Discussion Forum

If a=  $\cos \alpha +i \sin \alpha,b=\cos \beta+i \sin \beta$, $c=\cos \gamma+i \sin \gamma$ and $\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1$ then $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha- \beta)$  is equal to 

Answer:

Explanation:

Given, a=  $\cos \alpha +i \sin \alpha,b=\cos \beta+i \sin \beta$,

and  $c=\cos \gamma+i \sin \gamma$

 Now, $\frac{b}{c} =\frac{ \cos \beta+i \sin \beta}{\cos \gamma+i \sin \gamma} \times \frac{\cos \gamma-i \sin \gamma}{\cos \gamma-i \sin \gamma}$

$\Rightarrow$    $\frac{b}{c}= \cos (\beta-\gamma)+i \sin (\beta-\gamma)$ ....(i)

 Similarly, $\frac{c}{a}= \cos (\gamma-\alpha)+i \sin (\gamma-\alpha)$......(ii)

 and $\frac{a}{b}=\cos (\alpha-\beta)+i \sin (\alpha- \beta)$.....(iii)

 On adding Eqs.(i), (ii) and (iii), we get

 $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha+\beta)+i$

   $[(\sin (\beta-\gamma)+\sin(\gamma-\alpha)+ \sin (\alpha- \beta)]=1$

      $[ \because   \frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1]$

On equating  real parts  , we get

 $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha- \beta)$=1