If acos3α+3acosαsin2α = m and
asin3α+3acos2αsinα = n, Then (m+n)23+(m−n)23
is equal to
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.