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

[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