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

[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 … ? ​

 

[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! )​