If `cos^(3)x sin2x=sum_(n=1)^(n) a_(m)sinmx` is an identity in x, then which one of the following is in-correct?

Options

A. `a_(2)=0,a_(3)=3/8`
B. `a_(1)=1/2,n=6`
C. `a_(1)=1/4, n=5`
D. `suma_(m)=3/4`

Answer

B

Explanation

We have,
`cos^(3)x sin2x=((cos3x+3 cosx)/(4))sin2x`
`impliescos^(3)x sin2x=1/4(sin2x cos3x+3sin2xcosx)`
`impliescos^(3)x sin2x=1/8(sin2x cos 3x+6sin2x cosx)`
`impliescos^(3)x sin2x=1/8(sin 5xsinx+3 sin3x+3sinx)`
`thereforecos^(3)x sin2x=underset(m=1)overset(n)(sum)a_(n)sin mx`
`impliesa_(1)sinx+a_(2)sin2x+...+a_(n)sinnx`
`=1/4sinx+3/8sin 3x+1/8sin5x`
`impliesa_(1)=1/4,a_(2)=0,a_(3)=3/8,a_(4)=0,a_(5)=1/8`