If `cos^3xsin2x=sum_(r=0)^n a_xsin(r x),AAx in R` then

Options

A. `n=5,a_(1)=1//2`
B. `n=5,alpha_(1)=1//4`
C. `n=5,a_(2)=1//8`
D. `n=5,a_(2)=1//4`

Answer

B

Explanation

`cos^(3)xsin2x=cos^(2)x cos x sin 2x`
`=((1+cos 2x)/(2))(2sin2xcosx)/(2))`
`=(1)/(4)(1+cos 2x)(sin3x+sinx)`
`=(3)/(4)[sin3x+sinx+(1)/(2)(2sin 3xcos2x)`
`=(1)/(2)(2cos 2xsin x]`
`=(1)/(4)[sin3x+sinx+(1)/(2)(sin5x+sinx)+(1)/(2)(sin3x-sinx)]`
`=(!)/(4)[sinx+((3)/(2))sin3x+((1)/(2))sin5x]`
`rARr a_(1)=1//4,a_(3)=3//8,n=5`