TITLE: How do I show that $ \sin x, \cos x$ really are in $ [-1,1]$ using series notion?
QUESTION [5 upvotes]: I'm sorry to ask this question , but should ask it may help me to know more 
about series theory , It is well known that  $\cos x $ and $\sin x$  are 
represented  by alternative series  which hard for me to show wether sinx and 
cosx are in this range $[-1,1]$ .
My question here :Is there analytical proof show that $\sin x$ and $\cos x$ both are really in $[-1,1]$ using series theory if it is possible 
Note: I do not want to use geometric interpretation because it is standard at all 
Thank you for any help

REPLY [18 votes]: From the series is easy to prove that (using the fact that power series can be differentiated term by term inside the disk of convergence)
$$\frac{d}{dx} \cos(x) = -\sin(x)$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
If we define
$$F(x) := \cos^2(x) + \sin^2(x)$$
then
$F(0) = 1$ and
$$\frac{d}{dx}F(x) = 2\sin(x)\cos(x) - 2\cos(x)\sin(x)= 0\ \ \forall x\in \mathbb R$$
hence $F$ is constant
$$F(x) = \cos^2(x) + \sin^2(x) = 1 \ \ \forall x\in \mathbb R$$
From this is clear that
$$|\cos(x)|\leq 1 \ \ \forall x\in \mathbb R$$
$$|\sin(x)|\leq 1 \ \ \forall x\in \mathbb R$$