If is an identity, then a_2 + a_4 =

JEE MainMathematicsModerateTrigonometry

If $\sin^{3}xsin3x = \sum_{r = 0}^{n}\mspace{2mu} a_{r}cosrx$ is
an identity, then a_2 + a_4 =

Show answer and solution

Given

a_{0} + a_{1}cosx + \ldots. + a_{n}cosnx

= \sin^{3}xsin3x

= \frac{1}{4}(3sinx - sin3x)sin3x

= \frac{3}{8}(2sin3xsinx) - \frac{1}{8}(1 - cos6x)

= - \frac{1}{8} + \frac{3}{8}cos2x - \frac{3}{8}cos4x + \frac{1}{8}cos6x

Comparing we get

a_{2} = \frac{3}{8},a_{4} = - \frac{3}{8}

\therefore a_{2} + a_{4} = 0

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

Scan any doubt from your notes and get a whiteboard explanation.