# Prove that $\sin(x) + \cos(x) \geq 1$ [closed]

$\forall x\in[0,\pi/2]: \sin{x}+\cos{x} \ge 1.$

I am really bad at trigonometric functions, how could I prove it?

• @Lovsovs $\sin^2 (x) + \cos^2 (x) = 1$, you mean? Commented Mar 9, 2016 at 17:54
• @DylanSp Yep, read it wrong... my brain is porridge apparently! Commented Mar 9, 2016 at 17:55
– user137731
Commented Mar 9, 2016 at 18:00

If you look at the trigonometric unit circle, you will see that $\sin \theta$ and $\cos\theta$ are the legs of a right triangle with hypotenuse=$1$ and using triangle inequality you get that their sum is greater than or equal to the hypotenuse.

• Wow. That is much easier than what I tried :) Commented Mar 9, 2016 at 18:09
• Hey, that's cool! +1 (Note, this approach does not work if you for example go into the second quadrant) Commented Mar 9, 2016 at 18:58
• @imranfat Yes, it cannot work because the statement $\sin x+ \cos x\ge 1$ is also not true for $x\in[\pi/2,\pi]$ Commented Mar 9, 2016 at 19:02

Hint Since $0 \leq \sin(x) , \cos(x) \leq 1$ you have $$\sin(x) \geq \sin^2(x) \\ \cos(x) \geq \cos^2(x) \\ \sin(x) +\cos(x) \geq \sin^2(x)+\cos^2(x) =1$$

• terse, smart and simple +1 Commented Mar 9, 2016 at 18:06
• Wow. That is much easier than what I tried :) Commented Mar 9, 2016 at 18:25
• Yeah, much quicker indeed... Commented Mar 9, 2016 at 18:57

A straight line is the shortest path between two points, so the hypotenuse of a right-angled triangle is shorter than the sum of its legs. Hence the required inequality.

Here is a way: $\sin x+\cos x=\sqrt{2}(\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4})=\sqrt{2}\sin(x+\frac{\pi}{4})$ So you need to show that $\sqrt{2}\sin(x+\frac{\pi}{4})$ is greather or equal to $1$ on your given inteval. Divide both sides by $\sqrt{2}$ and see what you get...

• inserted a forgotten bracket... Commented Mar 9, 2016 at 18:58

squaring one times and observe that $$\sin(x)^2+\cos(x)^2=1$$ we get $$\sin(2x)<0$$ which is impossible. If we assume that $$\sin(x)+\cos(x)<1$$ Sonnhard.

• So are you implying that there are no solutions? Commented Mar 9, 2016 at 18:00
• but not in the given interval!!! with $$0\le x\le \frac{\pi}{2}$$ Commented Mar 9, 2016 at 18:02
• @imranfat No, he is saying that the cross-term must be equal to or greater than zero, and since the rest equals one, the whole thing must be equal to or greater than one. Commented Mar 9, 2016 at 18:05
• @imranfat he is proving by contradiction
– Mat
Commented Mar 10, 2016 at 1:14
• And explaining as badly as one possible.
– Did
Commented Mar 10, 2016 at 6:43

Re-arrange to get $\cos(x) < 1 - \sin(x)$

We know that $\sin^2 (x) + \cos^2(x) = 1$ and therefor that $|\cos(x)| = \sqrt{1-\sin^2(x)}$

Now we want to know if $\sqrt{1-\sin^2(x)} < 1 - \sin(x)$

Now we square both sides: $1-\sin^2(x) < (1 - \sin(x))^2 = 1 - 2\sin(x) + \sin^2(x)$

Re-arranging again we get $0 < 2\sin^2(x) - 2\sin(x)$

In order for $\sin^2(x) > \sin(x)$ we would need a $\sin(x)$ value greater than 1. This is not possible, and therefor neither is the original form of the inequality $\cos(x) < 1 - \sin(x)$.

you can consider the function $f(x) = \sin{x} +\cos{x} -1$ then see that in $[0,\pi /4)$ , $f'(x) >0$

and in $(\pi/4, \pi/2], f'(x) <0$

now see what happens at $0$ and $\pi/2$

you can write $$\sin x + \cos x = \sqrt 2 \left(\sin x \cos (\pi/4) + \cos x\sin(\pi/4)\right) = \sqrt 2 \sin(x+\pi/4)$$ now use the fact that $\frac{ \sqrt 2} 2 \le \sin(x + \pi/4) \le 1 \text{ for } 0 \le x \le \pi/2$ to conclude $$1 \le \sin x + \cos x \le \sqrt 2 \text{ for } 0 \le x \le \pi/2.$$

Let $E=\sin(x)+\cos(x)$

Multiply and divide with $\sqrt2$.

\begin{align} E&=\sqrt2\left(\frac{\sin(x)}{\sqrt2} +\frac{\cos(x)}{\sqrt2}\right)\\ \implies E&= \sqrt2\sin\left(\frac \pi 4 + x\right)\\ \end{align}

Let $(\frac \pi4 + x)$ be $y$ then range of $y\in \left(\frac\pi4,\frac {3\pi}4\right)$

So $E= \sqrt2\sin(y)$.

Since max value of $\sin(y)$ is $1$ and min value is $\frac1{\sqrt2}$. Therefore $\sqrt2 \ge E \ge 1$.