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Question

If acos3θ+3asin2θcosθ=m and asin3θ+3asinθcos2θ=n, find (m+n)2/3+(mn)2/3=

A
2a2/3.
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B
2a4/3.
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C
a2/3.
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D
None of these
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Solution

The correct option is A 2a2/3.
We have m=acos3θ+3asin2θcosθ
& n=asin3θ+3asinθcos2θ
Now, (m+n)2/3+(mn)2/3
=[(acos3θ+3asin2θcosθ)+(asin3θ+3asinθcos2θ)]2/3+[(acos3θ+3asin2θcosθ)(asin3θ+3asinθcos2θ)]2/3
=a2/3[(sin3θ+cos3θ)+3asinθcosθ(sinθ+cosθ)]2/3+a2/3[(cos3θsin3θ)3asinθcosθ(cosθsinθ)]2/3
[Now as we know, a3+b3+3ab(a+b)=(a+b)3 & (a3b33ab(ab)=(ab)3]
So, =a2/3{[(cosθ+sinθ)3]2/3+[(cosθsinθ)3]2/3}
=a2/3[(cosθ+sinθ)2+(cosθsinθ)2]
=a2/3(cos2θ+2sinθcosθ+sin2θ+cos2θ2sinθcosθ+sin2θ)
=a2/3(1+1)
=2a2/3
So, (m+n)2/3+(mn)2/3=2a2/3

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