Proof by contradiction: $\sin(x)-\cos(x)\ge1$ I am trying to prove that $\sin(x)-\cos(x)\ge 1$ for every $x$ in the interval $[\frac{\pi}2,\pi]$.
I started by assuming that it is false, i.e. there exists an $x$ for which $\sin(x)-\cos(x)<1$. In the next step I got stuck, since I wanted to take the square of each side of the inequality, so I can get $[\sin(x)-\cos(x)]^2<1$, but this is not true, since $a<b$ doesn't imply $a^2<b^2$. Can you assist me to prove this statement by contradiction?  Is there also a way to prove it without contradiction? Thank you.
 A: Just for a different tack:  $x$ is a 2nd-quadrant angle, so $\sin x$ and $-\cos x$ are the (positive) lengths of the legs of the corresponding triangle of hypotenuse $1$.  So by triangle inequality, $\sin x + (-\cos x) \geq 1.$
A: I think squaring was a good idea. Note that
$$
\sin x - \cos x \geq 1\\
(\sin x - \cos x)^2 \geq 1\\
\sin^2 x -2\sin x \cos x + \cos^2x \geq 1\\
(\sin^2 x + \cos^2 x) - 2\sin x \cos x \geq 1\\
1 - 2\sin x\cos x \geq 1\\
-2\sin x \cos x \geq 0
$$
which is true as long as $\sin x$ and $\cos x$ have opposite signs (or one of them is zero), which is fulfilled on the interval $[\pi/2, \pi]$.
Now, we may have introduced additional solutions while squaring. However, since at the domain in question we clearly have $\sin(x) - \cos x \geq 0$, it is fine.
A: Avoid squaring as it immediately introduces extraneous solution(s) which deserves elimination.
Let $x=2y,\dfrac\pi2\le2y\le\pi\iff\dfrac\pi4\le y\le\dfrac\pi2$
$\sin2y=2\sin y\cos y$ will be $\ge\cos2y+1=2\cos^2y$
$\iff2\sin y(\cos y-\sin y)\ge0$
Now for $\dfrac\pi4\le y\le\dfrac\pi2,\sin y>0$
So, we need $\cos y-\sin y\ge0$
But $\cos y-\sin y=\sqrt2\sin\left(\dfrac\pi4-y\right)$ and $\dfrac\pi4\le y\le\dfrac\pi2$
Can you take it from here?
A: $$\sin x - \cos x\ge1\text{ for }x\in[\pi/2,\pi]$$
$$ \iff \sin (x+\pi/2) - \cos (x+\pi/2)\ge 1\text{ for }(x+\pi/2)\in[\pi/2,\pi]$$
$$\iff \cos y + \sin y \ge 1 \text{ for }y\in[0,\pi/2]$$
Now, $\cos,\sin$ are both positive on this interval, so squaring doesn't lose any information:
$$\iff (\cos y + \sin y)^2=1+2\cos y\sin y \ge 1 \text{ for }y\in[0,\pi/2]$$
$$\iff \sin 2y\ge 0 \text{ for }y\in[0,\pi/2]$$
after noting that $\sin 2\theta = 2\sin\theta\cos\theta$
Then:
$$\iff \sin 2y\ge 0 \text{ for }2y\in[0,\pi]$$
$$\iff \sin z\ge 0 \text{ for }z\in[0,\pi]$$
and, $\sin$ is $\ge0$ on this interval, so all of the statements above are true, and we're done!
A: Since you're over $[\pi/2,\pi]$, you have that $\sin x\ge0$; moreover $1+\cos x\ge0$, so it's legal to square $\sin x<1+\cos x$, getting
$$
\sin^2x<1+2\cos x+\cos^2x
$$
In other words, recalling that $1-\sin^2x=\cos^2x$,
$$
2\cos^2x+2\cos x>0
$$
and so
$$
\cos x(1+\cos x)>0
$$
which is false, because in the given interval, $\cos x\le0$ and $1+\cos x\ge0$.
For a direct proof, write $x=2y$, so you have
$$
\sin x-\cos x=2\sin y\cos y-\cos^2y+\sin^2y
$$
You have to prove
$$
2\sin y\cos y-\cos^2y+\sin^2y\ge 1=\sin^2y+\cos^2y
$$
for $y\in[\pi/4,\pi/2]$, that becomes
$$
\sin y\cos y-\cos^2y\ge0
$$
This is true for $y=\pi/2$, so we can look at $[\pi/4,\pi/2)$, where the inequality becomes
$$
\cos y(\tan y-1)\ge0
$$
which is true.
