Question

If $cos\alpha +cos\beta +cos\gamma =sin\alpha +sin\beta +sin\gamma =0$, then
$cos3\alpha +cos3\beta +cos3\gamma =3cos(\alpha +\beta +\gamma ).$

• A. True
• B. False

Answer 1

The correct option is A True

Given $cos\alpha +cos\beta +cos\gamma =0$

$sin\alpha +sin\beta +sin\gamma =0$ (1)

Now, Let $z_1=cos\alpha +isin\alpha$

$z_2=cos\beta +isin\beta$

$z_3=cos\gamma +isin\gamma$

1) $z_1+z_2+z_3=(cos\alpha +isin\alpha )+(cos\beta +isin\beta )+(cos\gamma +isin\gamma )$

$\therefore z_1+z_2+z_3=(cos\alpha +cos\beta +cos\gamma )+i(sin\alpha +sin\beta +sin\gamma )$4

$\therefore z_1+z_2+z_3=0+i\left(0\right)$ (from equation (1))

$\therefore z_1+z_2+z_3=0$ (2)

2) $({z_1}^3+{z_2}^3)+{z_3}^3={(z_1+z_2)}^3-3{z_1}^2{z}_2-3z_1{z_2}^2+{z_3}^3$

Now, from equation (2), we can write,

$z_1+z_2=-z_3$

$\therefore ({z_1}^3+{z_2}^3)+{z_3}^3={(-z_3)}^3-3{z_1}^2{z}_2-3z_1{z_2}^2+{z_3}^3$

$\therefore ({z_1}^3+{z_2}^3)+{z_3}^3=-{z_3}^3-3z_1{z}_2(z_1+z_2)+{z_3}^3$

$\therefore ({z_1}^3+{z_2}^3)+{z_3}^3=-3z_1{z}_2(-z_3)$

$\therefore ({z_1}^3+{z_2}^3)+{z_3}^3=3z_1{z}_2{z}_3$

$\therefore {(cos\alpha +isin\alpha )}^3+{(cos\beta +isin\beta )}^3+{(cos\gamma +isin\gamma )}^3=3(cos\alpha +isin\alpha )(cos\beta +isin\beta )(cos\gamma +isin\gamma )$

By De-Moivre's theorem,

$\therefore {\left(e^{i\alpha }\right)}^3+{\left(e^{i\beta }\right)}^3+{\left(e^{i\gamma }\right)}^3=3\left(e^{i\alpha }\right)\left(e^{i\beta }\right)\left(e^{i\gamma }\right)$

$\therefore e^{3i\alpha }+e^{3i\beta }+e^{3i\gamma }=3e^{i(\alpha +\beta +\gamma )}$

$\therefore (cos3\alpha +isin3\alpha )+(cos3\beta +isin3\beta )+(cos3\gamma +isin3\gamma )=3[cos(\alpha +\beta +\gamma )+isin(\alpha +\beta +\gamma )]$

$\therefore (cos3\alpha +cos3\beta +cos3\gamma )+i(sin3\alpha +sin3\beta +sin3\gamma )=3cos(\alpha +\beta +\gamma )+3isin(\alpha +\beta +\gamma )$

Equating real parts on both the sides, we get,

$cos3\alpha +cos3\beta +cos3\gamma =3cos(\alpha +\beta +\gamma )$