Maclaurin Series of Sinx

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

f\left(  x \right) = \sin x


Put x = 0, the given equation function becomes

f\left(  0 \right) = \sin \left( 0 \right) = 0


Now take derivatives of the given function and put 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 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}

Comments

comments