If acos3θ+3acosθsin²θ=m, asin3θ+2acos²θsinθ=n, prove that
(m+n)2/3+(m-n)2/3=2a2/3.
Answers
(∛(m + n))² + (∛(m - n))² = 2 (∛a )²
Step-by-step explanation:
Correct Question is
m = aCos³θ + 3aCosθSin²θ
n = aSin³θ + 3aCos²θSinθ
m + n = aCos³θ + 3aCosθSin²θ + aSin³θ + 3aCos²θSinθ
= a (Cos³θ + Sin³θ) + 3aCosθSinθ(Cosθ + Sinθ)
using x³ + y³ = ( x + y)(x² + y² - xy)
= a(Cosθ + Sinθ)(Cos²θ + Sin²θ - CosθSinθ) + 3aCosθSinθ(Cosθ + Sinθ)
= a(Cosθ + Sinθ) (Cos²θ + Sin²θ - CosθSinθ + 3CosθSinθ)
= a(Cosθ + Sinθ) (Cos²θ + Sin²θ + 2CosθSinθ)
= a(Cosθ + Sinθ)(Cosθ + Sinθ)²
= a(Cosθ + Sinθ)³
=> m + n = a(Cosθ + Sinθ)³
∛(m + n) = ∛a ( Cosθ + Sinθ)
=> (∛(m + n))² = (∛a )² ( Cosθ + Sinθ)²
m - n = aCos³θ + 3aCosθSin²θ - aSin³θ - 3aCos²θSinθ
= a (Cos³θ - Sin³θ) - 3aCosθSinθ(Cosθ - Sinθ)
using x³ - y³ = ( x - y)(x² + y² + xy)
= a(Cosθ - Sinθ)(Cos²θ + Sin²θ + CosθSinθ) - 3aCosθSinθ(Cosθ - Sinθ)
= a(Cosθ - Sinθ) (Cos²θ + Sin²θ + CosθSinθ - 3CosθSinθ)
= a(Cosθ - Sinθ) (Cos²θ + Sin²θ - 2CosθSinθ)
= a(Cosθ - Sinθ)(Cosθ - Sinθ)²
= a(Cosθ - Sinθ)³
=> m - n = a(Cosθ - Sinθ)³
∛(m - n) = ∛a ( Cosθ - Sinθ)
=> (∛(m - n))² = (∛a )² ( Cosθ - Sinθ)²
(∛(m + n))² + (∛(m - n))² = (∛a )² ( Cosθ + Sinθ)² + (∛a )² ( Cosθ - Sinθ)²
= (∛a )² (Cos²θ + Sin²θ + 2CosθSinθ + Cos²θ + Sin²θ - 2CosθSinθ)
= (∛a )² (1 + 2CosθSinθ + 1 - 2CosθSinθ)
= (∛a )² (2)
= 2 (∛a )²
= 2 a ^(2/3)
(∛(m + n))² + (∛(m - n))² = 2 (∛a )²
QED
Proved
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