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I wanted to know, how can I derive:

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots$$

m0nhawk
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3 Answers3

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Let $f(x)$ and its derivative $f'(x),f''(x),...,f^{(n)}(x)$ exist and are continous on $[a,b]$ and suppose that $f^{(n+1)}(x)$ exists in $(a,b)$. Then we can have $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)(x-a)^n}{n!}+R_n$$

This fact, sometimes, is called Taylor's Theorem of the mean

Michael Hardy
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Mikasa
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  • Moreover, if for all $x$ and $\xi$ in $[a,b]$; we can see that the remainder is approaching zero then we can write $f(x)$ as follows: $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots $$In fact, as above, we have a Power series and this is what you are looking for. Now set $a=0$. – Mikasa Jul 22 '13 at 17:06
  • Why not put that comment in the answer? – Pedro Jul 22 '13 at 17:21
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    @PeterTamaroff: I didn't want Michael's edit removed. :-) – Mikasa Jul 22 '13 at 17:24
  • @BabakS. It needs double +. One for the answer and one for generosity. GOOD JOB MAN – Mahdi Khosravi Jul 22 '13 at 18:29
  • Well written! $\ddot\smile$ – amWhy Jul 22 '13 at 19:12
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Use Mclaren series : $f(x) = f(0)+f^{'}(0)\frac{x}{1!}+f^{''}(0)\frac{x^2}{2!}+f^{'''}(0)\frac{x^3}{3!}+\cdots$

for $f(x)=\sin x$

Shobhit
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One could use Euler's Theorem to get $$\cos x + i\sin x=e^{ix}=\sum_{k=0}^\infty \frac{(ix)^k}{k!}=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \right)$$ Now equating real and imaginary parts you will get the series expansion for both $\cos x$ and $\sin x$.

pritam
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