Why did this guy not post another trig question?

This is a calculus approach to the problem:

Let $f(x)=x\sqrt{1-x^2}$ . The maximum value of this function will be equal to $K$ , since by definition, a maximum point is where $f(x)\leq f(max)$ , for all $x$ .

$f'(x)=\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}$ .

Setting $f'(x)=0$ gives $2x^2=1$ after simplification, so the extrema are at $x=\pm\frac{1}{\sqrt{2}}$ .

As the square root term is positive, the maximum value must be achieved when $x$ is positive, so $x=\frac{1}{\sqrt{2}}$ is a maximum. This maximum is equal to $f(\frac{1}{\sqrt{2}})=\frac{1}{2}$ , so $K=\frac{1}{2}$ .

First , multiply by $2x$ : $2x\sqrt {1 - x^2} <= 2K$ . Now substitute $x = \sin(\theta)$ and utilise the pythagorean identity for sine to attain $2\sin(\theta)\cos(\theta) <= 2K$ .This is the double angle identity for sine, meaning that the above expression is equal to: $\sin(2\theta) <= 2K$ . Obviously the maximum value of the sine function is $1$ , so $K = 0.5$ . As a side note, equality occurs when $\theta = \frac {\pi}{4} , x = \frac {\sqrt {2}}{2}$