Find smallest k for which the inequality holds The smallest positive number $K$ for which the inequality
$|\sin^2 x  - \sin^2 y|  \le K|x-y|$ holds for all $x,y$ is  

Hint:
Here is sufficient to find the maximum absolute value of the derivative. For this function $2\sin x \cos x$ is $1$. This means that from Lagrange's theorem implies that $K = 1$ is suitable. A lower value will not work, if you take the number of $x, y$ is close to the point $\frac \pi{4}$, where the derivative is equal to $1$.
Addition:
$$
\sin^2 x - \sin^2 y = (\sin x - \sin y)\cdot (\sin x + \sin y) = \\
= 2\cdot\sin \frac{x-y}{2}\cdot \cos \frac{x+y}{2}\cdot 2\cdot\sin \frac{x+y}{2}\cdot \cos \frac{x-y}{2} = \sin(x-y)\cdot\sin(x+y)
$$
But $|\sin(x-y)|\le|x-y|$ and $|\sin(x+y)|\le1$
