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

In summary, Harry tried to solve the problem of proving that in any triangle ABC with a sharp angle at the peak C, apply inequality (a^2+b^2)cos(α-β)<=2ab but couldn't find the angle α-β. He then suggested making one, and when he did so, he found that equality occurred.
  • #71
soo I see that it apply but I don't know mathematically explain :(
 
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  • #72
ok, then explain it in ordinary English first …

what makes you think that it applies? :smile:
 
  • #73
I see from definition, better said I believe that apply :(
 
  • #74
or c²≤ a²+b²
c²/c² ≤ (a²+b²)/c²
sin²α + sin²β ≥ 1
 
  • #75
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 ?
 
  • #76
i don't understand … what is c ? :confused:

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°
 
  • #77
I don't know how :(
 
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  • #78
oh, I have feeling that I never finish this prove :(
sin²α + sin²β = 1 when α+β=90 so inequality doesn't apply I think
 
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  • #79
harry654 said:
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 ? :smile:
 
  • #80
certainly β or α >45° so sin²α or sin²β > 0,5 so sin²α+sin²β >1 but how explain it mathematically
 
  • #81
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° :smile:
 
  • #82
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)?
 
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  • #83
Could someone help me? I am desperate:(
 
  • #84
harry654 said:
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 :wink:)

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
 
  • #85
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
 
  • #86
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 :wink:)
 
  • #87
How can I prove sin²α + sin²β > 1 when α+β > 90°?
 
  • #88
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 :smile:

(btw, i'll be out soon, for the rest of the evening)
 
  • #89
Yes I know what do you mean, but when I use picture so it isn't correct mathematical proof so I don't know ...
 
  • #90
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 :smile:
 
  • #91
time to tidy-up …

hi harry654! :smile:

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 …
harry654 said:
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! :wink:) 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! :smile:)​
 
  • #92
Yes THANK YOU tiny-tim! I understand this and I thank you for your patience and assistance.
 
  • #93
Anyway I have question. Does it apply when sin(α-β) isn't 0 so cos(α-β) is positive or no?
 
  • #94
sorry, i don't understand :redface:
 
  • #95
Assume that apply sin(α-β) is not 0 then question is : Is cos(α-β) positive?
 
  • #96
harry654 said:
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 :wink:

why are you asking? :confused:
 
  • #97
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:)
 
  • #98
Hi tiny-tim!
If α=135 and β=30 then sin²α + sin²β < 1 and apply α+β>90 so what isn't correct?
 
  • #99
hi harry654! :smile:
harry654 said:
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 …
tiny-tim said:
so (tidying-up time! :wink:) 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! :smile:)​
 
<h2>1. What is the inequality (a^2+b^2)cos(α-β)<=2ab used for?</h2><p>The inequality (a^2+b^2)cos(α-β)<=2ab is used to determine when equality occurs in a trigonometric expression involving two variables, a and b, and two angles, α and β.</p><h2>2. How do you solve for equality in the inequality (a^2+b^2)cos(α-β)<=2ab?</h2><p>To solve for equality, you must first simplify the expression by expanding the trigonometric function using the cosine difference formula. Then, you can solve for the values of a and b that make the inequality true.</p><h2>3. Can equality occur in the inequality (a^2+b^2)cos(α-β)<=2ab for any values of a and b?</h2><p>Yes, equality can occur for any values of a and b as long as they satisfy the trigonometric expression. However, there may be multiple solutions or no solution at all.</p><h2>4. How does equality in the inequality (a^2+b^2)cos(α-β)<=2ab relate to the Pythagorean theorem?</h2><p>The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In this inequality, the left side represents the sum of the squares of the two sides, while the right side represents the square of the hypotenuse. Therefore, equality occurs when the triangle is a right triangle.</p><h2>5. Are there any real-world applications of the inequality (a^2+b^2)cos(α-β)<=2ab?</h2><p>Yes, this inequality can be used in various fields such as engineering, physics, and navigation to calculate the distance between two points or the angle between two objects. It can also be used in trigonometric identities and proofs.</p>

1. What is the inequality (a^2+b^2)cos(α-β)<=2ab used for?

The inequality (a^2+b^2)cos(α-β)<=2ab is used to determine when equality occurs in a trigonometric expression involving two variables, a and b, and two angles, α and β.

2. How do you solve for equality in the inequality (a^2+b^2)cos(α-β)<=2ab?

To solve for equality, you must first simplify the expression by expanding the trigonometric function using the cosine difference formula. Then, you can solve for the values of a and b that make the inequality true.

3. Can equality occur in the inequality (a^2+b^2)cos(α-β)<=2ab for any values of a and b?

Yes, equality can occur for any values of a and b as long as they satisfy the trigonometric expression. However, there may be multiple solutions or no solution at all.

4. How does equality in the inequality (a^2+b^2)cos(α-β)<=2ab relate to the Pythagorean theorem?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In this inequality, the left side represents the sum of the squares of the two sides, while the right side represents the square of the hypotenuse. Therefore, equality occurs when the triangle is a right triangle.

5. Are there any real-world applications of the inequality (a^2+b^2)cos(α-β)<=2ab?

Yes, this inequality can be used in various fields such as engineering, physics, and navigation to calculate the distance between two points or the angle between two objects. It can also be used in trigonometric identities and proofs.

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