Answer to Question #291195 in Calculus for shreyash

Question

prove that xcosecx=1+(x)^2/6+(7/360)*x^4

Answer 1

The Maclaurin series for $\sin x$ is

$$\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$$

Then

$$x\cosec x=\dfrac{x}{\sin x}=\dfrac{x}{x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...}$$

$$=\dfrac{1}{1-(\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+\dfrac{x^6}{7!}+...)}$$

The Maclaurin series for $\dfrac{1}{1-y}$ is

$$\dfrac{1}{1-y}=1+y+y^2+y^3+..., -1<y<1$$

Then

$$\dfrac{1}{1-(\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+\dfrac{x^6}{7!}+...)}$$ $$=1+(\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+\dfrac{x^6}{7!}+...)$$

$$+(\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+\dfrac{x^6}{7!}+...)^2$$

$$+(\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+\dfrac{x^6}{7!}+...)^3+...$$

$$=1+\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+\dfrac{x^4}{3!3!}+\dfrac{x^6}{7!}-\dfrac{2x^6}{3!5!}+\dfrac{x^6}{3!3!3!}+...$$

$$=1+\dfrac{x^2}{6}+\dfrac{7x^4}{360}+...$$

Therefore

$$x\cosec x=1+\dfrac{x^2}{6}+\dfrac{7x^4}{360}+...$$