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Can I prove " For every $x$ $\in$ $[\frac{\pi}{2},\pi]$, $\sin(x)-\cos(x) $ $\geq$ $1$ " like this?

Proof For the sake of contradiction suppose $x$ $\in$ $[\frac{\pi}{2},\pi]$ for which $\sin(x)-\cos(x) $ $<$ $1$

When $x$ $\in$ $[\frac{\pi}{2},\pi]$. We know that $0\leq\sin(x)\leq1$ and $-1\leq\cos(x)\leq0$

So $0\leq\sin(x)-cos(x)\leq2$.

We have a contradiction. $\sin(x)-\cos(x) $ $<$ $1$ and $\sin(x)-\cos(x) $ $\leq$ $2$

Just_A_User
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Mathskid
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    How is it a contradiction? – Devansh Kamra Aug 09 '20 at 05:54
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    $2< 3$ and $2 \leq 5$ are not a contradiction. The first inequality is just better. – Lucas Resende Aug 09 '20 at 05:56
  • I suppose sin(x)-cos(x) < 1 but I get sin(x)-cos(x) <=2 in the last. So sin(x)-cos(x) can equal to 1.5 which greater than 1. It's a contradiction. Is it true? – Mathskid Aug 09 '20 at 05:57
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    But does it ever equal $1.5$? Maybe your bound isn't tight enough. You have achieved an inequality, but you haven't ensured any value inbetween is ever achieved. – PrincessEev Aug 09 '20 at 05:58
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    but that case is not guaranteed. You can also have $\sin x-\cos x<1$ (hypothetically) always and still satisfying the second inequality – Devansh Kamra Aug 09 '20 at 05:58

3 Answers3

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You can directly prove this:

Let $f(x)=\sin x-\cos x$ where $D_f=[\frac{\pi}{2},\pi]$

$$f'(x)=\cos x +\sin x$$ Now if $f'(x_0)=0$ for some $x_0$, then: $$\cos x_0+\sin x_0=0\Rightarrow \cos x_0=-\sin x_0$$

Now $f(x_0)=\sin x_0-\cos x_0=2\sin x_0$

Also $x_0\in D_f$

Thus the only $x_0\in D_f$ such that $\cos x_0=-\sin x_0$ is $x_0=\dfrac{3\pi}{4}$ $$\therefore f(x_0)=\sqrt{2}$$

Now since we have a single maxima here, the minimum value of the function must be exhibited at one of the end points.

$$f(\frac{\pi}{2})=f(\pi)=1$$ $\therefore f(x)\in [1,\sqrt{2}]$ which means that $\sin x-\cos x\geq 1 \space \forall\ x\in [\frac{\pi}{2}, \pi]$

Devansh Kamra
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Let $f(x) = \sin(x)-\cos(x)$. Then $f'(x)=\cos(x)+\sin(x)$. On $[\tfrac{\pi}{2},\pi]$, $f'(x) = 0$ if and only if $\cos(x)=-\sin(x)$ if and only if $\tan(x)=-1$. So $f'(x) = 0$ if and only if $x=3\pi/4$. Then the critical points (where possible extrema of $f$ are attained) are at the endpoints and at $3\pi/4$. We see that $f(x) =1$ at the endpoints and $f(3\pi/4) = \sqrt{2}$. Thus, $f([\tfrac{\pi}{2},\pi]) = [1,\sqrt{2}]$. This completes the proof.

By the way, $\sqrt{2}<1.5$ so $f(x)<1.5$.

Jared S.
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A quick solution :

Writing on $[\pi/2,\pi]$

$$\sin(x)-\cos(x)=\frac{2} {\sqrt{2}} \sin(x)\sin(\pi/4)- \cos(x)\cos(\pi/4)=-\sqrt{2}\cos(x+\pi/4)\geq 1$$

EDX
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