Proof verification: $\forall x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) \geq 1$. Would someone be willing to verify the following proof by contradiction?
Theorem: $\forall x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) \geq 1$.
Suppose, for the sake of contradiction, that this statement is not true.  Then, $\exists x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) < 1$.
$sin(x) - cos(x) < 1 \Longrightarrow sin^2(x) - 2sin(x)cos(x) + cos^2(x) < 1$
$\Longrightarrow sin^2(x) + cos^2(x) < 1 + 2sin(x)cos(x) \Longrightarrow 0 < sin(x)cos(x)$
On the interval $(\frac{\pi}{2}, \pi), sin(x) > 0$, and $cos(x) < 0$.  Therefore, $sin(x)cos(x) < 0$, which is a contradiction.
 A: In order to square both sides of an equality  we need to make sure that both sides are positive.
Thus if you  mention that $ \forall x \in [\frac{\pi}{2}, \pi]$ ,   $$ 0< sin(x) - cos(x) < 1 \Longrightarrow sin^2(x) - 2sin(x)cos(x) + cos^2(x) < 1$$ Then your proof is flawless.  
A: I prefer the following approach: We know $$\sin(a+b)=\sin a\cos b+\cos a\sin b.$$ Look for $c$ such that $c\sin x-c\cos x=\sin(x+a)$ for some $a$. You need $c=\pm\sqrt2/2$. To make life easier, take $c>0$. This gives us $a=-\pi/4$, and $x+a\in[\pi/4,3\pi/4]$. Proceed from here.
A: Use that
$$\cos x=\sin \left( \frac{\pi}{2} -x\right)$$
and the sum-to-product formulas.
Notably from
$$\sin \theta - \sin \varphi = 2 \sin\left( \frac{\theta - \varphi}{2} \right) \cos\left( \frac{\theta + \varphi}{2} \right)$$
we obtain
$$\sin x - cos x=\sin x -\sin \left( \frac{\pi}{2} -x\right)=2\sin\left(x-\frac{\pi}{4}\right) \cdot\cos \frac{\pi}{4}=\sqrt2\cdot \sin\left(x-\frac{\pi}{4}\right) \geq 1$$
whic is true since
$$\frac{\pi}{2}\le x\le\pi\iff\frac{\pi}{4}\le x-\frac{\pi}{4}\le\frac{3\pi}{4}\implies \sin\left(x-\frac{\pi}{4}\right)\ge\frac{\sqrt2}{2} $$
