If acos3α+3acosαsin2α=m and
asin3α+3acos2αsinα=n, Then (m+n)23+(m−n)23
is equal to
2a23
Adding and subtracting the given relation,
we get (m+n)=acos3α+3acosαsin2α
+ 3acos2α.sinα+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.