# Maclaurin Series of Sinx

In this tutorial we shall derive the series expansion of the trigonometric function sine by using Maclaurin’s series expansion function.

Consider the function of the form
$f\left( x \right) = \sin x$

Using $x = 0$, the given equation function becomes
$f\left( 0 \right) = \sin \left( 0 \right) = 0$

Now taking the derivatives of the given function and using $x = 0$, we have
$\begin{gathered} f’\left( x \right) = \cos x,\,\,\,\,\,\,\,\,\,\,f’\left( 0 \right) = \cos \left( 0 \right) = 1 \\ f”\left( x \right) = – \sin x,\,\,\,\,\,\,\,\,\,\,f”\left( 0 \right) = – \sin \left( 0 \right) = 0 \\ f”’\left( x \right) = – \cos x,\,\,\,\,\,\,\,\,\,\,f”’\left( 0 \right) = – \cos \left( 0 \right) = – 1 \\ {f^{\left( {{\text{iv}}} \right)}}\left( x \right) = \sin x,\,\,\,\,\,\,\,\,\,\,{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) = \sin \left( 0 \right) = 0 \\ {f^{\left( {\text{v}} \right)}}\left( x \right) = \cos x,\,\,\,\,\,\,\,\,\,\,{f^{\left( {\text{v}} \right)}}\left( 0 \right) = \cos \left( 0 \right) = 1 \\ {f^{\left( {{\text{vi}}} \right)}}\left( x \right) = – \sin x,\,\,\,\,\,\,\,\,\,\,{f^{\left( {\text{v}} \right)}}\left( 0 \right) = – \sin \left( 0 \right) = 0 \\ \cdots \cdots \cdots \; \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{gathered}$

Now using Maclaurin’s series expansion function, we have
$f\left( x \right) = f\left( 0 \right) + xf’\left( 0 \right) + \frac{{{x^2}}}{{2!}}f”\left( 0 \right) + \frac{{{x^3}}}{{3!}}f”’\left( 0 \right) + \frac{{{x^4}}}{{4!}}{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) + \cdots$

Putting the values in the above series, we have
$\begin{gathered}\sin x = 0 + x\left( 1 \right) + \frac{{{x^2}}}{{2!}}\left( 0 \right) + \frac{{{x^3}}}{{3!}}\left( { – 1} \right) + \frac{{{x^4}}}{{4!}}\left( 0 \right) + \frac{{{x^5}}}{{5!}}\left( 1 \right) + \cdots \\ \sin x = x – \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} – \frac{{{x^7}}}{{7!}} + \cdots \\ \end{gathered}$