Question

If $Cos\alpha +Cos\beta +Cos\gamma =0=Sin\alpha +Sin\beta +Sin\gamma$ prove that
$Co{s}^2\alpha +Co{s}^2\beta +Co{s}^2\gamma =\frac{3}{2}=Si{n}^2\alpha +Si{n}^2\beta +Si{n}^2\gamma$

Answer 1

Given that:

$Cos\alpha +Cos\beta +Cos\gamma =0=Sin\alpha +Sin\beta +Sin\gamma$

To prove:

$Co{s}^2\alpha +Co{s}^2\beta +Co{s}^2\gamma =\frac{3}{2}=Si{n}^2\alpha +Si{n}^2\beta +Si{n}^2\gamma$

Solution:

$Cos\alpha +Cos\beta +Cos\gamma =0=Sin\alpha +Sin\beta +Sin\gamma$

Let, $Z=e^{(\alpha +\beta +\gamma )}=0$

Therefore, $Cos\alpha +Cos\beta +Cos\gamma =0=Sin\alpha +Sin\beta +Sin\gamma$

or, $Z^2=0$ $(\because Z=0)$

or, $Z^2=e^{2(\alpha +\beta +\gamma )}=0$

or, $Z^2=e^{(2\alpha +2\beta +2\gamma )}=0$

or, $Cos2\alpha +Cos2\beta +Cos2\gamma =0=Sin2\alpha +Sin2\beta +Sin2\gamma$

or, $Cos2\alpha +Cos2\beta +Cos2\gamma =0$

or, $2Co{s}^2\alpha -1+2Co{s}^2\beta -1+2Co{s}^2\gamma -1=0$ $(\because Cos2\theta =2Co{s}^2\theta -1)$

or, $2(Co{s}^2\alpha +Co{s}^2\beta +Co{s}^2\gamma )=3$

or, $Co{s}^2\alpha +Co{s}^2\beta +Co{s}^2\gamma =\frac{3}{2}$

Similarly,

or, $Sin2\alpha +Sin2\beta +Sin2\gamma =0$

or, $1-2Si{n}^2\alpha +1-2Si{n}^2\beta +1-2Si{n}^2\gamma =0$ $(\because Cos2\theta =1-2Si{n}^2\theta )$

or, $2(Si{n}^2\alpha +Si{n}^2\beta +Si{n}^2\gamma )=3$

or, $Si{n}^2\alpha +Si{n}^2\beta +Si{n}^2\gamma =\frac{3}{2}$

Therefore, $Co{s}^2\alpha +Co{s}^2\beta +Co{s}^2\gamma =\frac{3}{2}=Si{n}^2\alpha +Si{n}^2\beta +Si{n}^2\gamma$