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

Question

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

Answer 1

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

Answer 2

Using series, you can prove (by defining $\exp$, $\sin$ and $\cos$ as series) that

$$\forall x\in\Bbb C,e^{ix} = \cos x + i \sin x$$

$$\forall x \in \Bbb R, |e^{ix}|=1$$

And from there, it's easy to see that

$$|\sin x|\le |e^{ix}|=1$$

$$|\cos x|\le |e^{ix}|=1$$