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

• MHB
• Yankel
In summary, your proof is wrong because you assumed that sin(x) in the given interval is between 1 and 0 which is not true.
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.

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?

Yankel said:
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?

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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?

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.

Yankel said:
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 $a<b \implies a^2>b^2$.

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Rido12 said:
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.

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:
$$\sin x + (-\cos x) \ge 1$$​

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

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

Subject to the constraint:

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

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

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

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

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

On the given domain, this gives us:

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

Putting this into the constraint, we find:

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

Thus, the objective function at this critical point is:

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

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

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

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

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

$$\displaystyle f_{\min}=1$$

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

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:
$$1 \le \sin x - \cos x \le \sqrt 2 \quad\text{ when } x\in [\frac\pi 2,\pi]$$

## 1) What is the proof for sin(x)+cos(x) ≥ 1?

The proof for sin(x)+cos(x) ≥ 1 involves using the Pythagorean identity and basic algebraic manipulation. First, we can write sin(x)+cos(x) as (√(1-sin^2(x)))+cos(x). Then, using the Pythagorean identity sin^2(x)+cos^2(x)=1, we can substitute (√(1-sin^2(x)))+cos(x) as (√(cos^2(x)))+cos(x). Simplifying, we get (√(cos^2(x))+cos(x)) ≥ 1. Since cos(x) is always equal to or greater than 0, this inequality holds true and thus sin(x)+cos(x) is equal to or larger than 1.

## 2) How can we graphically visualize sin(x)+cos(x) ≥ 1?

To graphically visualize sin(x)+cos(x) ≥ 1, we can plot the points on a Cartesian plane. We know that the points (0,1) and (1,0) lie on the unit circle, which represents the values of sin(x) and cos(x) respectively. So, at any point on the unit circle, the sum of sin(x) and cos(x) will be equal to or larger than 1. This can be seen on the graph as the points on the unit circle always lie above or on the line y=1.

## 3) Does the inequality sin(x)+cos(x) ≥ 1 hold true for all values of x?

Yes, the inequality sin(x)+cos(x) ≥ 1 holds true for all values of x. As mentioned before, sin(x)+cos(x) can be rewritten as (√(cos^2(x))+cos(x)) ≥ 1. Since cos^2(x) is always equal to or less than 1, the square root will always yield a value equal to or greater than 1, which when added to cos(x) will always result in a value equal to or larger than 1.

## 4) Can you prove that sin(x)+cos(x) ≥ 1 using calculus?

Yes, we can prove that sin(x)+cos(x) ≥ 1 using calculus. We can take the derivative of sin(x)+cos(x) with respect to x, which results in cos(x)-sin(x). We know that the maximum value for cos(x) is 1 and the minimum value for sin(x) is -1. So, the maximum value for cos(x)-sin(x) is 1-(-1)=2. This means that the minimum value for sin(x)+cos(x) is 2, which is greater than or equal to 1.

## 5) How does the inequality sin(x)+cos(x) ≥ 1 relate to real-world applications?

The inequality sin(x)+cos(x) ≥ 1 has various real-world applications, especially in physics and engineering. For example, in alternating current (AC) circuits, the voltage and current can be represented by sine and cosine functions respectively. The sum of these two functions, sin(x)+cos(x), represents the total power in the circuit and is always equal to or greater than 1, ensuring that there is enough power to run the circuit. Additionally, this inequality is also used in trigonometric proofs and in solving optimization problems in mathematics.

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