Question

Expansion of sin x cos x in ascending powers of x is​

Answer 1

$\textbf{Given:}$

$\mathsf{f(x)=sinx\;cosx}$

$\textbf{To find:}$

$\textsf{Maclaurin series expansion of sinx cosx}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{f(x)=sinx\;cosx=\dfrac{2sinx\;cosx}{2}=\dfrac{1}{2}sin2x}$

$\mathsf{f(x)=\dfrac{1}{2}sin2x\;\implies\;f(0)=\dfrac{1}{2}sin(0)=0}$

$\mathsf{f^1(x)=\dfrac{1}{2}(cos2x)2=cos2x\;\implies\;f^1(0)=cos0=1}$

$\mathsf{f^2(x)=(-sin2x)2\;\implies\;f^2(0)=2(-sin0)=0}$

$\mathsf{f^3(x)=(-cos2x)2^2\;\implies\;f^3(0)=2^2(-cos0)=-4}$

$\mathsf{Now,}$

$\textsf{The Maclaurin series expansion of f(x) is}$

$\mathsf{f(x)=f(0)+\dfrac{f^1(0)}{1!}x^1+\dfrac{f^2(0)}{2!}x^2+\;.\;.\;.\;.\;.}$

$\mathsf{sinx\;cosx=0+\dfrac{1}{1!}x^1+\dfrac{0}{2!}x^2+\dfrac{(-4)}{3!}x^3+\;.\;.\;.\;.\;.}$

$\mathsf{sinx\;cosx=x^1-\dfrac{4}{1.2.3}x^3+\;.\;.\;.\;.\;.}$

$\implies\boxed{\mathsf{sinx\;cosx=x^1-\dfrac{2}{3}x^3+\;.\;.\;.\;.\;.}}$

$\textbf{Find more:}$

Expansion of sinX is ascending power of X​

https://brainly.in/question/28857947

Expansion of sinX is ascending power of X​

https://brainly.in/question/28857947