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Question

If acos3α+3acosαsin2α=m and asin3α+3acos2αsinα=n, then (m+n)23+(mn)23=

A
2a2
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B
2a13
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C
2a23
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D
2a3
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Solution

The correct option is C 2a23

Adding and subtracting the given relation,

we get (m+n)=acos3α+3acosαsin2α+3acos2αsinα+asin3α

=a(cosα+sinα)3

and similarly (mn)=a(cosαsinα)3

Thus, (m+n)23+(mn)23

=a23{(cosα+sinα)2+(cosαsinα)2}

=a23{2(cos2α+sin2α)}=2a23


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