Mixing
Inequality relation between sines of angles in a triangle.
Question: Given ${CD}^2=AD.BD$, then prove that, $sin A.sin B le sin^2{Cover2}$.
I got 2 approaches to the question. Both landing nowhere.
(i) Using sine rule I got that $sin A.sin B=sin alpha.sin beta$ but can't proceed.
(ii) On extending $CD$ further to $D'$ such that $CD=DD'={CD'over2}$ which makes $ABCD'$ a cyclic quadrilateral again ending where the first step ends.
Help me please.
Thanks for any hints or solution. Hope that question is decently put.
trigonometry triangle
asked Nov 17 '18 at 21:47
Love Invariants
86115
add a comment |
Question: Given ${CD}^2=AD.BD$, then prove that, $sin A.sin B le sin^2{Cover2}$.
I got 2 approaches to the question. Both landing nowhere.
(i) Using sine rule I got that $sin A.sin B=sin alpha.sin beta$ but can't proceed.
(ii) On extending $CD$ further to $D'$ such that $CD=DD'={CD'over2}$ which makes $ABCD'$ a cyclic quadrilateral again ending where the first step ends.
Help me please.
Thanks for any hints or solution. Hope that question is decently put.
trigonometry triangle
asked Nov 17 '18 at 21:47
Love Invariants
86115
How could I get inequality using law of sines?
– Love Invariants
Nov 17 '18 at 21:58
add a comment |
Question: Given ${CD}^2=AD.BD$, then prove that, $sin A.sin B le sin^2{Cover2}$.
I got 2 approaches to the question. Both landing nowhere.
(i) Using sine rule I got that $sin A.sin B=sin alpha.sin beta$ but can't proceed.
(ii) On extending $CD$ further to $D'$ such that $CD=DD'={CD'over2}$ which makes $ABCD'$ a cyclic quadrilateral again ending where the first step ends.
Help me please.
Thanks for any hints or solution. Hope that question is decently put.
trigonometry triangle
asked Nov 17 '18 at 21:47
Love Invariants
86115
Question: Given ${CD}^2=AD.BD$, then prove that, $sin A.sin B le sin^2{Cover2}$.
I got 2 approaches to the question. Both landing nowhere.
(i) Using sine rule I got that $sin A.sin B=sin alpha.sin beta$ but can't proceed.
(ii) On extending $CD$ further to $D'$ such that $CD=DD'={CD'over2}$ which makes $ABCD'$ a cyclic quadrilateral again ending where the first step ends.
Help me please.
Thanks for any hints or solution. Hope that question is decently put.
trigonometry triangle
trigonometry triangle
asked Nov 17 '18 at 21:47
Love Invariants
86115
asked Nov 17 '18 at 21:47
Love Invariants
86115
asked Nov 17 '18 at 21:47
Love Invariants
86115
asked Nov 17 '18 at 21:47
Love Invariants
86115
asked Nov 17 '18 at 21:47
Love Invariants
86115
86115
How could I get inequality using law of sines?
– Love Invariants
Nov 17 '18 at 21:58
add a comment |
How could I get inequality using law of sines?
– Love Invariants
Nov 17 '18 at 21:58
How could I get inequality using law of sines?
– Love Invariants
Nov 17 '18 at 21:58
How could I get inequality using law of sines?
– Love Invariants
Nov 17 '18 at 21:58
add a comment |
2 Answers
2
active
oldest
votes
DeepSea's answer is nice, but it seems that you don't get it.
So, let me add more details.
You've already got
$$sin Asin B=sinalphasinbetatag1$$
Since
$$frac{(x+y)^2}{4}-xy=frac{x^2+2xy+y^2-4xy}{4}=frac{(x-y)^2}{4}ge 0$$
holds for any $x,yinmathbb R$, we see that
$$xylefrac{(x+y)^2}{4}$$
holds for any $x,yinmathbb R$.
So, we have
$$sinalphasinbetale frac{(sinalpha+sinbeta)^2}{4}tag2$$
Using the following sum-to-product identity
$$sinalpha+sinbeta=2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)$$
one gets
$$begin{align}frac{(sinalpha+sinbeta)^2}{4}&
=frac{1}{4}left(2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)right)^2
\\&=sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)tag3end{align}$$
For any $thetainmathbb R$, we have $|costheta|le 1$, so $cos^2thetale 1$.
So, we have
$$cos^2bigg(frac{alpha-beta}{2}bigg)le 1$$
Multiplying the both sides by $sin^2(frac{alpha+beta}{2}) (gt 0)$ gives
$$sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)le sin^2bigg(frac{alpha+beta}{2}bigg)=sin^2frac{C}{2}tag4$$
It follows from $(1)(2)(3)(4)$ that
$$sin Asin Blesin^2frac C2$$
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
add a comment |
Taking from where you left off,and using the well-known fact: $xy le dfrac{(x+y)^2}{4}$, we have : $ sin Acdot sin B = sin alpha cdot sin beta le dfrac{ (sin alpha + sin beta )^2}{4} = dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} = sin^2(frac{alpha+beta}{2})cdot cos^2(frac{alpha-beta}{2}) le sin^2 (frac{alpha+beta}{2}) = sin^2 (C/2)$ because $cos^2(frac{alpha - beta}{2}) le 1$. This establishes the claim.
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
DeepSea's answer is nice, but it seems that you don't get it.
So, let me add more details.
You've already got
$$sin Asin B=sinalphasinbetatag1$$
Since
$$frac{(x+y)^2}{4}-xy=frac{x^2+2xy+y^2-4xy}{4}=frac{(x-y)^2}{4}ge 0$$
holds for any $x,yinmathbb R$, we see that
$$xylefrac{(x+y)^2}{4}$$
holds for any $x,yinmathbb R$.
So, we have
$$sinalphasinbetale frac{(sinalpha+sinbeta)^2}{4}tag2$$
Using the following sum-to-product identity
$$sinalpha+sinbeta=2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)$$
one gets
$$begin{align}frac{(sinalpha+sinbeta)^2}{4}&
=frac{1}{4}left(2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)right)^2
\\&=sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)tag3end{align}$$
For any $thetainmathbb R$, we have $|costheta|le 1$, so $cos^2thetale 1$.
So, we have
$$cos^2bigg(frac{alpha-beta}{2}bigg)le 1$$
Multiplying the both sides by $sin^2(frac{alpha+beta}{2}) (gt 0)$ gives
$$sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)le sin^2bigg(frac{alpha+beta}{2}bigg)=sin^2frac{C}{2}tag4$$
It follows from $(1)(2)(3)(4)$ that
$$sin Asin Blesin^2frac C2$$
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
add a comment |
DeepSea's answer is nice, but it seems that you don't get it.
So, let me add more details.
You've already got
$$sin Asin B=sinalphasinbetatag1$$
Since
$$frac{(x+y)^2}{4}-xy=frac{x^2+2xy+y^2-4xy}{4}=frac{(x-y)^2}{4}ge 0$$
holds for any $x,yinmathbb R$, we see that
$$xylefrac{(x+y)^2}{4}$$
holds for any $x,yinmathbb R$.
So, we have
$$sinalphasinbetale frac{(sinalpha+sinbeta)^2}{4}tag2$$
Using the following sum-to-product identity
$$sinalpha+sinbeta=2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)$$
one gets
$$begin{align}frac{(sinalpha+sinbeta)^2}{4}&
=frac{1}{4}left(2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)right)^2
\\&=sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)tag3end{align}$$
For any $thetainmathbb R$, we have $|costheta|le 1$, so $cos^2thetale 1$.
So, we have
$$cos^2bigg(frac{alpha-beta}{2}bigg)le 1$$
Multiplying the both sides by $sin^2(frac{alpha+beta}{2}) (gt 0)$ gives
$$sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)le sin^2bigg(frac{alpha+beta}{2}bigg)=sin^2frac{C}{2}tag4$$
It follows from $(1)(2)(3)(4)$ that
$$sin Asin Blesin^2frac C2$$
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
add a comment |
DeepSea's answer is nice, but it seems that you don't get it.
So, let me add more details.
You've already got
$$sin Asin B=sinalphasinbetatag1$$
Since
$$frac{(x+y)^2}{4}-xy=frac{x^2+2xy+y^2-4xy}{4}=frac{(x-y)^2}{4}ge 0$$
holds for any $x,yinmathbb R$, we see that
$$xylefrac{(x+y)^2}{4}$$
holds for any $x,yinmathbb R$.
So, we have
$$sinalphasinbetale frac{(sinalpha+sinbeta)^2}{4}tag2$$
Using the following sum-to-product identity
$$sinalpha+sinbeta=2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)$$
one gets
$$begin{align}frac{(sinalpha+sinbeta)^2}{4}&
=frac{1}{4}left(2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)right)^2
\\&=sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)tag3end{align}$$
For any $thetainmathbb R$, we have $|costheta|le 1$, so $cos^2thetale 1$.
So, we have
$$cos^2bigg(frac{alpha-beta}{2}bigg)le 1$$
Multiplying the both sides by $sin^2(frac{alpha+beta}{2}) (gt 0)$ gives
$$sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)le sin^2bigg(frac{alpha+beta}{2}bigg)=sin^2frac{C}{2}tag4$$
It follows from $(1)(2)(3)(4)$ that
$$sin Asin Blesin^2frac C2$$
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
DeepSea's answer is nice, but it seems that you don't get it.
So, let me add more details.
You've already got
$$sin Asin B=sinalphasinbetatag1$$
Since
$$frac{(x+y)^2}{4}-xy=frac{x^2+2xy+y^2-4xy}{4}=frac{(x-y)^2}{4}ge 0$$
holds for any $x,yinmathbb R$, we see that
$$xylefrac{(x+y)^2}{4}$$
holds for any $x,yinmathbb R$.
So, we have
$$sinalphasinbetale frac{(sinalpha+sinbeta)^2}{4}tag2$$
Using the following sum-to-product identity
$$sinalpha+sinbeta=2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)$$
one gets
$$begin{align}frac{(sinalpha+sinbeta)^2}{4}&
=frac{1}{4}left(2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)right)^2
\\&=sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)tag3end{align}$$
For any $thetainmathbb R$, we have $|costheta|le 1$, so $cos^2thetale 1$.
So, we have
$$cos^2bigg(frac{alpha-beta}{2}bigg)le 1$$
Multiplying the both sides by $sin^2(frac{alpha+beta}{2}) (gt 0)$ gives
$$sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)le sin^2bigg(frac{alpha+beta}{2}bigg)=sin^2frac{C}{2}tag4$$
It follows from $(1)(2)(3)(4)$ that
$$sin Asin Blesin^2frac C2$$
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
answered Nov 21 '18 at 11:55
mathlove
91.7k881214
91.7k881214
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
add a comment |
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
Thanks. Well DeepSea didn't mention how he got that third equation. I thought he wrote some random equation.
– Love Invariants
Nov 22 '18 at 16:51
add a comment |
Taking from where you left off,and using the well-known fact: $xy le dfrac{(x+y)^2}{4}$, we have : $ sin Acdot sin B = sin alpha cdot sin beta le dfrac{ (sin alpha + sin beta )^2}{4} = dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} = sin^2(frac{alpha+beta}{2})cdot cos^2(frac{alpha-beta}{2}) le sin^2 (frac{alpha+beta}{2}) = sin^2 (C/2)$ because $cos^2(frac{alpha - beta}{2}) le 1$. This establishes the claim.
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
add a comment |
Taking from where you left off,and using the well-known fact: $xy le dfrac{(x+y)^2}{4}$, we have : $ sin Acdot sin B = sin alpha cdot sin beta le dfrac{ (sin alpha + sin beta )^2}{4} = dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} = sin^2(frac{alpha+beta}{2})cdot cos^2(frac{alpha-beta}{2}) le sin^2 (frac{alpha+beta}{2}) = sin^2 (C/2)$ because $cos^2(frac{alpha - beta}{2}) le 1$. This establishes the claim.
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
add a comment |
Taking from where you left off,and using the well-known fact: $xy le dfrac{(x+y)^2}{4}$, we have : $ sin Acdot sin B = sin alpha cdot sin beta le dfrac{ (sin alpha + sin beta )^2}{4} = dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} = sin^2(frac{alpha+beta}{2})cdot cos^2(frac{alpha-beta}{2}) le sin^2 (frac{alpha+beta}{2}) = sin^2 (C/2)$ because $cos^2(frac{alpha - beta}{2}) le 1$. This establishes the claim.
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
Taking from where you left off,and using the well-known fact: $xy le dfrac{(x+y)^2}{4}$, we have : $ sin Acdot sin B = sin alpha cdot sin beta le dfrac{ (sin alpha + sin beta )^2}{4} = dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} = sin^2(frac{alpha+beta}{2})cdot cos^2(frac{alpha-beta}{2}) le sin^2 (frac{alpha+beta}{2}) = sin^2 (C/2)$ because $cos^2(frac{alpha - beta}{2}) le 1$. This establishes the claim.
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
edited Nov 18 '18 at 17:25
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
answered Nov 17 '18 at 22:20
DeepSea
70.9k54487
70.9k54487
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
add a comment |
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Can you explain it? This second inequality. Ok dude wait I'm searching for Jensen inequality. Read about it somewhere though its not in my book
– Love Invariants
Nov 17 '18 at 22:24
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Well what if one of $alpha ,beta$ is obtuse? I think Jensen inequality says only about increasing graphs which is what I thought of and didn't use AM-GM.
– Love Invariants
Nov 17 '18 at 22:28
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
Bro if you solve the last inequality: $dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} le sin^2 (frac{alpha+beta}{2})$. You would get $cos (frac{alpha-beta}{2}) le sin (frac{alpha+beta}{2})$ which isn't $cos^2 (frac{alpha-beta}{2}) le 1$
– Love Invariants
Nov 18 '18 at 16:51
add a comment |
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How could I get inequality using law of sines?
– Love Invariants
Nov 17 '18 at 21:58