Question

If sin x + sin y = $\sqrt{3}$ (cos y − cos x), then sin 3x + sin 3y =
(a) 2 sin 3x
(b) 0
(c) 1
(d) none of these

Answer 1

We have,
sin x + sin y = $\sqrt{3}$ (cos y − cos x)

$$\begin{gathered} \Rightarrow 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=2\sqrt{3}\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right) \\ \Rightarrow 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)-2\sqrt{3}\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)=0 \\ \Rightarrow 2\sin \left(\frac{x+y}{2}\right)\left[\cos \left(\frac{x-y}{2}\right)-\sqrt{3}\sin \frac{x-y}{2}\right]=0 \\ \Rightarrow \sin \left(\frac{x+y}{2}\right)\left[\cos \left(\frac{x-y}{2}\right)-\sqrt{3}\sin \frac{x-y}{2}\right]=0 \\ \Rightarrow \sin \frac{x+y}{2}=0\mathrm{or},\cos \left(\frac{x-y}{2}\right)-\sqrt{3}\sin \left({\displaystyle \frac{x-y}{2}}\right)=0 \\ \Rightarrow \frac{x+y}{2}=0\mathrm{or},\tan \left({\displaystyle \frac{x-y}{2}}\right)=\frac{1}{\sqrt{3}}=\tan \frac{\mathrm{\pi }}{6} \\ \Rightarrow x=-y\mathit{}\mathrm{or},\frac{x-y}{2}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow x=-y\mathit{}\mathrm{or},x-y=\frac{\mathrm{\pi }}{3} \end{gathered}$$
$\mathrm{Case}-\mathrm{I}$
$\mathrm{When}x=-y$
$\mathrm{In}\mathrm{this}\mathrm{case},$
$\sin 3x+\sin 3y=\sin \left(-3y\right)+\sin 3y=-\sin 3y+\sin 3y=0$
$\mathrm{Case}-\mathrm{II}$
$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}x-y=\frac{\mathrm{\pi }}{3}$
$$\begin{gathered} \mathrm{or},3x=\mathrm{\pi }+3\mathrm{y} \\ \mathrm{So},\sin 3\mathrm{x}+\sin 3\mathrm{y}=\sin \left(\mathrm{\pi }+3y\right)+\sin 3y \\ =-\sin 3y+\sin 3y \\ =0 \end{gathered}$$