Finding bounds for $\sin(x) \cos(x)$? I need help with the following:
How do you find the lower and upper bounds for:
$$\sin(x)\cos(x)$$
I have $-1\lt\sin(x)\cos(x)\lt1$ by first principles.
 A: $$|\sin(x)\cos(x)| = |\frac{1}{2}\sin (2x)| \le \frac{1}{2}$$ or,
$$-\frac{1}{2} \le \sin(x)\cos(x) \le \frac{1}{2}$$
Therefore upper bound for $\sin(x)\ cos(x)$ is $\frac{1}{2}$
A: $\sin(x)\cos(x)=\sin(2x)/2$ so its maximum is $1/2$.
A: Let $f(x)=\sin(x)\cdot \cos(x)$. For a calculus approach, solving $f'(x)=0$ will give the critical points of the function. $$f'(x)=\sin^2(x)-\cos^2(x)=\sin^2(x)-(1-\sin^2(x))=2\sin^2(x)-1=0$$ So we have $$\sin^2(x)=\frac{1}{2}$$ $$\sin(x)=\frac{\sqrt{2}}{2}$$ $$x=\frac{\pi}{4}$$
$f(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{1}{2}$ and this is indeed a maximum since the minimum is obviously 0. 
Another way to think of this problem is by imagining a rectangle with height $\sin(x)$ and length $\cos(x)$. So, the area of the rectangle would be f(x). It is well known that the area of a rectangle is maximized when the height equals the length. Therefore, f(x) is maximized when $\sin(x)=\cos(x)$. This implies that $x=\frac{\sqrt{2}}{2}$ which gives us $f(\frac{\sqrt{2}}{2})=\frac{1}{2}$.
A: $(\sin(x)-\cos(x))^2 \geq 0$ implies $ 2\sin(x)\cdot\cos(x)\leq \sin^2(x) + \cos^2(x)=1$ or $\sin(x)\cdot\cos(x) \leq 1/2 <1$
A: Fix $x$. Then one of $\sin(x)$ or $\cos(x)$ must be $\le{1\over \sqrt{2}}$. Meanwhile, the other is $\le1$. So their product is $\le{1\over \sqrt{2}}$.

EDIT: Note that you can refine this, with work: e.g. cut $[0, {\pi\over 2}]$ (I'm assuming WLOG that $0\le x\le{\pi\over 2}$) into four pieces, $I_1=[0, {\pi\over 6}]$, $I_2=[{\pi\over 6}, {\pi\over 4}]$, $I_3=[{\pi\over 4}, {\pi\over 3}]$, $I_4=[{\pi\over 3}, {\pi\over 2}]$. Then working by cases we get the upper bound of ${\sqrt{3}\over 4}$, which is significantly better than ${1\over\sqrt{2}}$. If we take the limit of this process, we get the actually best possible upper bound, ${1\over 2}$; it's a good exercise to go through this and understand why.
