Proof by contradiction how to show is properly For every
$x \in \left[\pi/2,\pi\right]\,,\ \sin\left(x\right) − \cos\left(x\right) \geq 1$.
I have drawn the graph and can clearly see that A is true however how do I prove it correctly.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\sin\pars{x} - \cos\pars{x}&
=\sin\pars{x} - \tan\pars{\pi/4}\cos\pars{x}
={\sin\pars{x}\cos\pars{\pi/4}  - \sin\pars{\pi/4}\cos\pars{x}\over \cos\pars{\pi/4}}
\\[5mm]&=\root{2}\sin\pars{x - {\pi \over 4}}
\end{align}

Also, with $\quad\ds{\xi\ \in\ \bracks{{\pi \over 4},{3\pi \over 4}}}$:
  $$
\sin\pars{\xi} \geq \sin\pars{\pi \over 4} = {1 \over \root{2}}
$$

Then,
$$
\color{#66f}{\large\sin\pars{x} - \cos\pars{x}} \geq \root{2}\,{1 \over \root{2}}
=\color{#66f}{\Large 1}\,,\qquad x\ \in\ \bracks{{\pi \over 2},\pi}
$$
A: Let’s assume the contrary. For simplicity, I have changed the region of x to an open interval.
That is, there exists an $x ∈ (π/2, π)$, such that $\sin x − \cos x \lt 1$.
[Please skip the following line and continue from its next.]
Then, try some values for x in that range to see the contradiction. An example is x = 3π/4.

Then, $\sin x \lt 1 + \cos x$
For x belongs to the given region, $0 \lt \sin x$.
∴ $0 \lt sin x \lt 1 + \cos x$
Squaring, we have $\sin ^2 x \lt 1 + 2\cos x + \cos ^2 x$
Simplifying, we have $0 \lt \cos x(\cos + 1)$
Then, $0 \lt \cos x(\cos + 1) = (negative)(positive)$; for x lying in the given region
i.e. , $0 \lt \cos x(\cos + 1) \lt 0$, which is a contradiction.
