Solution of triangles.

**mathsisalwayscool** · November 25th 2011, 09:45 PM

Prove That
(Sin^3 A).(Sin(B-C)) + (Sin^3 B).(Sin(C-A)) + (Sin^3 C).(Sin(A-B)) = 0

**sbhatnagar** · November 26th 2011, 08:10 AM

Prove $\sin^3{A}\sin(B-C)+\sin^3{B}\sin(C-A)+\sin^3{C}\sin(A-B)=0$

Assuming that $A,B \hspace{3} \text{and}\hspace{3} C$ are angles of a triangle:

$A+B+C=180^{\circ}$

This means:

$A=180^{\circ}-(B+C) \iff \sin{A}=\sin{(B+C)}$
$B=180^{\circ}-(A+C) \iff \sin{B}=\sin{(A+C)}$
$C=180^{\circ}-(A+B) \iff \sin{C}=\sin{(B+A)}$

Also:

$\begin{align*}\sin^3{C}\sin(A-B)&= \sin^2{C}[\sin(A-B)\sin(A+B)] \\ &=\sin^2{C}[\frac{1}{2}(\cos{2B}-\cos{2A})] \\ &= \sin^2{C}[\frac{1}{2}(\cos^2{B}-\sin^2{B}-\cos^2{A}+\sin^2{A})] \\ &= \sin^2{C}[\sin^2{A}-\sin^2{B}] \\ &= \sin^2{C}\sin^2{A}-\sin^2{C}\sin^2{B} \end{align*}$
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$\begin{align*}\sin^3{B}\sin(C-A)&= \sin^2{B}[\sin(C-A)\sin(A+C)] \\ &=\sin^2{B}[\frac{1}{2}(\cos{2A}-\cos{2C})] \\ &= \sin^2{B}[\frac{1}{2}(\cos^2{A}-\sin^2{A}-\cos^2{C}+\sin^2{C})] \\ &= \sin^2{B}[-\sin^2{A}+\sin^2{C}] \\ &= -\sin^2{B}\sin^2{A}+\sin^2{B}\sin^2{C} \end{align*}$

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$\begin{align*}\sin^3{A}\sin(B-C)&= \sin^2{A}[\sin(B-C)\sin(B+C)] \\ &=\sin^2{A}[\frac{1}{2}(\cos{2C}-\cos{2B})] \\ &= \sin^2{A}[\frac{1}{2}(\cos^2{C}-\sin^2{C}-\cos^2{B}+\sin^2{B})] \\ &= \sin^2{A}[\sin^2{B}-\sin^2{C}] \\ &= \sin^2{A}\sin^2{B}-\sin^2{A}\sin^2{C} \end{align*}$

So:

$\\ \sin^3{A}\sin(B-C)+\sin^3{B}\sin(C-A)+\sin^3{C}\sin(A-B) \\ ={ \color{red} \sin^2{C}\sin^2{A} }{\color{blue} -\sin^2{C}\sin^2{B}}{\color{green} +\sin^2{A}\sin^2{B}}{\color{red} -\sin^2{A}\sin^2{C}}{\color{green} -\sin^2{B}\sin^2{A}}{\color{blue}+ \sin^2{B}\sin^2{C}} \\ =0$

I guess there's a much easier way.

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