How to Prove $a^3 cos (B-C) + b^3 cos (C-A)+ c^3 cos (A-B)=3abc$?

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How to Prove $a^3 cos (B-C) + b^3 cos (C-A)+ c^3 cos (A-B)=3abc$?



If $A,$ $B,$ and $C$ are the vertices of the Triangle ABC, and $a,$ $b$ and $c$ is the side opposite to the sides of the triangle.



You may use -



Projection Formula:



$$a=b cos C + c cos B$$



and Sine and Cosine Relation too.



My try



$$a^3 cos (B - C)+ b^3 cos (C-A) + c^3 cos (A-B)=3abc$$
LHS =
$$ 3(b cos C + c cos B)(a cos C + c cos A)(a cos B + b cos A)$$
$$ 3[ ab cos B cos ^2 C + abc cos A cos B cos C +a^2c cos ^2 B cos C +ac^2 cos A cos ^2B ]+[b^2 cos A cos^2 C + b^2c cos^2 A cos A+abc cos A cos B cos C + ac^2 cos A cos^2 B] $$



Now I can't solve it further because there seems no way to reduce that further.



Thanks :)




trigonometry triangle




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  • askiitians.com/forums/Trigonometry/…
    – lab bhattacharjee
    Jul 24 at 15:54










  • @labbhattacharjee How do others know about this Question? Who made them? source? if you know?
    – Abhas Kumar Sinha
    Jul 24 at 15:59










  • Your last line has several errors. I suggest you redo the calculation again before trying to proceed further.
    – herb steinberg
    Jul 24 at 16:30










  • @herbsteinberg Oh, I see
    – Abhas Kumar Sinha
    Jul 24 at 16:45










  • sine, cosine rule and $R=fracabcsqrt(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$
    – Takahiro Waki
    Jul 28 at 8:38














up vote
1
down vote

favorite
1












How to Prove $a^3 cos (B-C) + b^3 cos (C-A)+ c^3 cos (A-B)=3abc$?



If $A,$ $B,$ and $C$ are the vertices of the Triangle ABC, and $a,$ $b$ and $c$ is the side opposite to the sides of the triangle.



You may use -



Projection Formula:



$$a=b cos C + c cos B$$



and Sine and Cosine Relation too.



My try



$$a^3 cos (B - C)+ b^3 cos (C-A) + c^3 cos (A-B)=3abc$$
LHS =
$$ 3(b cos C + c cos B)(a cos C + c cos A)(a cos B + b cos A)$$
$$ 3[ ab cos B cos ^2 C + abc cos A cos B cos C +a^2c cos ^2 B cos C +ac^2 cos A cos ^2B ]+[b^2 cos A cos^2 C + b^2c cos^2 A cos A+abc cos A cos B cos C + ac^2 cos A cos^2 B] $$



Now I can't solve it further because there seems no way to reduce that further.



Thanks :)




trigonometry triangle




share|cite|improve this question





















  • askiitians.com/forums/Trigonometry/…
    – lab bhattacharjee
    Jul 24 at 15:54










  • @labbhattacharjee How do others know about this Question? Who made them? source? if you know?
    – Abhas Kumar Sinha
    Jul 24 at 15:59










  • Your last line has several errors. I suggest you redo the calculation again before trying to proceed further.
    – herb steinberg
    Jul 24 at 16:30










  • @herbsteinberg Oh, I see
    – Abhas Kumar Sinha
    Jul 24 at 16:45










  • sine, cosine rule and $R=fracabcsqrt(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$
    – Takahiro Waki
    Jul 28 at 8:38












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





How to Prove $a^3 cos (B-C) + b^3 cos (C-A)+ c^3 cos (A-B)=3abc$?



If $A,$ $B,$ and $C$ are the vertices of the Triangle ABC, and $a,$ $b$ and $c$ is the side opposite to the sides of the triangle.



You may use -



Projection Formula:



$$a=b cos C + c cos B$$



and Sine and Cosine Relation too.



My try



$$a^3 cos (B - C)+ b^3 cos (C-A) + c^3 cos (A-B)=3abc$$
LHS =
$$ 3(b cos C + c cos B)(a cos C + c cos A)(a cos B + b cos A)$$
$$ 3[ ab cos B cos ^2 C + abc cos A cos B cos C +a^2c cos ^2 B cos C +ac^2 cos A cos ^2B ]+[b^2 cos A cos^2 C + b^2c cos^2 A cos A+abc cos A cos B cos C + ac^2 cos A cos^2 B] $$



Now I can't solve it further because there seems no way to reduce that further.



Thanks :)




trigonometry triangle




share|cite|improve this question













How to Prove $a^3 cos (B-C) + b^3 cos (C-A)+ c^3 cos (A-B)=3abc$?



If $A,$ $B,$ and $C$ are the vertices of the Triangle ABC, and $a,$ $b$ and $c$ is the side opposite to the sides of the triangle.



You may use -



Projection Formula:



$$a=b cos C + c cos B$$



and Sine and Cosine Relation too.



My try



$$a^3 cos (B - C)+ b^3 cos (C-A) + c^3 cos (A-B)=3abc$$
LHS =
$$ 3(b cos C + c cos B)(a cos C + c cos A)(a cos B + b cos A)$$
$$ 3[ ab cos B cos ^2 C + abc cos A cos B cos C +a^2c cos ^2 B cos C +ac^2 cos A cos ^2B ]+[b^2 cos A cos^2 C + b^2c cos^2 A cos A+abc cos A cos B cos C + ac^2 cos A cos^2 B] $$



Now I can't solve it further because there seems no way to reduce that further.



Thanks :)





trigonometry triangle





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edited Jul 24 at 15:54









saulspatz

10.4k21323




10.4k21323









asked Jul 24 at 15:50









Abhas Kumar Sinha

10011




10011











  • askiitians.com/forums/Trigonometry/…
    – lab bhattacharjee
    Jul 24 at 15:54










  • @labbhattacharjee How do others know about this Question? Who made them? source? if you know?
    – Abhas Kumar Sinha
    Jul 24 at 15:59










  • Your last line has several errors. I suggest you redo the calculation again before trying to proceed further.
    – herb steinberg
    Jul 24 at 16:30










  • @herbsteinberg Oh, I see
    – Abhas Kumar Sinha
    Jul 24 at 16:45










  • sine, cosine rule and $R=fracabcsqrt(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$
    – Takahiro Waki
    Jul 28 at 8:38
















  • askiitians.com/forums/Trigonometry/…
    – lab bhattacharjee
    Jul 24 at 15:54










  • @labbhattacharjee How do others know about this Question? Who made them? source? if you know?
    – Abhas Kumar Sinha
    Jul 24 at 15:59










  • Your last line has several errors. I suggest you redo the calculation again before trying to proceed further.
    – herb steinberg
    Jul 24 at 16:30










  • @herbsteinberg Oh, I see
    – Abhas Kumar Sinha
    Jul 24 at 16:45










  • sine, cosine rule and $R=fracabcsqrt(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$
    – Takahiro Waki
    Jul 28 at 8:38















askiitians.com/forums/Trigonometry/…
– lab bhattacharjee
Jul 24 at 15:54




askiitians.com/forums/Trigonometry/…
– lab bhattacharjee
Jul 24 at 15:54












@labbhattacharjee How do others know about this Question? Who made them? source? if you know?
– Abhas Kumar Sinha
Jul 24 at 15:59




@labbhattacharjee How do others know about this Question? Who made them? source? if you know?
– Abhas Kumar Sinha
Jul 24 at 15:59












Your last line has several errors. I suggest you redo the calculation again before trying to proceed further.
– herb steinberg
Jul 24 at 16:30




Your last line has several errors. I suggest you redo the calculation again before trying to proceed further.
– herb steinberg
Jul 24 at 16:30












@herbsteinberg Oh, I see
– Abhas Kumar Sinha
Jul 24 at 16:45




@herbsteinberg Oh, I see
– Abhas Kumar Sinha
Jul 24 at 16:45












sine, cosine rule and $R=fracabcsqrt(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$
– Takahiro Waki
Jul 28 at 8:38




sine, cosine rule and $R=fracabcsqrt(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$
– Takahiro Waki
Jul 28 at 8:38










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










It can be proved with sine-rule and cosine rule but it is really ugly.



$$rm LHS - rm RHS =
sum_cyc ( a^3cos(B-C) - abc )
= sum_cyc( a^3(cos Bcos C + sin Bsin C) - abc)$$
By sine rule,



$$a : b : c = sin A : sin B : sin C
quadimpliesquad begincases
a^3sin Bsin C = abcsin A^2\
b^3sin Csin A = abcsin B^2\
c^3sin Asin B = abcsin C^2
endcases$$
This leads to



$$beginalignrm LHS - rm RHS
&= sum_cyc( a^3 cos B cos C - abccos^2 A)\
&= sum_cycleft[a^3left(fraca^2+c^2-b^22acright)left(fraca^2+b^2-c^22abright) - abcleft(fracb^2+c^2-a^22bcright)^2right]\
&= frac14abcsum_cyca^2(underbracea^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2_I)
endalign
$$
Notice what's in the square bracket equals to
$$requirecancelI =cancela^4 - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - cancela^4
=2(a^2(b^2+c^2) - b^4-c^4)$$
We obtain



$$beginalignrm LHS - rm RHS
&= frac12abcsum_cyc a^4b^2 + colorreda^4colorgreenc^2 - a^2 b^4 - colorbluea^2colormagentac^4\
&= frac12abcsum_cyc a^4b^2 + colorredb^4colorgreena^2 - a^2 b^4 - colorblueb^2colormagentaa^4\
&= 0
endalign$$






share|cite|improve this answer





















  • A beautiful equation hiding under an Ugly Proof!
    – Abhas Kumar Sinha
    Jul 24 at 16:43










  • @AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
    – achille hui
    Jul 24 at 16:44










  • I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
    – Abhas Kumar Sinha
    Jul 24 at 16:50










  • But how you Broke that all cosine ratios to convert difference of angles to single angle?
    – Abhas Kumar Sinha
    Jul 24 at 16:54






  • 1




    $sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
    – achille hui
    Jul 24 at 17:00











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










It can be proved with sine-rule and cosine rule but it is really ugly.



$$rm LHS - rm RHS =
sum_cyc ( a^3cos(B-C) - abc )
= sum_cyc( a^3(cos Bcos C + sin Bsin C) - abc)$$
By sine rule,



$$a : b : c = sin A : sin B : sin C
quadimpliesquad begincases
a^3sin Bsin C = abcsin A^2\
b^3sin Csin A = abcsin B^2\
c^3sin Asin B = abcsin C^2
endcases$$
This leads to



$$beginalignrm LHS - rm RHS
&= sum_cyc( a^3 cos B cos C - abccos^2 A)\
&= sum_cycleft[a^3left(fraca^2+c^2-b^22acright)left(fraca^2+b^2-c^22abright) - abcleft(fracb^2+c^2-a^22bcright)^2right]\
&= frac14abcsum_cyca^2(underbracea^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2_I)
endalign
$$
Notice what's in the square bracket equals to
$$requirecancelI =cancela^4 - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - cancela^4
=2(a^2(b^2+c^2) - b^4-c^4)$$
We obtain



$$beginalignrm LHS - rm RHS
&= frac12abcsum_cyc a^4b^2 + colorreda^4colorgreenc^2 - a^2 b^4 - colorbluea^2colormagentac^4\
&= frac12abcsum_cyc a^4b^2 + colorredb^4colorgreena^2 - a^2 b^4 - colorblueb^2colormagentaa^4\
&= 0
endalign$$






share|cite|improve this answer





















  • A beautiful equation hiding under an Ugly Proof!
    – Abhas Kumar Sinha
    Jul 24 at 16:43










  • @AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
    – achille hui
    Jul 24 at 16:44










  • I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
    – Abhas Kumar Sinha
    Jul 24 at 16:50










  • But how you Broke that all cosine ratios to convert difference of angles to single angle?
    – Abhas Kumar Sinha
    Jul 24 at 16:54






  • 1




    $sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
    – achille hui
    Jul 24 at 17:00















up vote
1
down vote



accepted










It can be proved with sine-rule and cosine rule but it is really ugly.



$$rm LHS - rm RHS =
sum_cyc ( a^3cos(B-C) - abc )
= sum_cyc( a^3(cos Bcos C + sin Bsin C) - abc)$$
By sine rule,



$$a : b : c = sin A : sin B : sin C
quadimpliesquad begincases
a^3sin Bsin C = abcsin A^2\
b^3sin Csin A = abcsin B^2\
c^3sin Asin B = abcsin C^2
endcases$$
This leads to



$$beginalignrm LHS - rm RHS
&= sum_cyc( a^3 cos B cos C - abccos^2 A)\
&= sum_cycleft[a^3left(fraca^2+c^2-b^22acright)left(fraca^2+b^2-c^22abright) - abcleft(fracb^2+c^2-a^22bcright)^2right]\
&= frac14abcsum_cyca^2(underbracea^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2_I)
endalign
$$
Notice what's in the square bracket equals to
$$requirecancelI =cancela^4 - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - cancela^4
=2(a^2(b^2+c^2) - b^4-c^4)$$
We obtain



$$beginalignrm LHS - rm RHS
&= frac12abcsum_cyc a^4b^2 + colorreda^4colorgreenc^2 - a^2 b^4 - colorbluea^2colormagentac^4\
&= frac12abcsum_cyc a^4b^2 + colorredb^4colorgreena^2 - a^2 b^4 - colorblueb^2colormagentaa^4\
&= 0
endalign$$






share|cite|improve this answer





















  • A beautiful equation hiding under an Ugly Proof!
    – Abhas Kumar Sinha
    Jul 24 at 16:43










  • @AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
    – achille hui
    Jul 24 at 16:44










  • I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
    – Abhas Kumar Sinha
    Jul 24 at 16:50










  • But how you Broke that all cosine ratios to convert difference of angles to single angle?
    – Abhas Kumar Sinha
    Jul 24 at 16:54






  • 1




    $sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
    – achille hui
    Jul 24 at 17:00













up vote
1
down vote



accepted







up vote
1
down vote



accepted






It can be proved with sine-rule and cosine rule but it is really ugly.



$$rm LHS - rm RHS =
sum_cyc ( a^3cos(B-C) - abc )
= sum_cyc( a^3(cos Bcos C + sin Bsin C) - abc)$$
By sine rule,



$$a : b : c = sin A : sin B : sin C
quadimpliesquad begincases
a^3sin Bsin C = abcsin A^2\
b^3sin Csin A = abcsin B^2\
c^3sin Asin B = abcsin C^2
endcases$$
This leads to



$$beginalignrm LHS - rm RHS
&= sum_cyc( a^3 cos B cos C - abccos^2 A)\
&= sum_cycleft[a^3left(fraca^2+c^2-b^22acright)left(fraca^2+b^2-c^22abright) - abcleft(fracb^2+c^2-a^22bcright)^2right]\
&= frac14abcsum_cyca^2(underbracea^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2_I)
endalign
$$
Notice what's in the square bracket equals to
$$requirecancelI =cancela^4 - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - cancela^4
=2(a^2(b^2+c^2) - b^4-c^4)$$
We obtain



$$beginalignrm LHS - rm RHS
&= frac12abcsum_cyc a^4b^2 + colorreda^4colorgreenc^2 - a^2 b^4 - colorbluea^2colormagentac^4\
&= frac12abcsum_cyc a^4b^2 + colorredb^4colorgreena^2 - a^2 b^4 - colorblueb^2colormagentaa^4\
&= 0
endalign$$






share|cite|improve this answer













It can be proved with sine-rule and cosine rule but it is really ugly.



$$rm LHS - rm RHS =
sum_cyc ( a^3cos(B-C) - abc )
= sum_cyc( a^3(cos Bcos C + sin Bsin C) - abc)$$
By sine rule,



$$a : b : c = sin A : sin B : sin C
quadimpliesquad begincases
a^3sin Bsin C = abcsin A^2\
b^3sin Csin A = abcsin B^2\
c^3sin Asin B = abcsin C^2
endcases$$
This leads to



$$beginalignrm LHS - rm RHS
&= sum_cyc( a^3 cos B cos C - abccos^2 A)\
&= sum_cycleft[a^3left(fraca^2+c^2-b^22acright)left(fraca^2+b^2-c^22abright) - abcleft(fracb^2+c^2-a^22bcright)^2right]\
&= frac14abcsum_cyca^2(underbracea^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2_I)
endalign
$$
Notice what's in the square bracket equals to
$$requirecancelI =cancela^4 - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - cancela^4
=2(a^2(b^2+c^2) - b^4-c^4)$$
We obtain



$$beginalignrm LHS - rm RHS
&= frac12abcsum_cyc a^4b^2 + colorreda^4colorgreenc^2 - a^2 b^4 - colorbluea^2colormagentac^4\
&= frac12abcsum_cyc a^4b^2 + colorredb^4colorgreena^2 - a^2 b^4 - colorblueb^2colormagentaa^4\
&= 0
endalign$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 24 at 16:31









achille hui

91k5127246




91k5127246











  • A beautiful equation hiding under an Ugly Proof!
    – Abhas Kumar Sinha
    Jul 24 at 16:43










  • @AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
    – achille hui
    Jul 24 at 16:44










  • I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
    – Abhas Kumar Sinha
    Jul 24 at 16:50










  • But how you Broke that all cosine ratios to convert difference of angles to single angle?
    – Abhas Kumar Sinha
    Jul 24 at 16:54






  • 1




    $sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
    – achille hui
    Jul 24 at 17:00

















  • A beautiful equation hiding under an Ugly Proof!
    – Abhas Kumar Sinha
    Jul 24 at 16:43










  • @AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
    – achille hui
    Jul 24 at 16:44










  • I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
    – Abhas Kumar Sinha
    Jul 24 at 16:50










  • But how you Broke that all cosine ratios to convert difference of angles to single angle?
    – Abhas Kumar Sinha
    Jul 24 at 16:54






  • 1




    $sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
    – achille hui
    Jul 24 at 17:00
















A beautiful equation hiding under an Ugly Proof!
– Abhas Kumar Sinha
Jul 24 at 16:43




A beautiful equation hiding under an Ugly Proof!
– Abhas Kumar Sinha
Jul 24 at 16:43












@AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
– achille hui
Jul 24 at 16:44




@AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it.
– achille hui
Jul 24 at 16:44












I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
– Abhas Kumar Sinha
Jul 24 at 16:50




I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school
– Abhas Kumar Sinha
Jul 24 at 16:50












But how you Broke that all cosine ratios to convert difference of angles to single angle?
– Abhas Kumar Sinha
Jul 24 at 16:54




But how you Broke that all cosine ratios to convert difference of angles to single angle?
– Abhas Kumar Sinha
Jul 24 at 16:54




1




1




$sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
– achille hui
Jul 24 at 17:00





$sum_cyc$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $sum_cycf(a,b,c) = sum_cycf(b,c,a) = sum_cyc f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case).
– achille hui
Jul 24 at 17:00













 

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