Question

If ${\sin }^n\alpha +{\sin }^n\beta +{\sin }^n\gamma =\frac{3}{2}$ and $\sum \cos (\alpha +\beta )=\lambda$ then the ordered pair $(\lambda ,n)$ is

• A. (0,2)
• B. (0,3)
• C. (2,3)
• D. (3,2)

Answer 1

The correct option is A (0,2)

$\because {\sin }^n\alpha +{\sin }^n\beta +{\sin }^n\gamma =\frac{3}{2}$ and $\sum \cos (\alpha +\beta )=\lambda$
then for $a=b=c=1$ we have
$\sum \cos \alpha =0=\sum \sin \alpha$ for
$\alpha ,\beta ,\gamma ,-\pi <\alpha ,\beta ,\gamma ,\le \pi$
$\therefore A+B+C=0$ and $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=0$
$\Rightarrow A+B+C=0$ and $AB+BC+CA=0$
$\Rightarrow A^2+B^2+C^2=0$ and $AB+BC+CA=0$
$\Rightarrow e^{i2\alpha }+e^{i2\beta }+e^{i2\gamma }=0$ and $e^{i(\alpha +\beta )}+e^{i(\beta +\gamma )}+e^{i(\gamma +\alpha )}=0$
Now
by equating real & imaginary parts we get $\Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =0$ and
$\cos (\alpha +\beta )+\cos (\beta +\gamma )+\cos (\gamma +\alpha )=0\Rightarrow 1-2{\sin }^2\alpha +1-2{\sin }^2\beta +1-2{\sin }^2\gamma =0$ and $\cos (\alpha +\beta )+\cos (\beta +\gamma )+\cos (\gamma +\alpha )=\lambda =0$
$\Rightarrow {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =\frac{3}{2}$ & $\lambda =0$ $\Rightarrow {\sin }^n\alpha +{\sin }^n\beta +{\sin }^n\gamma =\frac{3}{2}$
$\therefore n=2,\lambda =0$
$\therefore (\lambda ,n)=(0,2)$