Question

$\frac{x}{sinx}$

Short answer

$\frac{x}{\sin \left(x\right)}=1+\frac{x^2}{6}+\frac{7x^4}{360}+\dots$

Step-by-step solution

Given information

The Maclaurin expansion of the function$\frac{x}{\sin x}$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$sin\left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

$(1+x{)}^{-1}=1-x+x^2-x^3+\dots$

Part-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series and evaluate.

$\sin \left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

$(1+x{)}^{-1}=1-x+x^2-x^3+\dots$

Part-B Step 3: Use the series to create an expression

Use the Maclaurin series to create the expression.

$\frac{x}{\sin \left(x\right)}=\frac{1}{\frac{\sin \left(x\right)}{x}}$

$\frac{\sin \left(x\right)}{x}=\frac{x-\frac{x^3}{6}+\frac{x^5}{120}+\dots }{x}$

$=1-\frac{x^2}{6}+\frac{x^4}{120}+\dots$

Continue to evaluate.

$\frac{x}{\sin \left(x\right)}=\frac{1}{\frac{\sin \left(x\right)}{x}}$

$=\frac{1}{1-\frac{x^2}{6}+\frac{x^4}{120}+\dots }={\left(1+\left[-\frac{x^2}{6}+\frac{x^4}{120}+\dots \right]\right)}^{-1}$

Part-C Step 3: Use the Maclaurin series to evaluate

Replace $x$with$\left[-\frac{x^2}{6}+\frac{x^4}{120}+\dots \right]$and evaluate the Maclaurin series.

${\left(1+\left[-\frac{x^2}{6}+\frac{x^4}{120}+\dots \right]\right)}^{-1}=1-\left[-\frac{x^2}{6}+\frac{x^4}{120}+\dots \right]+{\left[-\frac{x^2}{6}+\dots \right]}^2+\dots$

$\frac{x}{\sin \left(x\right)}={\left(1+\left[-\frac{x^2}{6}+\frac{x^4}{120}+\dots \right]\right)}^{-1}$

$\frac{x}{\sin \left(x\right)}=1+\frac{x^2}{6}+\frac{7x^4}{360}+\dots$

Thus, the final answer is$\frac{x}{\sin \left(x\right)}=1+\frac{x^2}{6}+\frac{7x^4}{360}+\dots$