Question

If sin x + sin y= root3 (cos y - cos x) then sin 3x + sin 3y is equal to

Answer 1

Answer:

solution of your question given below:

sinx+siny = root3 ( cosx-cosy)

2sin(x+y/2)cos(x-y/2) = root3 {2sin(x+y/2)sin(y-x/2)}

sin(x+y/2)[cos(x-y/2)-root3. sin(y-x/2)]=0

sin(x+y/2)= 0

x= -y.......................1

or cos(x-y/2)=root3. sin(y-x/2)

tan(x-y/2)=- 1/root3

then x-y/2 = - pie/6

x = - pie/3 +y...........2

putting the value of x from 1 & 2 in sin3x+sin3y we get the required answer

Answer 2

$\huge\sf\blue {GIVEN \: :-}$

$Sin x + sin y = \sqrt 3(cos y - cos)x$

$\huge\sf\red {TO \: FIND \: :-}$

$sin 3x + sin 3y = ?$

$\huge\sf\purple {SOLUTION \: :-}$ → $2 sin \bigg( \dfrac {x + y}{2} \bigg) cos \bigg ({x - y}{2} \bigg)\: = \: \sqrt 3$

→ $2 sin \bigg(\dfrac {x + y}{2} \bigg) cos \bigg ({x - y}{2} \bigg)$

→ $2 sin \bigg(\dfrac {x + y}{2} \bigg) cos \bigg ({x - y}{2}\bigg) \: = \: 0$

→ $sin \bigg(\dfrac {x + y}{2} \bigg) \bigg[cos \bigg(\dfrac {x - y}{2} \bigg) - \sqrt 3 sin \bigg(\dfrac {x - y}{2} \bigg) \bigg] \: = \: 0$

So,

→ $sin \bigg( \dfrac {x + y}{2} \: = \: 0 \bigg)$

→ $\dfrac {x + y}{2} \: = \: 0$

Now,

→ $sin 3x + sin 3y$

→ $2 sin \bigg( \dfrac {3(x+y)}{2} cos \bigg( 3 \dfrac {(x-y)}{2} \bigg)$

= 0.