Q:

An inequality for $\cos$ and $\sin$

Let $\alpha, x,y\in\mathbb{R}$. Do the inequalities $$|\sin(\alpha x)-\sin(\alpha y)|\le \alpha|x-y|$$ and
$$|\cos(\alpha x)-\cos(\alpha y)|\le \alpha|x-y|$$ hold?

A:

No. Because $\alpha$ in your question can be negative.
However, if you assume $\alpha>0$, then use
\begin{align}
|\sin x -\sin y|= \left|2\sin\frac{x-y}{2}\cos\frac{x+y}{2}\right|\leq \left|2\sin\frac{x-y}{2}\right|\leq |x-y|,
\end{align}
which shows the first inequality.
Similarly,
\begin{align}
|\cos x -\cos y|= \left|2\sin\frac{x-y}{2}\sin\frac{x+y}{2}\right|\leq \left|2\sin\frac{x-y}{2}\right|\leq |x-y|,
\end{align}
which shows the second one.

A:

Your inequalities hold only if $\alpha\geq 0$
$\sin(\alpha x)-\sin(\alpha y)=2\cos\left(\dfrac{\alpha x+\alpha y}{2}\right)\sin\left(\dfrac{\alpha x-\alpha y}{2}\right)$
Therefore, $|\sin(\alpha x)-\sin(\alpha y)|=2\left|\cos\left(\dfrac{\alpha x+\alpha y}{2}\right)\sin\left(\dfrac{\alpha x-\alpha y}{2}\right)\right|\leq2\left|\sin\left(\dfrac{\alpha x-\alpha y}{2}\right)\right|\leq2\left|\dfrac{\alpha(x-y)}{2}\right|=|\alpha||x-y|$
Similarly, you can prove the second one using the identity:
$\cos(\alpha x)-\cos(\alpha y)=-2\sin\left(\dfrac{\alpha x+\alpha y}{2}\right)\sin\left(\dfrac{\alpha x-\alpha y}{2}\right)$

