# Sinx+cosx> = 1

1. Apr 29, 2010

### estro

Is there any non geometrical way to proof this fact?
sinx+cosx>=1 for every x in [0,Pi/2]

Last edited: Apr 29, 2010
2. Apr 29, 2010

### tiny-tim

Welcome to PF!

Hi estro! Welcome to PF!
i] it's not true for 90º < x < 360º

ii] you can prove non-geometrically that sinx + cosx = sin(x + 45º)/sin45º

3. Apr 29, 2010

### estro

Re: Welcome to PF!

Oh, sorry I corrected my mistake.

4. Apr 30, 2010

### Anonymous217

Re: sinx+cosx>=1

$$f(x) = sinx + cosx$$

$$f'(x) = cosx - sinx = 0$$

$$cosx = sinx$$

$$x = \frac{\pi}{4}\\$$
For values < pi/4, f'(x) = +, and for values >pi/4, f'(x) = -. This means at x=pi/4, there is a local maximum on the interval [0,pi/2]. Therefore, x=0 and x=pi/2 must be local minimums. We can then say that for all values x on the interval [0,pi/2]

$$f(0 or \frac{\pi}{2}) \le f(x) \le f(\frac{\pi}{4})\\$$
$$1 \le f(x) \le \sqrt{2}\\$$

I sort of just made this up off the spot.

5. Apr 30, 2010

### nicksauce

Re: sinx+cosx>=1

Well if you square both sides it's just

sin^2 + cos^x + 2sinxcosx >= 1

2sinxcosx >= 0

Which is obviously true in the first quadrant.

6. Apr 30, 2010

### estro

Re: sinx+cosx>=1

I think from obvious reasons your idea is wrong...

7. Apr 30, 2010

### mathman

Re: sinx+cosx>=1

Since 1 ≥ (sinx, cosx) ≥ 0 in the interval, sinx ≥ sin2x and cosx ≥ cos2x.
Therefore sinx + cosx ≥ sin2x + cos2x = 1.

Last edited: Apr 30, 2010
8. Apr 30, 2010

### D H

Staff Emeritus
Re: sinx+cosx>=1

What obvious reason? nicksauce's argument is correct regarding the first quadrant.

That said, his argument is not correct throughout. $\sin x \cos x \ge 0$ for the third quadrant as well. In that quadrant, however, $\sin x + \cos x \le -1$.

9. Apr 30, 2010

### estro

Re: sinx+cosx>=1

From nicksauce's argument, we can't conclude sinx+cosx >=1 for x in [0,Pi/2].

10. Apr 30, 2010

### LCKurtz

Re: sinx+cosx>=1

You can get both from nick's argument. For x in quadrant I or III:

2 sin x cos x ≥ 0
sin2(x) + 2 sin(x)cos(x) + cos2x ≥ 1
((sin(x) + cos(x))2 ≥ 1
|sin(x) + cos(x)| ≥ 1

11. Apr 30, 2010

### Live2Learn

Re: sinx+cosx>=1

Is nick's argument weak because he assumed the premiss is true, then deduced his conclusion on that assumption?

12. May 1, 2010

### Anonymous217

Re: sinx+cosx>=1

That's what I was thinking: that you can't subtract 1 from or square both sides unless the equation's true in the first place.

13. May 1, 2010

### rock.freak667

Re: sinx+cosx>=1

I'd think to write sinx+cosx in the form Rsin(x+β) and then show what happens in the range [0,π/2]

14. May 1, 2010

### tiny-tim

Why is almost everyone questioning nicksauce's proof?

It's perfectly valid in the first quadrant ([0,π/2]), which is all the original question asks for!!

(and it's simpler than my proof!)

(and as LCKurtz says, it can be adapted to cover the whole circle)
see my first post

Last edited: May 1, 2010
15. May 1, 2010

### nicksauce

Re: sinx+cosx>=1

True. I wasn't trying to provide a rigorous proof, but rather roughly sketching out the means for a direction by which the OP could arrive at a rigorous proof.

16. May 1, 2010

### D H

Staff Emeritus
Re: sinx+cosx>=1

Which is a good thing. We aren't supposed to solve the students problems for them here. We (me included) came a little too close to that in this thread.

nicksauce's argument is perfectly valid given the right circumstances. The OP needs to show that these circumstances apply here.

17. May 3, 2010

### n1person

Re: sinx+cosx>=1

Sorry if someone else presented the same argument in a way i couldn't understand:

(x+y)^2 >= x^2+y^2

so x+y >= (x^2+y^2)^(1/2)

and so sin(x)+cos(x) >= (sin^2(x)+cos^2(x))^(1/2)=1
so sin(x)+cos(x) >= 1

or do you consider the triangle inequality "geometric"? (which, by the name of "triangle", i would not begrudge you for :P)

18. May 3, 2010

### Char. Limit

Re: sinx+cosx>=1

Well, the first part of the equation is only true if...

$$(x+y)^2 \geq x^2 + y^2$$

$$x^2+2xy+y^2 \geq x^2+y^2$$

$$2xy \geq 0$$

So sin(x)+cos(x) >= 1 only holds true if both sin(x) and cos(x) are positive, I suppose.

19. May 3, 2010

### D H

Staff Emeritus
Re: sinx+cosx>=1

Exactly.

20. May 3, 2010

### estro

Re: sinx+cosx>=1

I hope you understand that I've already solved this problem before posting this question on PF.
And I never asked for a formal proof, I indeed asked for an alternative idea to what I've used in my geometrical based proof.
And still I couldn't figure out pure "calculus" approach to solving this problem.

It's a little pity that this thread was flooded with unclear algebra rather ideas...

Anyway thank you all very much...

Last edited: May 3, 2010