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:
$\cos(\alpha x)-\cos(\alpha y)=-2\sin\left(\dfrac{\alpha x+\alpha y}{2}\right)\sin\left(\dfrac{\alpha x-\alpha y}{2}\right)$
A: Yes, as long as $\alpha \geq 0$. Both $\sin$ and $\cos$ are Lipschitz-continuous with constant $1$:
$$\sup_{x \in \mathbb{R}} |\sin'x| = 1 = \sup_{x \in \mathbb{R}} |\cos' x|.$$
However, if $f$ is $\kappa$-Lipschitz-continuous, then $x \mapsto f(\alpha x)$ is $(\kappa|\alpha|)$-Lipschitz-continuous
$$\Big|f(\alpha x) - f(\alpha y)\Big| \leq \kappa\Big|\alpha x - \alpha y\Big| = \kappa|\alpha|\cdot|x-y|.$$
Hence, 
$$\Big|\sin(\alpha x)-\sin(\alpha y)\Big| \leq |\alpha|\cdot|x-y|$$
and
$$\Big|\cos(\alpha x)-\cos(\alpha y)\Big| \leq |\alpha|\cdot|x-y|$$
which imply your formulas for any $\alpha = |\alpha|$, that is $\alpha \geq 0$.
I hope this helps ;-)
A: The only thing you should show is that it is true for $\alpha=1$. And this is geometrically obvious: imagine some angles $x$ and $y$ on the unit circle. $|\sin(x)-\sin(y)|$ is the distance between proections of $x$ and $y$ on $oY$ and is sure smaller than the distance between $x$ and $y$. And $|x-y|$ is the length of the arc between $x$ and $y$, which is larger then the distance between them. So, $|\sin(x)-\sin(y)|<|x-y|$. Same with cosines. The inequality obviously holds for any $\alpha\ge0$.
