Question

If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =

 

विकल्प

• 2 sin 3x

• 0

• 1

• none of these

Answer 1

We have,
sin x + sin y = \[\sqrt{3}\] (cos y − cos x)
\[\Rightarrow 2\sin\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right) = 2\sqrt{3}\sin\left( \frac{x + y}{2} \right) \sin\left( \frac{x - y}{2} \right)\]
\[ \Rightarrow 2\sin\left( \frac{x + y}{2} \right)\cos\left( \frac{x - y}{2} \right) - 2\sqrt{3}\sin\left( \frac{x + y}{2} \right)\sin\left( \frac{x - y}{2} \right) = 0\]
\[ \Rightarrow 2\sin\left( \frac{x + y}{2} \right)\left[ \cos\left( \frac{x - y}{2} \right) - \sqrt{3}\sin\frac{x - y}{2} \right] = 0\]
\[ \Rightarrow \sin\left( \frac{x + y}{2} \right)\left[ \cos\left( \frac{x - y}{2} \right) - \sqrt{3}\sin\frac{x - y}{2} \right] = 0\]
\[ \Rightarrow \sin\frac{x + y}{2} = 0 \text{ or }, \cos\left( \frac{x - y}{2} \right)-\sqrt{3}\sin\left( \frac{x - y}{2} \right)=0\]
\[\Rightarrow\frac{x + y}{2}=0\text{ or },\tan\left( \frac{x - y}{2} \right)=\frac{1}{\sqrt{3}}=\tan\frac{\pi}{6}\]
\[\Rightarrow x=-y\text{ or },\frac{x - y}{2}=\frac{\pi}{6}\]
\[\Rightarrow x=-y\text{ or },x-y=\frac{\pi}{3}\]

Case - I
Where x = -y

In this case,
sin3x + sin3y = sin(-3y) + sin3y = - sin3y + sin3y = 0
Case - II
Where x - y = `pi/3`
or, \[ 3x = \pi + 3y\]
\[\text{So,} \sin 3x + \sin 3y = \sin\left( \pi + 3y \right) + \sin 3y\]
\[ = - \sin 3y + \sin 3y\]
\[ = 0\]