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If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$, prove that: $\cos 3A+\cos 3B+ \cos 3C=3\cos(A+B+C)$.

My Attempt;

Here,

$$e^{iA}=\cos A+i\sin A$$ $$e^{iB}=\cos B+i\sin B$$ $$e^{iC}=\cos C+i\sin C$$

Then, $$e^{iA}+e^{iB}+e^{iC}=0$$

Now, what should I do further. Please help.

pi-π
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2 Answers2

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Hint: Use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ by taking $a=e^{iA},b=e^{iB}$ and $c=e^{iC}$.

  • I did not get the answer. Could you please elaborate a bit more? – pi-π Oct 16 '16 at 05:36
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    @user354073 the left side is $0$ means that you have equality $a^3+b^3+c^3=3abc$ – Nick Oct 16 '16 at 05:58
  • ##Nick Liu, Yea. I tried by using that too, but did not get. Could you please show me the expression of $R.H.S$ only? I have done $L.H.S$ but not $R.H.S$. – pi-π Oct 16 '16 at 06:12
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    $3(e^{iA}e^{iB}e^{iC})=3e^{i(A+B+C)}=3\cos(A+B+C)+i3\sin(A+B+C)$ – Nick Oct 16 '16 at 07:29
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let $a = e^{iA}, b = e^{iB}, c = e^{iC}$. Also note that $a+b+c = 0$ $$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca) = 0$$ $$a^3 + b^3 + c^3 = 3abc$$ $$\mathrm{cis}\ 3A + \mathrm{cis}\ 3B + \mathrm{cis}\ 3C = 3\ \mathrm{cis}\ (a+b+c)$$ equating the real parts will give you the required expression: $$\mathrm{cos}\ 3A + \mathrm{cos}\ 3B + \mathrm{cos}\ 3C = 3\ \mathrm{cos}\ (a+b+c)$$

Nanoputian
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