If $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$, prove that
- $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$
- $\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$
Given $\;\;$ $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$ $\;\;\; \cdots \; (1)$
Let
$a = \cos \alpha + i \sin \alpha = e^{i \alpha}$ $\;\;\; \cdots \; (2a)$
$b = \cos \beta + i \sin \beta = e^{i \beta}$ $\;\;\; \cdots \; (2b)$
$c = \cos \gamma + i \sin \gamma = e^{i \gamma}$ $\;\;\; \cdots \; (2c)$
$\therefore \; a + b + c = \left(\cos \alpha + \cos \beta + \cos \gamma\right) + i \left(\sin \alpha + \sin \beta + \sin \gamma\right)$
i.e. $\;$ $a + b + c = 0$ $\;\;$ [from equation $(1)$] $\;\;\; \cdots \; (3)$
$\therefore$ $\;$ We have from equation $(3)$, $\;$ $a^3 + b^3 + c^3 = 3 abc$ $\;\;\; \cdots \; (4)$
[Note:
$a + b + c = 0$ $\implies$ $a + b = -c$ $\;\;\; \cdots \; (A)$
$\therefore$ $\;$ $\left(a + b\right)^3 = \left(-c\right)^3$
i.e. $\;$ $a^3 + b^3 + 3ab \left(a + b\right) = - c^3$
i.e. $\;$ $a^3 + b^3 + c^3 = - 3 ab \left(a + b\right)$
i.e. $\;$ $a^3 + b^3 + c^3 = 3abc$ $\;$ (from equation $A$)
]
From equation $(2a)$,
$a^3 = \left(\cos \alpha + i \sin \alpha\right)^3 = \cos 3 \alpha + i \sin 3 \alpha$ $\;\;\; \cdots \; (5a)$
[By De Moivre's theorm: $\;$ $\left(\cos \theta + i \sin \theta \right)^n = \cos n \theta + i \sin n \theta$]
From equation $(2b)$,
$b^3 = \left(\cos \beta + i \sin \beta\right)^3 = \cos 3 \beta + i \sin 3 \beta$ $\;\;\; \cdots \; (5b)$
From equation $(2c)$,
$c^3 = \left(\cos \gamma + i \sin \gamma\right)^3 = \cos 3 \gamma + i \sin 3 \gamma$ $\;\;\; \cdots \; (5c)$
Adding equations $(5a)$, $(5b)$ and $(5c)$ we get,
$a^3 + b^3 + c^3 = \left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right)$
$\hspace{3cm}$ $+ i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$ $\;\;\; \cdots \; (6a)$
We have from equations $(2a)$, $(2b)$ and $(2c)$,
$\begin{aligned}
3abc & = 3 e^{i \alpha} \cdot e^{i \beta} \cdot e^{i \gamma} \\\\
& = 3 e^{i \left(\alpha + \beta + \gamma\right)} \\\\
& = 3 \left[\cos \left(\alpha + \beta + \gamma\right) + i \sin \left(\alpha + \beta + \gamma\right)\right] \\\\
& = 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right) \;\;\; \cdots \; (6b)
\end{aligned}$
$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(4)$ becomes
$\left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right) + i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$
$\hspace{4cm}$ $= 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right)$ $\;\;\; \cdots \; (7)$
- Equating the real parts on either side of equation $(7)$ we have,
$\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$ - Equating the imaginary parts on either side of equation $(7)$ we have,
$\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$