Is there a shorter or more trivial way to prove that $ x > \cos (x)-\cos (2 x) $ holds for all $x>0$?

Question

I want to prove that the inequality

$$ x > \cos (x)-\cos (2 x) $$

holds for all $x>0$.

My attempt: Since the function on the RHS is periodic, we can find the position of extrema (on first period $2\pi$) by computing the first derivative; comparing them with the function $x$ which is increasing, we can prove that it holds.

Question Is there a shorter or more trivial/obvious way to prove this?

Answer 1

This can be seen geometrically.

Let $P = (\cos x, \sin x)$ and $Q = (\cos 2x, \sin 2x)$; then arc $PQ$ on the unit circle has length $x$. Its projection onto the horizontal axis is an interval containing both $\cos x$ and $\cos 2x$, and the length of that projection is always smaller, because arc $PQ$ is not perfectly horizontal and flat. Therefore $$ x > |{\cos x - \cos 2x}| $$ for all $x>0$, which implies the inequality we want.

Answer 2

Almost solution.
Mean value theorem. Note $\cos' = -\sin$. For $x>0$ we have $ 0 < x < 2x$ and $$ \frac{\cos(2x) - \cos(x)}{2x-x} = -\sin(\xi) $$ for some $\xi$ between $x$ and $2x$. Thus $$ \frac{\cos(2x) - \cos(x)}{x} \le 1 \quad\text{so}\quad \cos(2x)-\cos(x) \le x . $$ Why can we not have equality? For example, $\xi = 3\pi/2$?

Answer 3

If $x>2$, the statement is trivial. For $0<x\leq 2,$ one has $$x>2\sin\left(\frac x2\right)\geq 2\sin\left(\frac{3x}2\right)\sin\left(\frac x2\right)=\cos(x)-\cos(2x).$$

Answer 4

For $x > 0$ is $$ \cos(x)-\cos(2x) = \int_x^{2x} \sin(t) \, dt < \int_x^{2x} 1 \, dt = x \, . $$ Strict inequality between the integrals holds because $\sin(t) < 1$ for “almost all” $t$.