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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.

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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 $c=e^{iC}$.

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  • $\begingroup$ I did not get the answer. Could you please elaborate a bit more? $\endgroup$ – pi-π Oct 16 '16 at 5:36
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    $\begingroup$ @user354073 the left side is $0$ means that you have equality $a^3+b^3+c^3=3abc$ $\endgroup$ – Nick Oct 16 '16 at 5:58
  • $\begingroup$ ##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 $R.H.S$. $\endgroup$ – pi-π Oct 16 '16 at 6:12
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    $\begingroup$ $3(e^{iA}e^{iB}e^{iC})=3e^{i(A+B+C)}=3\cos(A+B+C)+i3\sin(A+B+C)$ $\endgroup$ – Nick Oct 16 '16 at 7:29
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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: $$\mathrm{cos}\ 3A + \mathrm{cos}\ 3B + \mathrm{cos}\ 3C = 3\ \mathrm{cos}\ (a+b+c)$$

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  • $\begingroup$ what is $cis$? Could you express it in terms of cos and sin? $\endgroup$ – pi-π Oct 16 '16 at 6:23
  • $\begingroup$ @user354073 $\mathrm{cis}(x)=e^{ix}$ $\endgroup$ – StubbornAtom Oct 16 '16 at 7:18
  • $\begingroup$ @user354073 $\mathrm{cis}\ x = \mathrm{cos}\ x + i\ \mathrm{sin}\ x$ $\endgroup$ – Nanoputian Oct 16 '16 at 9:42

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