If acos3α+3acosαsin2α=m and asin3α+3acos2αsinα=n, then (m+n)23+(m−n)23=
Adding and subtracting the given relation,
we get (m+n)=acos3α+3acosαsin2α+3acos2αsinα+asin3α
=a(cosα+sinα)3
and similarly (m−n)=a(cosα−sinα)3
Thus, (m+n)23+(m−n)23
=a23{(cosα+sinα)2+(cosα−sinα)2}
=a23{2(cos2α+sin2α)}=2a23