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A trigonometry question from STEP examination

3

3

$begingroup$

Show that if at least one of the four angles A ± B ± C is a multiple of π, then
$$sin^4A + sin^4 B + sin^4 C − 2 sin^2 B sin^2 C − 2 sin^2 C sin^2 A
− 2 sin^2 A sin^2 B + 4 sin^2 A sin^2 B sin^2 C = 0$$

I want to start with proving $sin(A+B+C)$ or $(sin(A)+sin(B)+sin(C))^2$, however, I failed in both cases.

trigonometry

share|cite|improve this question

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

asked 2 days ago

KevinKevin

466

$endgroup$

add a comment | 

3

3

$begingroup$

Show that if at least one of the four angles A ± B ± C is a multiple of π, then
$$sin^4A + sin^4 B + sin^4 C − 2 sin^2 B sin^2 C − 2 sin^2 C sin^2 A
− 2 sin^2 A sin^2 B + 4 sin^2 A sin^2 B sin^2 C = 0$$

I want to start with proving $sin(A+B+C)$ or $(sin(A)+sin(B)+sin(C))^2$, however, I failed in both cases.

trigonometry

share|cite|improve this question

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

asked 2 days ago

KevinKevin

466

$endgroup$

add a comment | 

$begingroup$

Show that if at least one of the four angles A ± B ± C is a multiple of π, then
$$sin^4A + sin^4 B + sin^4 C − 2 sin^2 B sin^2 C − 2 sin^2 C sin^2 A
− 2 sin^2 A sin^2 B + 4 sin^2 A sin^2 B sin^2 C = 0$$

I want to start with proving $sin(A+B+C)$ or $(sin(A)+sin(B)+sin(C))^2$, however, I failed in both cases.

trigonometry

share|cite|improve this question

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

asked 2 days ago

KevinKevin

466

$endgroup$

share|cite|improve this question

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

asked 2 days ago

KevinKevin

466

share|cite|improve this question

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

asked 2 days ago

KevinKevin

466

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

edited 2 days ago

Jean-Claude Arbaut

15.5k63865

asked 2 days ago

KevinKevin

466

asked 2 days ago

KevinKevin

466

2 Answers
2

active

oldest

votes

6

$begingroup$

the following information may help you towards a solution.

since all the sine terms are squared we may as well assume that A,B and C are the angles of a triangle. let the sides be $a,b,c$ in the usual configuration. then if $Delta$ represents the area of the triangle we have the two relations:
$$
Delta = sqrts(s-a)(s-b)(s-c)
$$
and
$$
R = fracabc4Delta
$$
where $s = fraca+b+c2$ is the semi-perimeter and $R$ is the circumradius.

if you eliminate $Delta$ and substitute for $s$ you have a polynomial relation in $a,b,c$ which will give you the required result after applying the sine rule:
$$
fracasin A = fracbsin B = fraccsin C = 2R
$$

share|cite|improve this answer

answered 2 days ago

David HoldenDavid Holden

15k21226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "we may as well assume that $A$,$B$ and $C$ are the angles of a triangle" But then you assume $A+B+C=pi$. It's stronger than the assumption in the question.
  $endgroup$
  – Jean-Claude Arbaut
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Jean-ClaudeArbaut . There IS another case. If $Apm Bpm C=pi$ then there are non-negative $A',B',C'$ with $A'+B'+C'=pi$ and $ |sin A|=|sin A'|,;|sin B|=|sin B'|,;|sin C|=sin C'|.$ And if none of $A',B',C'$ is zero then we do have a triangle. But there is a "degenerate" case, which is an easy case. E.g. if $C'=0 $ then $sin C'=0$ and the equation simplifies to $0=sin^4 A'+sin^4 B'-2sin^2 A' sin^2 B'=(sin^2 A'-sin^2 B')^2,$ and since $pi=A'+B',$ we also have $sin A'=sin B'.$
  $endgroup$
  – DanielWainfleet
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @David Holden the proof is splendid,but what make u think of constructing the triangle in terms of the semi-perimeter and R?
  $endgroup$
  – Kevin
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin well i stared at it for a while. brute force looked unappealing, so i felt stumped. then i thought of trying to get a simplification using the sine rule, but the slight 'inhomogeneity' seemed a problem (the term with three sine squared factors). after a bit more staring into space i vaguely remembered once getting the abc/4R formula for triangle area. this would dissolve the inhomogeneity, so the other terms must somehow resolve into the other well-known expression (Heron's formula). 1/2
  $endgroup$
  – David Holden
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  then i thought i was probably wishful thinking... but after a coffee break, a quick bit of scribbling on the back of an envelope suggested i might be on the right track after all. fortunately in this case i was. such positive outcomes are rare, as my logic circuits are rather unreliable
  $endgroup$
  – David Holden
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

Hint:

First of all writing $sin A=a$ etc.,

$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$

$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$

Now if $A+B+C=pi$

by this $sin A+sin B+sin C=4cosdfrac A2cosdfrac B2cosdfrac C2$

and by this $sin A+sin B-sin C=4sindfrac A2sindfrac B2cosdfrac C2$

Use $sin2x=2sin xcos x$

We shall same expressions in some order if $Apm Bpm C=pi$

share|cite|improve this answer

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280

$endgroup$

add a comment | 

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6

$begingroup$

the following information may help you towards a solution.

since all the sine terms are squared we may as well assume that A,B and C are the angles of a triangle. let the sides be $a,b,c$ in the usual configuration. then if $Delta$ represents the area of the triangle we have the two relations:
$$
Delta = sqrts(s-a)(s-b)(s-c)
$$
and
$$
R = fracabc4Delta
$$
where $s = fraca+b+c2$ is the semi-perimeter and $R$ is the circumradius.

if you eliminate $Delta$ and substitute for $s$ you have a polynomial relation in $a,b,c$ which will give you the required result after applying the sine rule:
$$
fracasin A = fracbsin B = fraccsin C = 2R
$$

share|cite|improve this answer

answered 2 days ago

David HoldenDavid Holden

15k21226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "we may as well assume that $A$,$B$ and $C$ are the angles of a triangle" But then you assume $A+B+C=pi$. It's stronger than the assumption in the question.
  $endgroup$
  – Jean-Claude Arbaut
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Jean-ClaudeArbaut . There IS another case. If $Apm Bpm C=pi$ then there are non-negative $A',B',C'$ with $A'+B'+C'=pi$ and $ |sin A|=|sin A'|,;|sin B|=|sin B'|,;|sin C|=sin C'|.$ And if none of $A',B',C'$ is zero then we do have a triangle. But there is a "degenerate" case, which is an easy case. E.g. if $C'=0 $ then $sin C'=0$ and the equation simplifies to $0=sin^4 A'+sin^4 B'-2sin^2 A' sin^2 B'=(sin^2 A'-sin^2 B')^2,$ and since $pi=A'+B',$ we also have $sin A'=sin B'.$
  $endgroup$
  – DanielWainfleet
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @David Holden the proof is splendid,but what make u think of constructing the triangle in terms of the semi-perimeter and R?
  $endgroup$
  – Kevin
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin well i stared at it for a while. brute force looked unappealing, so i felt stumped. then i thought of trying to get a simplification using the sine rule, but the slight 'inhomogeneity' seemed a problem (the term with three sine squared factors). after a bit more staring into space i vaguely remembered once getting the abc/4R formula for triangle area. this would dissolve the inhomogeneity, so the other terms must somehow resolve into the other well-known expression (Heron's formula). 1/2
  $endgroup$
  – David Holden
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  then i thought i was probably wishful thinking... but after a coffee break, a quick bit of scribbling on the back of an envelope suggested i might be on the right track after all. fortunately in this case i was. such positive outcomes are rare, as my logic circuits are rather unreliable
  $endgroup$
  – David Holden
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

6

$begingroup$

the following information may help you towards a solution.

since all the sine terms are squared we may as well assume that A,B and C are the angles of a triangle. let the sides be $a,b,c$ in the usual configuration. then if $Delta$ represents the area of the triangle we have the two relations:
$$
Delta = sqrts(s-a)(s-b)(s-c)
$$
and
$$
R = fracabc4Delta
$$
where $s = fraca+b+c2$ is the semi-perimeter and $R$ is the circumradius.

if you eliminate $Delta$ and substitute for $s$ you have a polynomial relation in $a,b,c$ which will give you the required result after applying the sine rule:
$$
fracasin A = fracbsin B = fraccsin C = 2R
$$

share|cite|improve this answer

answered 2 days ago

David HoldenDavid Holden

15k21226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "we may as well assume that $A$,$B$ and $C$ are the angles of a triangle" But then you assume $A+B+C=pi$. It's stronger than the assumption in the question.
  $endgroup$
  – Jean-Claude Arbaut
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Jean-ClaudeArbaut . There IS another case. If $Apm Bpm C=pi$ then there are non-negative $A',B',C'$ with $A'+B'+C'=pi$ and $ |sin A|=|sin A'|,;|sin B|=|sin B'|,;|sin C|=sin C'|.$ And if none of $A',B',C'$ is zero then we do have a triangle. But there is a "degenerate" case, which is an easy case. E.g. if $C'=0 $ then $sin C'=0$ and the equation simplifies to $0=sin^4 A'+sin^4 B'-2sin^2 A' sin^2 B'=(sin^2 A'-sin^2 B')^2,$ and since $pi=A'+B',$ we also have $sin A'=sin B'.$
  $endgroup$
  – DanielWainfleet
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @David Holden the proof is splendid,but what make u think of constructing the triangle in terms of the semi-perimeter and R?
  $endgroup$
  – Kevin
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin well i stared at it for a while. brute force looked unappealing, so i felt stumped. then i thought of trying to get a simplification using the sine rule, but the slight 'inhomogeneity' seemed a problem (the term with three sine squared factors). after a bit more staring into space i vaguely remembered once getting the abc/4R formula for triangle area. this would dissolve the inhomogeneity, so the other terms must somehow resolve into the other well-known expression (Heron's formula). 1/2
  $endgroup$
  – David Holden
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  then i thought i was probably wishful thinking... but after a coffee break, a quick bit of scribbling on the back of an envelope suggested i might be on the right track after all. fortunately in this case i was. such positive outcomes are rare, as my logic circuits are rather unreliable
  $endgroup$
  – David Holden
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

the following information may help you towards a solution.

since all the sine terms are squared we may as well assume that A,B and C are the angles of a triangle. let the sides be $a,b,c$ in the usual configuration. then if $Delta$ represents the area of the triangle we have the two relations:
$$
Delta = sqrts(s-a)(s-b)(s-c)
$$
and
$$
R = fracabc4Delta
$$
where $s = fraca+b+c2$ is the semi-perimeter and $R$ is the circumradius.

if you eliminate $Delta$ and substitute for $s$ you have a polynomial relation in $a,b,c$ which will give you the required result after applying the sine rule:
$$
fracasin A = fracbsin B = fraccsin C = 2R
$$

share|cite|improve this answer

answered 2 days ago

David HoldenDavid Holden

15k21226

$endgroup$

share|cite|improve this answer

answered 2 days ago

David HoldenDavid Holden

15k21226

answered 2 days ago

David HoldenDavid Holden

15k21226

answered 2 days ago

David HoldenDavid Holden

15k21226

3

$begingroup$

Hint:

First of all writing $sin A=a$ etc.,

$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$

$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$

Now if $A+B+C=pi$

by this $sin A+sin B+sin C=4cosdfrac A2cosdfrac B2cosdfrac C2$

and by this $sin A+sin B-sin C=4sindfrac A2sindfrac B2cosdfrac C2$

Use $sin2x=2sin xcos x$

We shall same expressions in some order if $Apm Bpm C=pi$

share|cite|improve this answer

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280

$endgroup$

add a comment | 

3

$begingroup$

Hint:

First of all writing $sin A=a$ etc.,

$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$

$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$

Now if $A+B+C=pi$

by this $sin A+sin B+sin C=4cosdfrac A2cosdfrac B2cosdfrac C2$

and by this $sin A+sin B-sin C=4sindfrac A2sindfrac B2cosdfrac C2$

Use $sin2x=2sin xcos x$

We shall same expressions in some order if $Apm Bpm C=pi$

share|cite|improve this answer

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280

$endgroup$

add a comment | 

$begingroup$

Hint:

First of all writing $sin A=a$ etc.,

$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$

$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$

Now if $A+B+C=pi$

by this $sin A+sin B+sin C=4cosdfrac A2cosdfrac B2cosdfrac C2$

and by this $sin A+sin B-sin C=4sindfrac A2sindfrac B2cosdfrac C2$

Use $sin2x=2sin xcos x$

We shall same expressions in some order if $Apm Bpm C=pi$

share|cite|improve this answer

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280

$endgroup$

share|cite|improve this answer

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280

answered 2 days ago

lab bhattacharjeelab bhattacharjee

229k15159280