How do I show that $ \sin x, \cos x$ really are in $ [-1,1]$ using series notion? 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

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
$$|\cos(x)|\leq 1 \ \ \forall x\in \mathbb R$$
$$|\sin(x)|\leq 1 \ \ \forall x\in \mathbb R$$
