# Thread: Prove that sin(x)+cos(x) is equal or larger than 1

1. Hello all

I am trying to prove that

$sin(x)-cos(x)\geq 1$

For each x in the interval $[\frac{\pi }{2},\pi ]$

I tried doing it by contradiction, what I did was:

Assume

$sin(x)-cos(x)< 1$

Then I used the power of 2 on each side of the inequality and got:

$sin^{2}(x)-2sin(x)cos(x)+cos^{2}(x)<1$

which led me to

$0<2sin(x)cos(x)$

$cos(\frac{\pi }{2})=0$

I am not sure that what I did is correct or complete. Can you please check my proof and give me your opinion on the matter?

2. What you did upto the last but one step is perfect, but your showing the contradiction is not that correct, as $\cos (\frac{\pi}{2})=0$ is true only for one value, i.e. at $\pi/2$. The correct contradiction comes from the fact that $2\sin x \cos x\le0 \ \forall x\in[\frac{\pi}{2},\pi]$ from the non-positive nature of $\cos$ function.

Thank you. One question if I may.

Logically, if I need to show that for all point in the interval this statement is true, and I showed that there exists a point for which it is not, isn't it enough?

4. Originally Posted by Yankel
Logically, if I need to show that for all point in the interval this statement is true, and I showed that there exists a point for which it is not, isn't it enough?
Your conclusion only shows that the original statement must be true at $x=\frac{\pi}{2}$. What about the remainder of the interval? Does your conclusion tell us anything about that?

I see what you mean ! You are right !

I know notice that my proof is wrong !

$a<b \sim \rightarrow a^{2}<b^{2}$

My first step is wrong, isn't it?

by the way, is there a way of proving it without using contradiction?

If I know that sin(x) in this interval is between 1 to 0, and cos(x) between 0 to -1, can I claim that this is enough to be considered a proof?

7. Your proof is ok, it's your conclusion that needs work. See vidyarth's post (post #2).

As an alternative, write $\sin(x)-\cos(x)$ as $\sqrt2\sin\left(x-\frac{\pi}{4}\right)$ and note the properties of this sine function over the given interval.

8. Originally Posted by Yankel
I see what you mean ! You are right !

I know notice that my proof is wrong !

$a<b \sim \rightarrow a^{2}<b^{2}$

My first step is wrong, isn't it?
Yes, $a<b$ does not imply $a^2<b^2$ in generality. Consider $a=-4$ and $b=2$. I haven't looked at the problem, but if you want to use that line of thinking, you will need to consider the cases when (a) $a, b>0$, (b) $a, b <0$, (c), ...etc. In particular, for the case of (b), we have $a<b \implies a^2>b^2$.

9. Originally Posted by Rido12
Yes, $a<b$ does not imply $a^2<b^2$ in generality. Consider $a=-4$ and $b=2$. I haven't looked at the problem, but if you want to use that line of thinking, you will need to consider the cases when (a) $a, b>0$, (b) $a, b <0$, (c), ...etc. In particular, for the case of (b), we have $a<b \implies a^2>b^2$.
Yes; quite right. My apologies for my blunder.

10. We can use the triangle inequality.
\begin{tikzpicture}
[blue, ultra thick]
\draw (0,0) -- node[above right] {1} (-3,4) -- node[left] {$\sin x$} (-3, 0) -- node[below] {$-\cos x$} cycle;

\end{tikzpicture}

The triangle inequality says that the sum of two sides of a triangle is greater than or equal to the third side.
So:
$$\sin x + (-\cos x) \ge 1$$