Question

expansion of sinX is ascending power of X​

Answer 1

$\textbf{Given:}$

$\textsf{Function is sinx}$

$\textbf{To find:}$

$\textsf{Series expansion of sinx}$

$\textbf{Solution:}$

$\textsf{Ler f(x)=sinx}$

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

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

$\mathsf{f^3(x)=-cosx\;\;\implies\;\;f^3(0)=-cos0=-1}$

$\mathsf{f^4(x)=sinx\;\;\implies\;\;f^4(0)=sin0=0}$

$\mathsf{f^5(x)=cosx\;\;\implies\;\;f^5(0)=cos0=1}$

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

$\mathsf{f(x)=f(0)+\dfrac{f^1(0)}{1!}x+\dfrac{f^2(0)}{2!}x^2+\dfrac{f^3(0)}{3!}x^3+.\;.\;.\;.\;.\;.\;.}$

$\mathsf{sinx=o+\dfrac{1}{1!}x+\dfrac{0}{2!}x^2+\dfrac{(-1)}{3!}x^3+\dfrac{0}{4!}x^4+\dfrac{1}{5!}x^5.\;.\;.\;.\;.\;.\;.}$

$\boxed{\mathsf{sinx=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\;.\;.\;.\;.\;.\;.}}$

$\textsf{This is the series expansion of sinx}$