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
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m + n = acos3x + 3a cosx sin2x + asin3x + 3a cos2xsinx = a(sinx + cosx)3
(m + n)2/3 = a2/3 (sinx + cosx)2
m – n = acos3x + 3a cosx sin2x – asin3x – 3a cos2xsinx = a(cosx + sinx)3
(m – n)2/3 = a2/3 (cosx – sinx)2
(m + n)2/3 + (m – n)2/3 = a2/3 (sinx + cosx)2+ a2/3 (cosx – sinx)2 =
a2/3[(sinx + cosx)2 + (cosx – sinx)2] = 2a2/3
m + n = acos3x + 3a cosx sin2x + asin3x + 3a cos2xsinx = a(sinx + cosx)3
(m + n)2/3 = a2/3 (sinx + cosx)2
m – n = acos3x + 3a cosx sin2x – asin3x – 3a cos2xsinx = a(cosx + sinx)3
(m – n)2/3 = a2/3 (cosx – sinx)2
(m + n)2/3 + (m – n)2/3 = a2/3 (sinx + cosx)2+ a2/3 (cosx – sinx)2 =
a2/3[(sinx + cosx)2 + (cosx – sinx)2] = 2a2/3
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