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

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Homework Help Overview

The discussion revolves around the inequality \((a^2+b^2)\cos(\alpha-\beta) \leq 2ab\) within the context of triangle geometry, specifically focusing on conditions for equality in this inequality.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the geometric interpretation of angles in triangle ABC, questioning how to construct the angle \(\alpha - \beta\) and its implications on the inequality.
  • There are attempts to apply the cosine law and relate it to the inequality, with participants expressing confusion about the steps and reasoning involved.
  • Some participants suggest drawing additional lines or reflecting triangles to visualize the problem better.
  • Questions arise regarding the conditions under which equality holds and the relationship between the areas of triangles involved.

Discussion Status

The discussion is ongoing, with participants sharing various insights and hints. Some have proposed methods to manipulate the inequality, while others express uncertainty about their progress. There is no clear consensus yet on the approach to take or the conditions for equality.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available and the methods they can use. There is also a language barrier affecting some participants' ability to communicate their ideas clearly.

  • #31
uhh? :confused:

just write out the inequality, with 2ab replaced by (a²+b²-h²)/cos(α-β)
 
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  • #32
yes I see, but what inequality do you think ?
 
  • #33
harry654 said:
yes I see, but what inequality do you think ?

what are you talking about?

there is only one inequality
 
  • #34
but I can't
in (a²+b²)cos(α-β)≤2ab
substitude 2ab because I must prove that inequality and when I substitude 2ab it isn't mathematical proof
 
  • #35
do it now to see where you're going …

you can tidy it up later! :biggrin:​
 
  • #36
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?
 
  • #37
oh, forget about later! :rolleyes:

ok, now simplify (a²+b²)cos²(α-β) ≤ a²+b²-h² :smile:
 
  • #38
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
 
  • #39
harry654 said:
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? :confused: … now you're going backwards :redface:

simplify (a²+b²)cos²(α-β) ≤ a²+b²-h²
 
  • #40
(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β) OK?
 
  • #41
harry654 said:
(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 :frown:
 
  • #42
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) :(
 
  • #43
well, for a start, you could try putting all similar terms on the same side
 
  • #44
similar terms on the same side? uff
 
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  • #45
I don't know how carry on again :(
 
  • #46
could you write me an inequality at which should I arrive?
 
  • #47
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:(
 
  • #48
good morning! :smile:
harry654 said:
(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) :smile:​
 
  • #49
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?
 
  • #50
harry654 said:
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! :rolleyes:

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 :redface:​
 
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  • #51
should I substitute 1=cos²(α-β)+sin²(α-β)
 
  • #52
yes, of course :smile:

get on with it!​
 
  • #53
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?
 
  • #54
how is that simpler?? :redface:
 
  • #55
(a²+b²)2cos(α-β)
and then?
 
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  • #56
harry654 said:
(a²+b²)2cos(α-β)

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

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 :redface:
 
  • #57
(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?
 
  • #58
harry654 said:
(a²+b²)(-sin²(α-β)) is that right?

at last! :smile:

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? :smile:​
 
  • #59
substitute h²=a²+b²-2abcos(α-β)
 
  • #60
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 … ? :smile:​
 

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