Question

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

• A. True
• B. False

Answer 1

The correct option is A True

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

$z_2=cos\beta +isin\beta$

$z_3=cos\gamma +isin\gamma$

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

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

From the given condition,

$z_1+z_2+z_3=\left(0\right)+i\left(0\right)$

$\therefore z_1+z_2+z_3=0$ (1)

2) $\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=\frac{1}{cos\alpha +isin\alpha }+\frac{1}{cos\beta +isin\beta }+\frac{1}{cos\gamma +isin\gamma }$

$\therefore \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=\frac{1}{e^{i\alpha }}+\frac{1}{e^{i\beta }}+\frac{1}{e^{i\gamma }}$

$\therefore \frac{z_2{z}_3+z_1{z}_3+z_1{z}_2}{z_1{z}_2{z}_3}=e^{-i\alpha }+e^{-i\beta }+e^{-i\gamma }$

$\therefore \frac{z_2{z}_3+z_1{z}_3+z_1{z}_2}{z_1{z}_2{z}_3}=(cos\alpha -isin\alpha )+(cos\beta -isin\beta )+(cos\gamma -isin\gamma )$

$\therefore \frac{z_2{z}_3+z_1{z}_3+z_1{z}_2}{z_1{z}_2{z}_3}=(cos\alpha +cos\beta +cos\gamma )-i(sin\alpha +sin\beta +sin\gamma )$

$\therefore \frac{z_2{z}_3+z_1{z}_3+z_1{z}_2}{z_1{z}_2{z}_3}=\left(0\right)-i\left(0\right)$

$\therefore \frac{z_2{z}_3+z_1{z}_3+z_1{z}_2}{z_1{z}_2{z}_3}=0$

$\therefore z_2{z}_3+z_1{z}_3+z_1{z}_2=0$ (2)

3) ${(z_1+z_2+z_3)}^2={z_1}^2+{z_2}^2+{z_3}^2+2z_1{z}_2+2z_2{z}_3+2z_1{z}_3$

${(z_1+z_2+z_3)}^2={z_1}^2+{z_2}^2+{z_3}^2+2(z_1{z}_2+z_2{z}_3+z_1{z}_3)$

From equations (1) and (2), we get,

${\left(0\right)}^2={z_1}^2+{z_2}^2+{z_3}^2+2\left(0\right)$

$\therefore {z_1}^2+{z_2}^2+{z_3}^2=0$

$\therefore {(cos\alpha +isin\alpha )}^2+{(cos\beta +isin\beta )}^2+{(cos\gamma +isin\gamma )}^2=0$

$\therefore {cos}^2\alpha -{sin}^2\alpha +isin2\alpha +{cos}^2\beta -{sin}^2\beta +isin2\beta +{cos}^2\gamma -{sin}^2\gamma +isin2\gamma =0$

$\therefore cos2\alpha +cos2\beta +cos2\gamma +i(sin2\alpha +sin2\beta +sin2\gamma )=0+i\left(0\right)$

Comparing real and imaginary parts of both sides, we get,

$cos2\alpha +cos2\beta +cos2\gamma =0$

$sin2\alpha +sin2\beta +sin2\gamma =0$

Now,

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

$sin\alpha +sin\beta +sin\gamma =0$

Squaring and taking difference of both equations,

${(cos\alpha +cos\beta +cos\gamma )}^2-{(sin\alpha +sin\beta +sin\gamma )}^2=0$

$\therefore {cos}^2\alpha +{cos}^2\beta +{cos}^2\gamma +2cos\alpha cos\beta +2cos\beta cos\gamma +2cos\alpha cos\gamma -{sin}^2\alpha -{sin}^2\beta -{sin}^2\gamma -2sin\alpha sin\beta -2sin\beta sin\gamma -2sin\alpha sin\gamma =0$

$\therefore ({cos}^2\alpha -{sin}^2\alpha )+({cos}^2\beta --{sin}^2\beta )+({cos}^2\gamma -{sin}^2\gamma )+2[(cos\alpha cos\beta -sin\alpha sin\beta )+(cos\beta cos\gamma --sin\beta sin\gamma )+(cos\alpha cos\gamma -sin\alpha sin\gamma )]=0$

$\therefore cos2\alpha +cos2\beta +cos2\gamma +2[cos(\alpha +\beta )+cos(\beta +\gamma )+cos(\alpha +\gamma )]=0$

$\therefore 0+2[cos(\alpha +\beta )+cos(\beta +\gamma )+cos(\alpha +\gamma )]=0$

$\therefore cos(\alpha +\beta )+cos(\beta +\gamma )+cos(\alpha +\gamma )=0$