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.

$\endgroup$
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: $$ \Delta = \sqrt{s(s-a)(s-b)(s-c)} $$ 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: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$

$\endgroup$
  • $\begingroup$ "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 Apr 21 at 7:05
  • 2
    $\begingroup$ @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 $\sin A'=\sin B'.$ $\endgroup$ – DanielWainfleet Apr 21 at 7:28
  • 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 Apr 21 at 7:44
  • 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 Apr 21 at 8:09
  • 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 Apr 21 at 8:09
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 $\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$

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 $A\pm B\pm C=\pi$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.