Maclaurin series of sinx | Taylor series of sinx

The Maclaurin series expansion of sinx or the Taylor series expansion of sinx at x=0 is given as follows:

$\sin x= \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}$

$=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots$

Taylor Series Expansion of Sinx at x=0

We know that the Maclaurin series expansion of $\sin x$ or the Taylor series of a function $f(x)$ at $x=0$ is given by the following series:

$f(x)=\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!} x^n$ $\cdots (\star)$

Thus, we will follow the below steps to find the Taylor series of sin(x) at x=0.

Step 1: At first, we calculate the n-th derivatives of $f(x)=\sin x$ at the point x=0 for every n=0, 1, 2, 3, …, that is, we will find

$[f^{(n)}(x)]_{x=0}$

where $f^{(n)}$ denotes the n-th derivative of $f(x)$. The 0-th derivative of $f(x)$ is the functional value of $f(x)$ at $x=0$, that is, $f(0)$.

Step 2: Note that the derivative of sinx is cosx and the derivative of cosx is -sinx.

$f(x)=\sin x$ $\Rightarrow f(0)=\sin 0 =0$

$f'(x)=\cos x$ $\Rightarrow f'(0)=\cos 0=1$

$f”(x)=-\sin x$ $\Rightarrow f”(0)=-\sin 0=0$

$f”'(x)=-\cos x$ $\Rightarrow f”'(0)=-\cos 0=-1$

$f^{(4)}(x)=\sin x$ $\Rightarrow f^{(4)}(0)=\sin 0=0$

$f^{(5)}(x)=\cos x$ $\Rightarrow f^{(5)}(0)=\cos 0=1$

$\vdots$

Step 3: Now, by the above definition $(\star)$, the series for $f(x)=\sin x$ will be equal to

$f(x)=f(0)+xf'(0)+\dfrac{x^2}{2!}f”(0)$ $+\dfrac{x^3}{3!}f”'(0)$ $+\dfrac{x^4}{4!}f^{(4)}(0)+\dfrac{x^5}{5!}f^{(5)}(0)+\cdots$

$\therefore \sin x$ $=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots$

Conclusion:  Writing the above series in sigma notation, we obtain the Maclaurin series expansion of $\sin x$ which is

$\sin x= \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}$.

Note that the above series for sin x converges for all real values, that is, the radius of converges of sinx series is the interval (-∞, ∞).

Also Read: 

Derivative of 2^x

Derivative of pi

Derivative of sin4x

Derivative of e^1/x

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