Question

What would be the expansion of \[\sin x\] in powers of x?

Answer 1

Hint: Before solving this question you should know about the Taylor series of a function and you can expand any function by this till an infinite sum of terms which are expressed in terms of the functions derivatives at a single Point. Mainly in most of the functions and the sum of its Taylor series are equal near this point.

Complete step by step answer:
In this question it is asked to expand the \[\sin x\] in power of x.
We know that the Taylor series about the point \[x=a\] is given by \[f\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+.......+\dfrac{{{f}^{n}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}\]
Now, from the Maclaurin series expansion we have, \[f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)\dfrac{\left( x \right)}{1!}+f''\left( 0 \right){{\dfrac{x}{2!}}^{2}}+.......+{{f}^{n}}\left( 0 \right)\dfrac{{{x}^{n}}}{n!}+......\]
According to out question we have \[f\left( x \right)=\sin x\]
So, the derivatives of \[\sin x=f\left( x \right)\];
\[f'\left( x \right)=\cos x,f''\left( x \right)=-\sin x,f'''\left( x \right)=-\cos x,.....\]
If we consider 0, then \[\sin 0=0\] all even powers of x will equal to zero in series expansion.
So, \[f\left( x \right)=\sin x=\cos \left( 0 \right)\dfrac{x}{1!}-\cos \left( 0 \right)\dfrac{{{x}^{3}}}{3!}+\cos \left( 0 \right)\dfrac{{{x}^{5}}}{5!}+.....\]
Now,
Since \[\cos 0=1\], which will reduce the series to \[f\left( x \right)=\sin x=\dfrac{x}{1!}-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}........\]
By this we can say that the series is infinite in odd powers of x with the alternating signs of plus and minus. So, it can be reduced till the last formula or the last step.
The Taylor series is defined at \[x=0\] to make the Maclaurin series and the Maclaurin series used to expand the function.
Further we can say that first we have to find Taylor expansion and then we find the Maclaurin series and then we expand to the function of any question asked.
So, the series if written as:
\[\sum\limits_{n=0\to \infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}}\]
Thus, we found the expansion of \[\sin x\] in powers of x and the expansion series can be written as:
\[\sum\limits_{n=0\to \infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}}\]

Note: For solving the expansion of any function you should know about to calculate the Maclaurin series. Because if the Maclaurin series is satisfied or formed then we can easily form the Taylor series and if it does not then you have to substitute any other value regarding that.