If #sinx+siny=sqrt3 (cosy-cosx)# Find the value of #sin3x+sin3y#?

Question

18747 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-25T14:09:49

Given

#sinx+siny=sqrt3 (cosy-cosx)#

#2sin((x+y)/2)cos((x-y)/2)=sqrt3* 2sin((x+y)/2)sin((x-y)/2)#

#=>sin((x+y)/2)cos((x-y)/2)-sqrt3sin((x+y)/2)sin((x-y)/2)=0#

#=>sin((x+y)/2)[cos((x-y)/2)-sqrt3sin((x-y)/2)]=0#

So

#sin((x+y)/2)=0#

#=>(x+y)/2=0#

Now

#sin3x+sin3y#

#=2sin((3(x+y))/2)cos((3(x-y))/2)#

#=2sin(3*0)cos((3(x-y))/2)#

#=0#