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.

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.

< Question 4 Question 6(i-ix) >