# How can you find the taylor expansion of sin x about x= 0?

Oct 6, 2015

The Taylor series formula is:

sum_(n=0)^N (f^((n))(a))/(n!)(x-a)^n

The Taylor series around $a = 0$ (not $x = 0$... the question is technically off) is also known as the Maclaurin series. You can write it then as:

sum_(n=0)^N (f^((n))(0))/(n!)x^n

= (f(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + (f''''(0))/(4!)x^4 + ...

So, you know you have to take some derivatives. $\sin x$ has cyclic derivatives that follow this pattern:

$\textcolor{g r e e n}{\sin x} = {f}^{\left(0\right)} \left(x\right) = \textcolor{g r e e n}{f \left(x\right)}$

$\frac{d}{\mathrm{dx}} \left[\sin x\right] = \textcolor{g r e e n}{\cos x = f ' \left(x\right)}$

$\frac{d}{\mathrm{dx}} \left[\cos x\right] = \textcolor{g r e e n}{- \sin x = f ' ' \left(x\right)}$

$\frac{d}{\mathrm{dx}} \left[- \sin x\right] = \textcolor{g r e e n}{- \cos x = f ' ' ' \left(x\right)}$

$\frac{d}{\mathrm{dx}} \left[- \cos x\right] = \textcolor{g r e e n}{\sin x = f ' ' ' ' \left(x\right)}$

Finally you can write the whole thing out, knowing that whenever $t r i g \left(0\right) = 0$, the whole term disappears. $\sin x$ appears in every even derivative. Hence, $f \left(x\right)$, $f ' ' \left(x\right)$, and every even derivative disappears.

You have only odd terms to worry about, and those are all just $1$ in the numerator and the signs alternate due to the alternating signs in front of $\cos x$.

=> cancel((f(0))/(0!)x^0) + (f'(0))/(1!)x^1 + cancel((f''(0))/(2!)x^2) + (f'''(0))/(3!)x^3 + cancel((f''''(0))/(4!)x^4) + ...

$= \cos \left(0\right) x + \frac{\left(- \cos \left(0\right)\right) {x}^{3}}{6} + \frac{\cos \left(0\right) {x}^{5}}{120} + \frac{\left(- \cos \left(0\right)\right) {x}^{7}}{5040} + \ldots$

$= \textcolor{b l u e}{x - {x}^{3} / 6 + {x}^{5} / 120 - {x}^{7} / 5040 + \ldots}$