====== Question 5, Exercise 1.4 ======
Solutions of Question 5 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 5=====
If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$, show that:

(i) $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$.

(ii) $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma)$.

** Solution. **

Given:
\begin{align}
\cos \alpha + \cos \beta + \cos \gamma &= 0 -- (1) \\
\sin \alpha + \sin \beta + \sin \gamma &= 0 -- (2)
\end{align}
Suppose $a=e^{i\alpha}$,  $b=e^{i\beta}$ and $c=e^{i\gamma}$,
then
\begin{align}
& a+b+c \\
=&e^{i\alpha}+e^{i\beta}+e^{i\gamma}\\
=& \cos\alpha +i\sin\alpha+\cos\beta +i\sin\beta+\cos\gamma +i\sin\gamma \\
=& (\cos \alpha + \cos \beta + \cos \gamma)+i(\sin \alpha + \sin \beta + \sin \gamma). \\
\end{align}
Using (1) and (2) above, we get
$$a+b+c=0 -- (3).$$
Since we know
\begin{align}
&a^3+b^3+c^3-3abc \\
=&(a+b+c)(a^2+b^2+c^2-ab-bc-ca),
\end{align}
Using (3), we have
$$a^3+b^3+c^3-3abc=0$$
$$\implies a^3+b^3+c^3=3abc.$$
Putting values of $a$, $b$ and $c$, we get
\begin{align}
&(e^{i\alpha})^3+(e^{i\beta})^3+(e^{i\gamma})^3=3e^{i\alpha}e^{i\beta}e^{i\gamma}\\
\implies &e^{3i\alpha}+e^{3i\beta}+e^{3i\gamma}=3e^{i(\alpha+\beta+\gamma)} \\
\implies & \cos 3\alpha +i\sin 3\alpha +\cos 3\beta +i\sin 3\beta+\cos 3\gamma +i\sin 3\gamma \\
&=3[\cos(\alpha+\beta+\gamma)+i\sin(\alpha+\beta+\gamma)] \\
\implies & \cos 3\alpha +\cos 3\beta +\cos 3\gamma +i(\sin 3\alpha+\sin 3\beta +\sin 3\gamma) \\
&=3\cos(\alpha+\beta+\gamma)+i3\sin(\alpha+\beta+\gamma).
\end{align}
Equating real and imaginary parts, we get
\begin{align}
\cos 3\alpha +\cos 3\beta +\cos 3\gamma=3\cos(\alpha+\beta+\gamma) \\
\sin 3\alpha+\sin 3\beta +\sin 3\gamma = 3\sin(\alpha+\beta+\gamma),
\end{align}
as required.







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