Show that $f(x)=\cos x-\sin x$ is bounded Show that $$f(x)=\cos x-\sin x$$ is bounded.
We know that $$-1\le\cos x\le1\\{-1}\le\sin x\le1,$$ so I tried to subtract these two inequalities, but it seems that it doesn't work as we get: $$0\le\cos x-\sin x\le0$$
I also noticed: $f(x)=\cos x-\sin x\ge\cos x-(-1)=\cos x+2$ and $f(x)=\cos x-\sin x\ge\cos x - 1$.

You can write
$$\begin{cases}-1\le\cos x\le1\\{-1}\le\sin x\le1\end{cases}\\
\implies \begin{cases}-1\le\cos x\le1\\{1}\ge-\sin x\ge -1\end{cases}
$$
We see that it is not possible to sum these inequalities directly from side by side.
We can try the following way to subtract the two inequalities.
Let $a≥x≥b \wedge c≥y≥d$, then we have
\begin{align*}&\begin{cases} a≥x≥b\\ c≥y≥d \end{cases}\\
\implies &\max x-\min y ≥x-y≥ \min x-\max y \\
\implies & a-d≥x-y≥b-c \end{align*}
Hence you get,
$$-2\le \cos x-\sin x\le 2$$
However, you can also write
$${1}\ge-\sin x\ge -1\\
\implies
2\ge1+\cos x\ge\cos x-\sin x\ge \cos x-1\ge-2\\
\implies -2\le \cos x-\sin x\le 2$$
But, more rigorous way is:
$$-\sqrt {a^2+b^2}≤a\cos x+b\sin x≤\sqrt {a^2+b^2}$$
Because, $\cos x-\sin x=-2$ and $\cos x-\sin x=2$ is not possible.
