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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.

Jean-Claude Arbaut
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Kevin
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2 Answers2

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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: $$ \Delta = \sqrt{s(s-a)(s-b)(s-c)} $$ and $$ R = \frac{abc}{4\Delta} $$ where $s = \frac{a+b+c}2$ 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: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$

David Holden
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  • "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. – Jean-Claude Arbaut Apr 21 '19 at 07:05
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    @Jean-ClaudeArbaut . There IS another case. If $A\pm B\pm C=\pi$ 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 $0=\sin^4 A'+\sin^4 B'-2\sin^2 A' \sin^2 B'=(\sin^2 A'-\sin^2 B')^2,$ and since $\pi=A'+B',$ we also have $\sin A'=\sin B'.$ – DanielWainfleet Apr 21 '19 at 07:28
  • @DanielWainfleet Ah, yes, we don't change the sines in absolute value, so we can indeed assume $A+B+C=\pi$. Nice! – Jean-Claude Arbaut Apr 21 '19 at 07:31
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    @David Holden the proof is splendid,but what make u think of constructing the triangle in terms of the semi-perimeter and R? – Kevin Apr 21 '19 at 07:44
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    @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 – David Holden Apr 21 '19 at 08:09
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    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 – David Holden Apr 21 '19 at 08:09
  • @DavidHolden thank a lot,btw did u try to use substituting sin(a+b) for sin(c), i was trying to prove in this way initially but i got stuck. – Kevin Apr 21 '19 at 08:16
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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 $\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$

and by this $\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos\dfrac C2$

Use $\sin2x=2\sin x\cos x$

We shall same expressions in some order if $A\pm B\pm C=\pi$

lab bhattacharjee
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