How to prove $|\sin x| \leq 1$, $|\cos x| \leq 1$ in a simplistic way without resorting to pictures? So my question is  looking at if there is a way to show in a less advanced way than this question:
Is it possible to prove $|\sin(x)| \leq 1$, $|\cos(x)| \leq 1$ and $|\sin(x)| \leq |x|$ algebraically? if it is possible to show these inequalities?
I ask because I'm working through Spivak's Calculus and it dawned on me that up to this point I haven't "shown" that $|\sin (x)| \leq 1$, $|\cos(x)| \leq 1$ are actually true. We've taken them as true, but I would've thought that since they appear to be "simple" expressions that we would've shown these properties at this point. This came up because I was trying to prove that $\cos(x)$ is continuous everywhere but the only idea I have of doing this is by using the mean value theorem which hasn't been addressed yet (I'm on Chapter 8: Least Upper Bound and Uniform Continuity).
 A: A rigorous way to define the trigonometric functions is to define $\cos$ on $[0, \pi]$ as the inverse function of
\begin{equation}
\arccos(x) = \int_x^1 \frac{1}{\sqrt{1-t^2}} dt
\end{equation}
This function $\cos$ is then extended to $[-\pi, +\pi]$ by parity and then to $\mathbb R$ by periodicity.
With this definition, $\cos(x)$ is necessarily in $[-1, 1]$ for all $x$.
A: This is usual high school stuffL first, those functions are defined as the fraction of one leg of a rught triangle over the hypotenuse. That's always less than $\;1\;$ . Then these functions are extended to the whole real numbers by means of the trigonometric circle (the canonical circle of radius one), and once again the absolute value of those functions is always less or equal$\;1\;$ ....that's all.
A: By Pythagore's Theorem and by definition, we have for all reals $ x$
$$\cos^2(x)+\sin^2(x)=1$$
$$\implies \cos^2(x)=1-\sin^2(x)\le 1$$
$$\implies -1\le \cos(x) \le 1$$
$$\implies |\cos(x)|\le 1.$$
A: As far as I remember, Spivak defines the sine as the solution of the differential equation with initial conditions
$$y''+y=0,\quad y(0)=0,\enspace y'(0)=1$$
and $\cos x=y'(x)$. You deduce instantly that
\begin{align}
(\sin^2x+\cos^2x)'&=(y(x)^2+y'(x)^2)'=2y(x)y'(x)+2y'(x)y''(x)\\
&=2\bigl(y(x)y'(x)-y'(x)y(x)\bigr)=0
\end{align}
so $\; \sin^2 x+\cos^2 x=\sin^20+\cos^20=1$, and consequently
$$\sin^2x,\:\cos^2x\le 1 \iff |\sin x|, \:|\cos x|\le 1.$$
