# prove this...

• Jan 25th 2006, 01:53 PM
mathAna1ys!5
prove this...

sin x + cos x is greater than or equal to 1,
if x is greater or equal to 0 degrees and less than or equal to 90 degrees (between 0 and 90 deg)

what about tan x and cot x is greater/equal to 2.. what identities do I refer to? I think what I lack is which identities to use and when. I don't like this chapter much.
• Jan 25th 2006, 02:26 PM
ThePerfectHacker
Quote:

Originally Posted by mathAna1ys!5

sin x + cos x is greater than or equal to 1,
if x is greater or equal to 0 degrees and less than or equal to 90 degrees (between 0 and 90 deg)

what about tan x and cot x is greater/equal to 2.. what identities do I refer to? I think what I lack is which identities to use and when. I don't like this chapter much.

Notice for any $\displaystyle 0\leq x\leq 90$ we have that,
$\displaystyle \sin 2x\geq 0$
Because the sin function is non-negative in the first two quadrants.
Now add 1 to both sides thus,
$\displaystyle \sin 2x+1\geq 1$
Use identity $\displaystyle \cos^2x+\sin^2x=1$
Thus,
$\displaystyle \sin^2x+\cos^2x+\sin 2x\geq 1$
Use identity $\displaystyle \sin 2x=2\sin x\cos x$
Thus,
$\displaystyle \sin^2x+2\sin x\cos x+\cos^2x\geq 1$
Notice that the LHS is a perfect square,
Namely that, $\displaystyle \sin^2x+2\sin x\cos x+\cos^2x=(\sin x+\cos x)^2$
Thus,
$\displaystyle (\sin x+\cos x)^2\geq 1\geq 0$
You take the square root of both sides you can do this because these expressions are non-negative thus,
$\displaystyle \sin x+\cos x\geq 1$
Q.E.D.

You may think how did I ever come up with such a proof, well let me teach you how to do these types of problems. I first assumed that $\displaystyle \sin x+\cos x\geq 1$ and ended up with, $\displaystyle \sin 2x\geq 0$. But that is not a proof, thus, I worked backwards to this result.
• Jan 25th 2006, 02:48 PM
ThePerfectHacker
Next for problem 2,
instead of $\displaystyle 0\leq x\leq 90$ we have that,
$\displaystyle 0<x<90$ because otherwise the tan and cot are undefined :eek:

Thus, if $\displaystyle 0<x<90$
Then,
$\displaystyle 0<\sin 2x<1$
Because the sin is AT MOST 1, but not on this interval.
Thus, take the reciprocal of both sides and switch the arrow :eek:
Thus,
$\displaystyle \frac{1}{\sin 2x}>1$
Thus, multiply by two,
$\displaystyle \frac{2}{\sin 2x}>2$
Thus, by identities,
$\displaystyle \frac{2}{2\sin x\cos x}>2$
Thus, simply LHS,
$\displaystyle \frac{1}{\sin x\cos x}>2$
Thus, write LHS as,
$\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}>2$
Thus, by identities,
$\displaystyle \tan x+\cot x>2$
Q.E.D.
• Jan 26th 2006, 12:37 AM
ticbol
Quote:

Originally Posted by mathAna1ys!5

sin x + cos x is greater than or equal to 1,
if x is greater or equal to 0 degrees and less than or equal to 90 degrees (between 0 and 90 deg)

what about tan x and cot x is greater/equal to 2.. what identities do I refer to? I think what I lack is which identities to use and when. I don't like this chapter much.

Here is one way.

sinX +cosX >= 1 ----------(i)
sinx >= 1 -cosX
Square both sides,
sin^2(X) >= 1 -2cosX +cos^2(X) --------(ii)

In getting (ii), the sense/direction of the inequality sign is not reversed because there is no possibility that sinX or cosX would be negative numbers. The sinX or cosX here are always positive numbers because angle X is in the 1st quadrant, or 0 <= X <= 90deg, and sine and cosine there are always positive.

Continuing from (ii),
Since sin^2(X) = 1 -cos^2(X), then,
1 -cos^2(X) >= 1 -2cosX +cos^2(X)
Transfer them all to the lefthand side,
1 -cos^2(X) -1 +2cosX -cos^2(X) >= 0
2cosX -2cos^2(X) >= 0
2cosX >= 2cos^2(X)
Divide both sides by 2cosX, ----which is positive always here,
1 >= cosX

Therefore, proven.
Cosine of any angle in the 1st quadrant is minimum zero and maximum 1.0

------
Another one, not to disturb the RHS,
sinX +cosX >= 1
Square both sides,
sin^2(X) +2sinXcosX +cos^2(X) >= 1
[sin^2(X) +cos^2(X)] +2sinXcosX >= 1
1 +2sinXcosX >= 1
Which is true because the "1" in the LFS has an additional "2sinXcosX" more than the "1" at the RFS. And 2sinXcosX is either positive or zero.

--------------------------
tanX +cotX >= 2

sinX/cosX +cosX/sinX >= 2
Combine the two fractions in the LFS into one fraction only,
(sinX*sinX +cosX*cosX) /(sinX*cosX) >= 2
[sin^2(X) +cos^2(X)] /[(1/2)(2sinXcosX)] >= 2
[1] /[(1/2)*sin(2X)] >= 2 -------------sin(2X) = 2sinXcosX.
2 /sin(2x) >= 2
Multiply both sides by sin(2X),
2 >= 2sin(2X) ---------------------(iii)

The sense or direction of the inequality sign did not reverse because sin(2x) here is always positive. Angle 2X here is from 2(0) to 2(90deg), or from 0 to 180deg. The sine of any angle from 0 to 180deg is minimum zero and maximum 1.0

Continuing from (iii),
1 >= sin(2X)

Therefore, proven.
Again, the sine of any angle from 0 to 180deg is minimum zero and maximum 1.
• Jan 26th 2006, 10:33 AM
ThePerfectHacker
ticbol you are just making one mistake.
You are first assuming, for example, $\displaystyle \sin x+\cos x>1$. Then you arrive at $\displaystyle \sin 2x>0$ which is true. But that is not a proof! Because you are taking the CONVERSE of the statement. You need to work backward to prove this.

This is my 3:):) Post!!!
• Jan 26th 2006, 10:04 PM
ticbol
Quote:

Originally Posted by ThePerfectHacker
ticbol you are just making one mistake.
You are first assuming, for example, $\displaystyle \sin x+\cos x>1$. Then you arrive at $\displaystyle \sin 2x>0$ which is true. But that is not a proof! Because you are taking the CONVERSE of the statement. You need to work backward to prove this.

This is my 3:):) Post!!!

What converse? What not a proof? What backwards?

I don'ty know what you mean.

I proved it, and that is it. And if you don't get it, well,....

ticbol
• Jan 27th 2006, 10:58 AM
Rich B.
Hi:

Now please don't get mad when I say this but, I agree with P.Hacker. Your arguments are well thought out, and you demonstrate a good working knowledge but, the argument seems to possess one flaw (and it is indeed possible that I am simply failing to comprehend something). Your opening statement is that which you are attempting to prove. That is, you have hypothesized the conclusion. Why go any further then? You were finished the moment you put the period at the end of the opening line.

Further, your closing statement for proof #1 is that 1>=cos(x). You seem to go to a lot of work in order to draw this conclusion, which you could have stated without argument, and no one would have batted an eye. Cos(x)<=1 is a simple extention of the definition of the cosine function. I.e., Cos(x) in [-1,1] for all real x implies Cos(x)<=1. Now, these things having been said, on what basis does cos(x)<=1 imply sin(x) + cos(x)>= 1? Please explain.

I close with this sincere and respectfully intended remark. P.Hacker sent his last post because he wanted to help you, not to elicit hostility and disrespect. Moreover, no, a proof is not complete until it works for the competent reader. We call it a proof because it establishes concurrence among the readership. ...for what it's worth.

Regards,

Rich B.
• Jan 27th 2006, 11:42 AM
ThePerfectHacker
Let me explain with this example,
2=6

1)Assume $\displaystyle 2=6$
2)Thus, $\displaystyle 2-4=6-4$
3)Thus, $\displaystyle -2=2$
4)Thus, $\displaystyle (-2)^2=2^2$
5)Thus, $\displaystyle 4=4$
6)Thus, $\displaystyle 2=6$
WHAT HAPPENED? :eek:

Everything was good until I reached step 6, I took the converse of the statement.
Let me explain let:
$\displaystyle p\mbox{ be the statement 2=6 }$
$\displaystyle q\mbox{ be the statement 4=4 }$
Now, $\displaystyle q$ is true.
And I am trying to prove $\displaystyle p$.
Then according to you ticbol,
Since, $\displaystyle p\rightarrow q$ is true,
then $\displaystyle q\rightarrow p$ is true also.
But that is called taking the converse of the statement.
Which is not necessarility true (it can be but not always)
Thus, you flawed in the proof, sorry but that is the truth.
• Jan 27th 2006, 12:03 PM
ticbol
Quote:

Originally Posted by Rich B.
Hi:

Now please don't get mad when I say this but, I agree with P.Hacker. Your arguments are well thought out, and you demonstrate a good working knowledge but, the argument seems to possess one flaw (and it is indeed possible that I am simply failing to comprehend something). Your opening statement is that which you are attempting to prove. That is, you have hypothesized the conclusion. Why go any further then? You were finished the moment you put the period at the end of the opening line.

Further, your closing statement for proof #1 is that 1>=cos(x). You seem to go to a lot of work in order to draw this conclusion, which you could have stated without argument, and no one would have batted an eye. Cos(x)<=1 is a simple extention of the definition of the cosine function. I.e., Cos(x) in [-1,1] for all real x implies Cos(x)<=1. Now, these things having been said, on what basis does cos(x)<=1 imply sin(x) + cos(x)>= 1? Please explain.

I close with this sincere and respectfully intended remark. P.Hacker sent his last post because he wanted to help you, not to elicit hostility and disrespect. Moreover, no, a proof is not complete until it works for the competent reader. We call it a proof because it establishes concurrence among the readership. ...for what it's worth.

Regards,

Rich B.

I don't know what you guys are trying to say, or prove. Too many words, or run around words do not work for me. And I do not need any help.

What does cosX < 1 imply sinX +copsX >= 1?
If you cannot understand my "computations" or "way" how I arrived at that, then I cannot explain it any further more for you.

I am not hostile. I talk straight. No running around words to confuse.
Perfecthacker's example below, or somewhere, re 2=6, is not trigonometry, so no good explanation.

I don't know how you guys understand Math, I have my "own" understanding of it. If you cannot "see" it my "way", then I am not forcing you guys to see it my "way". Maybe the seekers----those seeking help here---would see it it my "way" or would be guided by it.

Is that hostility? Then you have not seen yet when I really argue.

Again, I do not need any help in Math.

ticbol
• Jan 28th 2006, 12:22 PM
Rich B.
Greetings:

Claim: Let
x lie in [0,90]. Then sin(x) + cos(x) >= 1.

Proof: Let
f be a function on the closed interval [0,90] such that
f(x) = sin(x) + cos(x) . Then f '(x) = cos(x) - sin(x). Letting f '(x) = 0, we identify local extrema for
f as follows. 0 = cos(x) - sin(x) -> cos(x) = sin(x). This equation has exactly one solution in [0,90], i.e., x=45. Hence fhas an extreme value at (45, f(45)) = (45, sqrt(2)). Now, because cos(x) > sin(x) for all x in [0,45), f '(x) = cos(x) - sin(x) > 0 and we conclude f is increasing across this interval. Similarly, cos(x) < sin(x) in (45, 90] implies f '(x) < 0 which in turn implies fdecreases in (45, 90]. Hence f(45) = sqrt(2) is a maximum. Finally, noting f(0) = f(90) = 1, we conclude 1 <= f(x) <= sqrt(2) for all x in [0,90]. Therefore sin(x)+cos(x)>=1 across the entire interval, and the proof is complete. ÿ

Claim: Let x lie in
(0,90). Then tan(x) + cot(x) >= 2.

While I omit much detail here, the proof follows similarly to that for #1. If we let
f(x) = tan(x) + cot(x), then f '(x) = sec^2(x) - csc^2(x). Letting f '(x) = 0 gives, again, x = 45. Thus f (45) = 1 + 1 = 2 is an extremum which, in this case, is determined to be a minimum. Because the only local extremum, (45, 2), is a minimum, it follows that f(x) >= 2 across the entire interval and our work is complete. You may wish to show that Lim f(x) = positive infinity as x tends toward both zero from the right and 90 from the left. However, such is not a technical necessity.

I hope this helps.

Regards,

Rich B.

• Jan 28th 2006, 02:44 PM
ThePerfectHacker
Quote:

Originally Posted by ticbol
I don't know what you guys are trying to say, or prove. Too many words, or run around words do not work for me. And I do not need any help.

What does cosX < 1 imply sinX +copsX >= 1?
If you cannot understand my "computations" or "way" how I arrived at that, then I cannot explain it any further more for you.

I am not hostile. I talk straight. No running around words to confuse.
Perfecthacker's example below, or somewhere, re 2=6, is not trigonometry, so no good explanation.

I don't know how you guys understand Math, I have my "own" understanding of it. If you cannot "see" it my "way", then I am not forcing you guys to see it my "way". Maybe the seekers----those seeking help here---would see it it my "way" or would be guided by it.

Is that hostility? Then you have not seen yet when I really argue.

Again, I do not need any help in Math.

ticbol

There is a common mistake people make when making proofs you did the same mistake.
"You assume something is true, you reach the conclusion that is definitely true. Then you cannot conclude that you hypothesis was true".

1)You assume $\displaystyle \sin x+\cos x\geq 1$
2)You arrive at $\displaystyle \sin 2x\geq 1$
Now statement (1) was your hypothesis, you assumed it was true. And statement (2) was definitely true. You cannot conclude that (1) is true, it is inconclusive.

It was demonstated with my faulty proof $\displaystyle 2=6$.
1)I assumed that $\displaystyle 2=6$
2)I arrived at $\displaystyle 4=4$
Now statement (2) was definitely true. I cannot conclude that statement (1) is true. Namely, that $\displaystyle 2=6$ it clearly does not work here.

This is called taking the converse of a statement.

Everything you did was good and well-done but it had a problem based on mathematical logic. My proof that is exatly like you only backwards is not based on the assumption of the true validity of the hypothesis statement.
• Jan 28th 2006, 10:31 PM
ticbol
Quote:

Originally Posted by ThePerfectHacker
There is a common mistake people make when making proofs you did the same mistake.
"You assume something is true, you reach the conclusion that is definitely true. Then you cannot conclude that you hypothesis was true".

1)You assume $\displaystyle \sin x+\cos x\geq 1$
2)You arrive at $\displaystyle \sin 2x\geq 1$
Now statement (1) was your hypothesis, you assumed it was true. And statement (2) was definitely true. You cannot conclude that (1) is true, it is inconclusive.

It was demonstated with my faulty proof $\displaystyle 2=6$.
1)I assumed that $\displaystyle 2=6$
2)I arrived at $\displaystyle 4=4$
Now statement (2) was definitely true. I cannot conclude that statement (1) is true. Namely, that $\displaystyle 2=6$ it clearly does not work here.

This is called taking the converse of a statement.

Everything you did was good and well-done but it had a problem based on mathematical logic. My proof that is exatly like you only backwards is not based on the assumption of the true validity of the hypothesis statement.

Whoa, my reply to this was deleted?

Touche!
• Jan 29th 2006, 02:22 AM
CaptainBlack
Quote:

Originally Posted by ticbol
Whoa, my reply to this was deleted?

Touche!

It is poor practice for a moderator to exercise their power to remove a post
replying to one of their own posts even if it is not entirely :D constructive.

I suggest that we consider this thread closed.

RonL
• Jan 29th 2006, 08:18 AM
ThePerfectHacker