Discussion Forum : Trigonometric Functions
Question -


If acos3α+3acosαsin2α=m and


asin3α+3acos2αsinα=n, Then (m+n)23+(mn)23


is equal to

Options:
A .   2a2
B .   2a13
C .   2a23
D .   2a3
Answer: Option C
:
C

Adding and subtracting the given relation,


we get (m+n)=acos3α+3acosαsin2α


     + 3acos2α.sinα+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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