Show that $f(x)=\cos x-\sin x$ is bounded

Question

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 $f(x)=\cos x-\sin x\ge\cos x - 1$.

Answer 1

In this way you can find "exact bound" (least upper bound) $$\cos(x)-\sin(x)=a$$ $$\frac{\cos(x)}{\sqrt{2}}-\frac{\sin(x)}{\sqrt{2}}=\frac{a}{\sqrt{2}}$$ $$\cos\left(\frac{\pi}{4}\right)\cos(x)-\sin\left(\frac{\pi}{4}\right)\sin(x)=\frac{a}{\sqrt{2}}$$ $$\cos\left(\frac{\pi}{4}+x\right)=\frac{a}{\sqrt{2}}$$ $$a=\sqrt{2}\cos\left(\frac{\pi}{4}+x\right)$$ $$-\sqrt{2} \leq a \leq\sqrt{2} $$

Answer 2

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.

Answer 3

Just apply the triangle inequality:

$$\lvert \cos x-\sin x\rvert\leq\lvert \cos x\rvert +\lvert \sin x\rvert\leq 2.$$

Answer 4

Never subtract two inequalities which are in the same direction It some time you get right result and wrong other time. So subtraction of inequalities is a spurious exercise as the subtraction is non-commutative. Instead one can add two inequalities which are in the same direction to get correct statement all the times. As $-1 \le \sin x \le 1, -1 \le \cos x\le 1$, we can write $\sin x \le 1$ and $-1\le \cos x$ you can add these two to get $$\sin x -1 \le 1+\cos x \implies \sin x-\cos x \le 2.....(1)$$ Similarly, we have $\sin x \ge -1 $ and $1 \ge \cos x$ adding these two. we get $$\sin x+1 \ge \cos x-1 \implies \sin x-\cos x \ge -2.....(2)$$