If $sin x + sin y = \sqrt{3} (cos y - cos x)$ $sinx+siny=sqrt3 (cosy-cosx)$ Find the value of $sin 3 x + sin 3 y$ $sin3x+sin3y$ ?

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If $sin x + sin y = \sqrt{3} (cos y - cos x)$ $sinx+siny=sqrt3 (cosy-cosx)$ Find the value of $sin 3 x + sin 3 y$ $sin3x+sin3y$ ?

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Answer 1 · 2018-03-25T14:09:49

Given

$sin x + sin y = \sqrt{3} (cos y - cos x)$ $sinx+siny=sqrt3 (cosy-cosx)$

$2 sin (\frac{x + y}{2}) cos (\frac{x - y}{2}) = \sqrt{3} \cdot 2 sin (\frac{x + y}{2}) sin (\frac{x - y}{2})$ $2sin((x+y)/2)cos((x-y)/2)=sqrt3* 2sin((x+y)/2)sin((x-y)/2)$

$\Rightarrow sin (\frac{x + y}{2}) cos (\frac{x - y}{2}) - \sqrt{3} sin (\frac{x + y}{2}) sin (\frac{x - y}{2}) = 0$ $=>sin((x+y)/2)cos((x-y)/2)-sqrt3sin((x+y)/2)sin((x-y)/2)=0$

$\Rightarrow sin (\frac{x + y}{2}) [cos (\frac{x - y}{2}) - \sqrt{3} sin (\frac{x - y}{2})] = 0$ $=>sin((x+y)/2)[cos((x-y)/2)-sqrt3sin((x-y)/2)]=0$

So

$sin (\frac{x + y}{2}) = 0$ $sin((x+y)/2)=0$

$\Rightarrow \frac{x + y}{2} = 0$ $=>(x+y)/2=0$

Now

$sin 3 x + sin 3 y$ $sin3x+sin3y$

$= 2 sin (\frac{3 (x + y)}{2}) cos (\frac{3 (x - y)}{2})$ $=2sin((3(x+y))/2)cos((3(x-y))/2)$

$= 2 sin (3 \cdot 0) cos (\frac{3 (x - y)}{2})$ $=2sin(3*0)cos((3(x-y))/2)$

$= 0$ $=0$

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