If \( \sin x+\sin y=\sqrt3(\cos y-\cos x) \), then \( \sin3x+\sin3y \) is
Options:
(a) \(2\sin3x\)
(b) \(0\)
(c) \(1\)
(d) none of these
Solution:
Using identities,
\[
\sin x+\sin y
=
2\sin\frac{x+y}{2}\cos\frac{x-y}{2}
\]
and
\[
\cos y-\cos x
=
2\sin\frac{x+y}{2}\sin\frac{x-y}{2}
\]
Therefore,
\[
2\sin\frac{x+y}{2}\cos\frac{x-y}{2}
=
\sqrt3
\left(
2\sin\frac{x+y}{2}\sin\frac{x-y}{2}
\right)
\]
\[
\cos\frac{x-y}{2}
=
\sqrt3\sin\frac{x-y}{2}
\]
\[
\tan\frac{x-y}{2}
=
\frac1{\sqrt3}
\]
\[
\frac{x-y}{2}
=
30^\circ
\]
\[
x-y=60^\circ
\]
Now,
\[
\sin3x+\sin3y
=
2\sin\frac{3x+3y}{2}\cos\frac{3x-3y}{2}
\]
\[
=
2\sin\frac{3(x+y)}{2}\cos\frac{3(x-y)}{2}
\]
Since,
\[
x-y=60^\circ
\]
\[
=
2\sin\frac{3(x+y)}{2}\cos90^\circ
\]
\[
=0
\]
\[
\boxed{0}
\]
Correct option: (b)