Question

If $\cos \alpha +\cos \beta +\cos \alpha =0=\sin \alpha +\sin \beta +\sin \alpha$

prove that

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\alpha =\frac{3}{\alpha }={\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\alpha$

Answer 1

$Ifcos\alpha +cos\beta +cos\gamma =sin\alpha +sin\beta +sin\gamma cos\alpha +cos\beta +cos\gamma =0........\left(1\right)sin\alpha +sin\beta +sin\gamma =0..........\left(2\right)Multiplying\left(2\right)\times \left(i\right)andaddingto\left(1\right)(cos\alpha +isin\alpha )+(cos\beta +isin\beta )+(cos\gamma +isin\gamma )Let{z}_1+z_2+z_3=0\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=(cos\alpha -isin\alpha )+(cos\beta -isin\beta )+(cos\gamma -isin\gamma )=0\therefore z_2{z}_3+z_3{z}_1+z_1{z}_2=0And{z}_1+z_2+z_3=0,Squaringbothside,z_1^2+z_2^2+z_3^2+2(z_1{z}_2+z_2{z}_3+z_3{z}_1)=0Or{z}_1^2+z_2^2+z_3^2=0Or{(cos\alpha +isin\alpha )}^2+{(cos\beta +isin\beta )}^2+{(cos\gamma +isin\gamma )}^2=0Orcos2\alpha +isin2\alpha +cos2\beta +isin2\beta +cos2\gamma +isin2\gamma =0Or2{\cos }^2(\alpha -1)=1-2{\sin }^2\alpha =cos2\alpha \therefore Likewise,{\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma ={\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =\frac{3}{2}$