Question

if a cos3 theta+3 sin2theta costheta =m and a sin3theta+ 3a sintheta cos2theta = n , then prove that (m+n)2/3 + (m-n)2/3 =2a2/3

Answer 1

$m+n = a ( cos^3\theta+Sin^3\theta+3 sin\theta\ Cos\theta(cos\theta+sin\theta)) \\ \\ = a((cos\theta+sin\theta)(cos^2\thetea+sin^2\theta-sin\theta cos\theta)+3 sin\theta cos\theta(cos\theta+sin\theta)) \\ \\ = a (cos\theta+sin\theta)(cos^2\theta+sin^2\theta+2sin\theta\cos\theta)$

$m-n = a ( cos^3\theta-Sin^3\theta+3 sin\theta\ Cos\theta(sin\theta-cos\theta)) \\ \\ = a((cos\theta-sin\theta)(cos^2\thetea+sin^2\theta+sin\theta cos\theta)-3 sin\theta cos\theta(cos\theta-sin\theta)) \\ \\ = a (cos\theta-sin\theta)(cos^2\theta+sin^2\theta-2sin\theta\cos\theta) \\ \\ a (cos\theta - sin\theta)^3 \\ \\ (m+n)^{2/3} + (m-n)^{2/3} =$

$a^{2/3} ((cos\theta+sin\theta)^2+(cos\theta-sin\theta)^2) \\ \\ = a^{2/3} (1 + 2 sin\theta cos\theta + 1 - 2 sin\theta cos\theta ) \\ \\ 2a^{2/3}$

Answer 2

here's your answer..............