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

[Yankel]

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

\[0<2sin(x)cos(x)\]

which is a contradiction since

\[cos(\frac{\pi }{2})=0\]

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?

Thank you in advance.

 

[vidyarth]

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.

 

[Yankel]

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?

 

[Greg]

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?

 

[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?

 

[Yankel]

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?

 

[Greg]

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.

 

[Dethrone]

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 $a<b \implies a^2>b^2$.

 

[Greg]

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.

 

[I like Serena]

We can use the triangle inequality.
[TIKZ][blue, ultra thick]
\draw (0,0) -- node[above right] {1} (-3,4) -- node

{$\sin x$} (-3, 0) -- node[below] {$-\cos x$} cycle;
[/TIKZ]

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

 

[MarkFL]

We could use Lagrange Multipliers...consider the objective function:

$$f(x,y)=\sin(x)-\sin(y)$$

Subject to the constraint:

$$g(x,y)=x+y-\frac{\pi}{2}=0$$

This leads to:

$$\cos(x)=\lambda=-\cos(y)$$

$$\cos(x)+\cos(y)=0$$

Using a sum to product identity, and dividing through by $2$, we obtain:

$$\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=0$$

On the given domain, this gives us:

$$x\pm y=\pi\implies x=\pi\pm y$$

Putting this into the constraint, we find:

$$\pi\pm y+y=\frac{\pi}{2}\implies y=-\frac{\pi}{4}\implies x=\frac{3\pi}{4}$$

Thus, the objective function at this critical point is:

$$f\left(\frac{3\pi}{4},-\frac{\pi}{4}\right)=\sqrt{2}$$

At the end-points of the given domain, we find:

$$f\left(\frac{\pi}{2},0\right)=1$$

$$f\left(\pi,-\frac{\pi}{2}\right)=1$$

And so we conclude that on the given domain, we have:

$$f_{\min}=1$$

$$f_{\max}=\sqrt{2}$$

 

[I like Serena]

We can also do a classical function analysis for$f(x)=\sin x - \cos x$ on the domain $[\frac\pi 2,\pi]$.
To find the extrema:
$$f'(x)=0 \quad\Rightarrow\quad \cos x + \sin x = 0 \quad\Rightarrow\quad \sin x=-\cos x
\quad\Rightarrow\quad \tan x=-1\quad\Rightarrow\quad x=\frac{3\pi}{4}$$
So we have one extremum at $x=\frac{3\pi}{4}$ with value $\sqrt 2$.
And we have 2 boundary extrema at $x=\frac{\pi}{2}$ respectively $x=\pi$, both with value $1$.
Therefore:
$$1 \le \sin x - \cos x \le \sqrt 2 \quad\text{ when } x\in [\frac\pi 2,\pi]$$