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Question

If a cos​​​​3 theta + 3a cos theta sin​​​​2 theta = m and a sin 3 theta + 3a sin theta cos 2 = n prove that (m+n ) 2/3 +( m-n) 2/3 = 2a 2/3

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Solution

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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