prove that sin3α+8sin3β+27sin3¥= 18sin(α+¥+β) and cos3α+8cos3β +27cos3¥=18cos(α+¥+β) if, sinα+2sinβ+3sin¥=0, cosα+2cosβ+3cos¥=0

                Bs Grewal

        Exercise question

  Here is the solution of a complex number

 Question:-:-if sinα+2sinβ+3sin¥=0, cosα+2cosβ+3cos¥=0 prove that sin3α+8sin3β+27sin3¥= 18sin(α+¥+β) and cos3α+8cos3β +27cos3¥=18cos(α+¥+β)

Solution:- let a= cosα+isinα  =  cisα

         b   =2( cosβ+ isinβ  )=  2cisβ

     c    = 3(cos¥ + isin¥ )=  3cis¥

      and given sinα+2sinβ+3sin¥=0,

         cosα+2cosβ+3cos¥=0

    then  a+b+c = cisα+2cisβ+3cis¥

   => a+b+c = (cosα+2cosβ+3cos¥)+i(sinα+2sinβ+3sin¥)

   => a+b+c = 0+i0=0

   => now (a+b+c)^3 = a^3+ b^3+ c^3+3(a+b+c)(ab+bc+ca)-3abc

   since,  (a+b+c)=0

   => (cisα)^3+(2cisβ)^3+(3cis¥)^3 =3 cisα cisβ cis¥

   =>Cis3α+ 8Cis3β+27 Cis3¥=18cisα cisβ cis¥

  => Cos3 α +isin3 α +8cos3 β +8isin3 β +27cos3¥  +27isin3¥ =18cis(α+ β+¥ )

  => Cos3 α+8cos3 β+27cos3¥  +i(sin3α+8sinβ+27sin3¥)=18(cos(α+β+¥)+isin(α+β+¥))

    On comparing real and imaginary part

    cos3 α+8cos3 β+27cos3¥  =18cos(α+β+¥)
   (sin3α+8sinβ+27sin3¥)= 18sin(α+β+¥)

     Hence proved

solve (x^2 - 4xy -2y^2)dx+(y^2-4xy -2x^2)dy=0