$(\mathrm{r}, heta)$ denotes $r(\cos heta+i \sin heta)$. If $x=(1, \alpha), y=(1, \beta)$, $\mathrm{z}=(1, \gamma)$ and $x+y+z=0$ then $\sum \cos (2 \alpha-\beta-\gamma)=$

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$(\mathrm{r}, 	heta)$ denotes $r(\cos 	heta+i \sin 	heta)$. If $x=(1, \alpha), y=(1, \beta)$, $\mathrm{z}=(1, \gamma)$ and $x+y+z=0$ then $\sum \cos (2 \alpha-\beta-\gamma)=$

Aicharya · Accepted Answer

Since $(r, 	heta)$ denotes $r(\cos 	heta+i \sin 	heta)$ $\begin{aligned} & 	ext { So, } x=(1, \alpha)=\cos \alpha+i \sin \alpha=e^{i \alpha} \ & y=(1, \beta)=\cos \beta+i \sin \beta=e^{i \beta} \ & 	ext { and } z=(1, \gamma)=\cos \gamma+i \sin \gamma=e^{i \gamma} \ & 	ext { Now, } x+y+z=0 \ & \Rightarrow x^3+y^3+z^3=3 x y z \ & \Rightarrow \frac{x^2}{y z}+\frac{y^2}{y z}+\frac{z^2}{x y}=3 \Rightarrow \frac{e^{2 i \alpha}}{e^{i \beta} e^{i \gamma}}+\frac{e^{2 i \beta}}{e^{i \gamma} e^{i \alpha}}+\frac{e^{2 i \gamma}}{e^{i \alpha} e^{i \beta}}=3 \ & \Rightarrow e^{i(2 \alpha-\beta-\gamma)}+e^{i(2 \beta-\alpha-\gamma)}+e^{i(2 \gamma-\alpha-\beta)}=3 \end{aligned}$ On comparing both sides $\sum \cos (2 \alpha-\beta-\gamma)=3$

$(\mathrm{r}, \theta)$ denotes $r(\cos \theta+i \sin \theta)$. If $x=(1, \alpha), y=(1, \beta)$,…

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