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I'm trying to prove that $|\sin(x)| \le 1$, $|\cos(x)| \le 1$ and $|\sin(x)| \le |x|$ for all $x \in \mathbb{R}$ using the power series of sine and cosine :

$$\begin{align*} \sin(x) &= \sum_{k=0}^{+ \infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}\\ \cos(x) &= \sum_{k=0}^{+ \infty} (-1)^{k} \frac{x^{2k}}{(2k)!} \end{align*}$$

Does anyone have an idea ? I've tried to find an upper bound for the partial sums :

$$\left|\sum_{k=0}^{N} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}\right| $$

but it seems difficult.

Thanks :-)

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Homework ? If so, see this post on meta discussing how to ask a homework question –  Sasha Nov 2 '11 at 18:31
    
Take the expansion of $e^{ix}$ and separate the real and imaginary parts, now consider the absolute value of $e^{ix}$. –  Dinesh Nov 2 '11 at 18:32
    
@Dinesh, I guess the OP will ask how to prove the absolute value of $e^{ix}$ is $1$ using only its power series. Back at square one. –  Did Nov 2 '11 at 18:35
    
@Didier Piau Then I lose. –  Dinesh Nov 2 '11 at 18:36

1 Answer 1

Hint: Using the power series, show that $\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\cos(x)$ and $\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)=-\sin(x)$. Using those, show that $\sin^2(x)+\cos^2(x)=1$.

Added much later: This hint has been explained in more detail in this answer.

Furthermore, by the mean value theorem $\left|\frac{\sin(x)}{x}\right|=\left|\frac{\sin(x)-\sin(0)}{x-0}\right|=|\cos(\xi)|\le1$ for some $\xi$ between $0$ and $x$.

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It's not a homework, I am not a student :-). But ok, thank you for the trick ! –  Louis Nov 2 '11 at 19:00
    
@Selim: I gave a hint, just in case, but I think the hint is pretty easy to implement. If you still have trouble, I will be glad to expand. –  robjohn Nov 2 '11 at 19:02
    
No, using the hint it's ok, thanks a lot :-) –  Louis Nov 2 '11 at 19:07

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