²α ≤ (a²+b²)

 

[tiny-tim]

ok, from the sine rule you had a²/sin²α = b²/sin²β = K, say

and from that you got

a²/sin²α ≤ (a²+b²)

and

b²/sin²β ≤ (a²+b²)

but you'd like something with (a²+b²) on the left (as well as on the right) …

can you see how to do that?

(hint: use K )

 

[harry654]

so I try but ..
Ksin²α+Ksin²β= a²+b²

 

[tiny-tim]

ok , that gets you to

(a² + b²)/(sin²α + sin²β) ≤ (a²+b²),

which is the same as (sin²α + sin²β) ≥ 1 …

if you can prove that, you can prove the inequality in the question

how are you going to do that?

(in other words, what piece of information in the question haven't you used yet? )​

 

[harry654]

α+β > 90°

 

[tiny-tim]

that's the one!

sooo … ? ​