Question

If $a\sin \alpha +b\sin \beta +c\sin \gamma =0$ $=a\cos \alpha +b\cos \beta +c\cos \gamma$ where $a,b,c,\in R$ and $-\pi <\alpha ,\beta ,\gamma \le \pi .$

Then assume $A=e^{i\alpha },B=e^{i\beta },C=e^{i\gamma }$ where $\sqrt{-1}=i$ $\therefore aA+bB+cC=0$ and $\frac{a}{A}+\frac{b}{B}+\frac{c}{C}=0$, then we get $(aA{)}^3+(bB{)}^3+(cC{)}^3=3abcABC$

${\left(\frac{a}{A}\right)}^3+{\left(\frac{b}{B}\right)}^3+{\left(\frac{c}{C}\right)}^3=\frac{3abc}{ABC}$

If $\sin \alpha +2\sin \beta +3\sin \gamma =0,$ then value of $\alpha +\beta -\gamma$ is

• A. $-2\alpha$
• B. $-2\beta$
• C. $-2\gamma$
• D. 0

Answer 1

The correct option is C $-2\gamma$

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

$\therefore \sin (-\beta -\gamma )+2\sin (-\gamma -\alpha )+3\sin (-\alpha -\beta )=0$

$\therefore \sin (\beta +\gamma )+2\sin (\gamma +\alpha )+3\sin (\alpha +\beta )=0$ which is true $\alpha +\beta -\gamma =(\alpha +\beta )-\gamma =-\gamma -\gamma =-2\gamma$

Therefore, Answer is $-2\gamma$