Prove sinx+cosx≥1 for x in [0,π/2]

[estro]

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

 

[tiny-tim]

Welcome to PF!

Hi estro! Welcome to PF!

Is there any non geometrical way to proof this fact?
sinx+cosx>=1

i] it's not true for 90º < x < 360º

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

 

[estro]

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

Oh, sorry I corrected my mistake.

 

[Anonymous217]

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

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

$$cosx = sinx$$

$$x = \frac{\pi}{4}\\$$
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.

 

[nicksauce]

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.

 

[estro]

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.

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

 

[mathman]

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

 

[D H]

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.

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

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, $\sin x + \cos x \le -1$.

 

[estro]

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, $\sin x + \cos x \le -1$.

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

 

[LCKurtz]

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, $\sin x + \cos x \le -1$.

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

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

Both results follow considering the signs in those two quadrants.

 

[Live2Learn]

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

 

[Anonymous217]

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

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.

 

[rock.freak667]

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

 

[tiny-tim]

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

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.

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

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)

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

see my first post ​

 

[nicksauce]

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.

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.

 

[D H]

True. I wasn't trying to provide a rigorous proof, ...

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.

 

[n1person]

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)

 

[Char. Limit]

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)

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.

 

[D H]

Exactly.

 

[estro]

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.

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...

 

[LCKurtz]

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...

It's a pity? It looks to me like you received several alternative ideas.