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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Suppose $\sin^2x\sin 3x=\sum\limits_{m=0}^{n}C_m\cos mx$ is an identity in $x$ where $C_0,C_1.......C_n$ are constants and $C_n\neq 0$ then the value of n is

$(a)\;6\qquad(b)\;5\qquad(c)\;4\qquad(d)\;3$

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$\sin^3x.\sin 3x=\sum\limits_{m=0}^{n}C_m\cos mx$
$\sin^3x\sin 3x=\large\frac{1}{4}$$[3\sin x-\sin 3x]\sin 3x$
$\qquad\qquad\;\;\;=\large\frac{1}{4}$$[\large\frac{3}{2}$$.2\sin x.\sin 3x-\sin^23x]$
$\qquad\qquad\;\;\;=\large\frac{1}{4}$$\big[\large\frac{3}{2}$$(\cos 2x-\cos x)-\large\frac{1}{2}$$(1-\cos x)\big]$
$\qquad\qquad\;\;\;=\large\frac{1}{8}$$\big[\cos 6x+3\cos 2x-3\cos x-1\big]$
Hence we observe that on LHS 6 is the max value of $m$
$\therefore n=6$
Hence (a) is the correct answer.
answered Oct 11, 2013 by sreemathi.v
 

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