Question

In Maclaurin series of $si{n}^{2}x$, the coefficient of the third term is?

• A. $3$
• B. $\frac{3}{2}$
• C. $\frac{2}{45}$
• D. $\frac{2}{65}$

Answer 1

The correct option is C $\frac{2}{45}$

We have to find the coefficient of third term in Maclaurin series of $si{n}^{2}x$.
The Maclaurin series is given by $f\left(x\right)={\sum }_{k=0}^{\infty }\frac{f^{\left(k\right)}\left(a\right)}{k!}{x}^k$ where $a=0$
We have $f\left(x\right)=si{n}^{2}x$
Since we have to find the coefficient of the third term, let us take $n=8$.
$\therefore f\left(x\right)\approx {\sum }_{k=0}^8\frac{f^{\left(k\right)}\left(0\right)}{k!}{x}^k$
$f^{\left(0\right)}\left(x\right)=si{n}^{2}x,\Rightarrow f^{\left(0\right)}\left(0\right)=0$
$f^{\left(1\right)}\left(x\right)=2sinxcosx,\Rightarrow f^{\left(1\right)}\left(0\right)=0$
$f^{\left(2\right)}\left(x\right)=-2si{n}^{2}x+2co{s}^{2}x,\Rightarrow f^{\left(2\right)}\left(0\right)=2$
$f^{\left(3\right)}\left(x\right)=-8cosxsinx,\Rightarrow f^{\left(3\right)}\left(0\right)=0$
$f^{\left(4\right)}\left(x\right)=8si{n}^{2}x-8co{s}^{2}x,\Rightarrow f^{\left(4\right)}\left(0\right)=-8$
$f^{\left(5\right)}\left(x\right)=32sinxcosx,\Rightarrow f^{\left(5\right)}\left(0\right)=0$
$f^{\left(6\right)}\left(x\right)=-32si{n}^{2}x+32co{s}^{2}x,\Rightarrow f^{\left(6\right)}\left(0\right)=32$
$f^{\left(7\right)}\left(x\right)=-128sinxcosx,\Rightarrow f^{\left(7\right)}\left(0\right)=0$
$f^{\left(8\right)}\left(x\right)=128si{n}^{2}x-128co{s}^{2}x,\Rightarrow f^{\left(8\right)}\left(0\right)=-128$
$\therefore f\left(x\right)\approx 0x^0+0x^1+\frac{2}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{-8}{4!}{x}^4+\frac{0}{5!}{x}^5+\frac{32}{6!}{x}^6+\frac{0}{7!}{x}^7+\frac{-128}{8!}{x}^8$
$\Rightarrow f\left(x\right)\approx x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\frac{1}{135}{x}^5$
Thus the coefficient of third term is $\frac{2}{45}$.