Question

1.    If acos3Ɵ + 3acosƟsin2Ɵ = m, asin3Ɵ + 3acos2ƟsinƟ = n, prove that (m+n)2/3 + (m-n)2/3 = 2a2/3

Answer 1

m+n=a(cos³θ+3cos²θsinθ+3cosθsin²θ+sin³θ)=a(cosθ+sinθ)³;

cosθ+sinθ=∛[(m+n)/a];

similarly m-n=a(cosθ-sinθ)³; cosθ-sinθ=∛[(m-n)/a].

(cosθ+sinθ)²+(cosθ-sinθ)²=2cos²θ+2sinθcosθ-2sinθcosθ+2sin²θ=2.

Therefore (∛[(m+n)/a])²+(∛[(m-n)/a])²=2,

multiply through by a^⅔: (m+n)^⅔+(m-n)^⅔=2a^⅔ QED