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estro
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Is there any non geometrical way to proof this fact?
sinx+cosx>=1 for every x in [0,Pi/2]
sinx+cosx>=1 for every x in [0,Pi/2]
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estro said:Is there any non geometrical way to proof this fact?
sinx+cosx>=1
tiny-tim said: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
nicksauce said: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.
What obvious reason? nicksauce's argument is correct regarding the first quadrant.estro said:I think from obvious reasons your idea is wrong...nicksauce said: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.
D H said:What obvious reason? nicksauce's argument is correct regarding the first quadrant.
That said, his argument is not correct throughout. [itex]\sin x \cos x \ge 0[/itex] for the third quadrant as well. In that quadrant, however, [itex]\sin x + \cos x \le -1[/itex].
D H said:What obvious reason? nicksauce's argument is correct regarding the first quadrant.
That said, his argument is not correct throughout. [itex]\sin x \cos x \ge 0[/itex] for the third quadrant as well. In that quadrant, however, [itex]\sin x + \cos x \le -1[/itex].
estro said:From nicksauce's argument, we can't conclude sinx+cosx >=1 for x in [0,Pi/2].
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.Live2Learn said:Is nick's argument weak because he assumed the premiss is true, then deduced his conclusion on that assumption?
estro said:Is there any non geometrical way to proof this fact?
sinx+cosx>=1 for every x in [0,Pi/2]
nicksauce said: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.
Live2Learn said:Is nick's argument weak because he assumed the premiss is true, then deduced his conclusion on that assumption?
rock.freak667 said:I'd think to write sinx+cosx in the form Rsin(x+β) and then show what happens in the range [0,π/2]
Anonymous217 said: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.
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 said:True. I wasn't trying to provide a rigorous proof, ...
n1person said: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)
D H said: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.
estro said: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...
The inequality sinx+cosx≥1 means that the sum of the sine and cosine of any angle x in the interval [0,π/2] is greater than or equal to 1.
This inequality can be proved using the properties of sine and cosine functions and basic algebraic manipulations. We can also use geometric interpretations, such as the unit circle, to show that sinx+cosx≥1 for all values of x in [0,π/2].
For example, let's take x=π/4. In this case, sin(π/4)=√2/2 and cos(π/4)=√2/2. When we add these two values, we get √2/2+√2/2=√2, which is greater than 1. Therefore, sin(π/4)+cos(π/4)≥1, as required by the inequality.
Yes, there are several methods and strategies that can be used to prove this inequality. Some common approaches include using trigonometric identities, using the unit circle, and using the properties of sine and cosine functions. Each method may have its own advantages and disadvantages, so it is important to choose a method that suits the specific problem at hand.
Yes, this inequality can be extended to other intervals such as [0,2π] or [-π/2,π/2]. However, the proof may differ slightly depending on the specific interval. It is important to consider the properties of sine and cosine functions and the given interval when proving this inequality for other intervals.