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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 $acontradiction? Is there also a way to prove it without contradiction? Thank you.

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    Perhaps simpler is to consider that $\cos\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2}(\cos x - \sin x)$ and manipulate this expression a bit. – 2017-01-14
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    Just as a warning, squaring doesn't preserve inequality. $-2 < 1$ but $(-2)^2 > 1^2.$ – 2017-01-14
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    Exactly ! It doesn't preserve the inequality, and in order to preserve it, I assume the thing I am trying to prove ! – 2017-01-14

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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, $\sin x + (-\cos x) \geq 1.$

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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.

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    How can I square based on the assumption that sin(x)-cos(x)>=1 when this is what I am trying to prove in the first place? – 2017-01-14
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    @user2899944 When we square, we're assuming that $\sin x - \cos x \geq 0$. This is _not_ what we want to prove, but significantly weaker, and a lot easier to show, for instance by pointing out that $\sin x \geq 0$ and $0 \geq \cos x$. – 2017-01-14
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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?

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$$\sin x - \cos x\ge1\text{ for }x\in[\pi/2,\pi]$$

$$ \iff \sin (x+\pi/2) - \cos (x+\pi/2)\ge 1\text{ for }(x+\pi/2)\in[\pi/2,\pi]$$

$$\iff \cos y + \sin y \ge 1 \text{ for }y\in[0,\pi/2]$$

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!

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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.