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Topic: sin cos and cos sin (Read 14544 times) |
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BNC
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Re: sin cos and cos sin
« Reply #1 on: Jun 12th, 2004, 5:56am » |
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which goes to show that computer proofs have thei place: "solving" it graphically us very easy...
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Eigenray
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Re: sin cos and cos sin
« Reply #2 on: Jun 12th, 2004, 10:28pm » |
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First, we only need to consider x in [-[pi],[pi]] by periodicity.
Second, because sin(cos(x)) and cos(sin(x)) are both even, we only need to consider x in [0, [pi]].
If x is in [[pi]/2, [pi]], then sin(cos(x)) [le] sin(0) = 0 < cos(1) [le] cos(sin(x)).
So now suppose 0 < x < [pi]/2, and suppose sin(cos(x))=cos(sin(x)).
Let y = sin(x), z = cos(x), so that
0 < y,z < 1
sin(z) = cos(y) = sin(y+[pi]/2),
which means z = y+[pi]/2 +2n[pi], or z = [pi]-(y+[pi]/2) + 2n[pi], for some integer n.
In the first case,
-1 < z-y = [pi]/2 + 2n[pi] < 1,
which is impossible. In the second case,
0 < y+z = [pi]/2 + 2n[pi] < 2,
which is possible only for n=0, which means:
[pi]/2 = y+z = sin(x)+cos(x) = [sqrt]2 sin(x+[pi]/4),
which is impossible because [pi]/2 > [sqrt]2.
And by the intermediate value theorem, if the function sin(cos(x))-cos(sin(x)) is never 0, it's either always positive or always negative.
So to answer the question, cos(sin(x)).
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srn437
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Re: sin cos and cos sin
« Reply #3 on: Sep 16th, 2007, 4:16pm » |
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Also, cos(x)>sin(x). But what about sin2(x)+cos2(x) or cos4(x) + sin4(x) +2sin2(x)(cos2(x))?
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| « Last Edit: Sep 16th, 2007, 4:25pm by srn437 » |
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ThudnBlunder
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Re: sin cos and cos sin
« Reply #4 on: Sep 16th, 2007, 5:13pm » |
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on Sep 16th, 2007, 4:16pm, srn347 wrote:| Also, cos(x)>sin(x). But what about sin2(x)+cos2(x) ? |
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cos(pi/3) < sin(pi/3)
sin2x + cos2x = 1
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srn437
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Re: sin cos and cos sin
« Reply #5 on: Sep 16th, 2007, 5:51pm » |
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Check again. My calculator says otherwise about the pi/3 thing. That sin2(x)+cos2(x)=1 thing is true and in the question I asked, it is useful in finding the answer.
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ThudnBlunder
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Re: sin cos and cos sin
« Reply #6 on: Sep 16th, 2007, 6:01pm » |
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on Sep 16th, 2007, 5:51pm, srn347 wrote:| Check again. My calculator says otherwise about the pi/3 thing. |
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Change your setting from Degrees to Radians and we'll be on the same page, kid.
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srn437
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Re: sin cos and cos sin
« Reply #7 on: Sep 16th, 2007, 6:29pm » |
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It isn't that kind of calculator, but ok. How is that though? Shouldn't the series be the same? What about the other question I asked(which is higher:sin2(x)+cos2(x) or sin4(x)+cos4(x)+2sin2(x)cos2(x))?
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Aryabhatta
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Re: sin cos and cos sin
« Reply #8 on: Sep 16th, 2007, 9:04pm » |
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on Sep 16th, 2007, 6:29pm, srn347 wrote:| What about the other question I asked(which is higher:sin2(x)+cos2(x) or sin4(x)+cos4(x)+2sin2(x)cos2(x))? |
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Do you really think such a trivial question should be in medium?
One is the square of the other and both are 1.
Also, please don't use existing threads to post your questions, create new ones instead (and in the appropriate forum).
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srn437
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Re: sin cos and cos sin
« Reply #9 on: Sep 17th, 2007, 6:25am » |
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Once the question on a thread has been answered correctly, what else is there to do other than ask another question. Your answer was correct though.
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Barukh
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Re: sin cos and cos sin
« Reply #10 on: Sep 17th, 2007, 7:58am » |
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on Sep 17th, 2007, 6:25am, srn347 wrote:| Once the question on a thread has been answered correctly, what else is there to do other than ask another question |
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Incorrect. Please start a different thread if you want to ask a different question. You may supply a link to a relevant existing thread, if you wish.
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towr
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Re: sin cos and cos sin
« Reply #11 on: Sep 17th, 2007, 8:00am » |
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on Sep 17th, 2007, 6:25am, srn347 wrote:| Once the question on a thread has been answered correctly, what else is there to do other than ask another question. |
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If the new question is relevant to the topic, that's fine. Otherwise, it's better to leave the thread alone and ask the question in its own (and properly named) thread, so people that are interested in finding it can find it.
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Wikipedia, Google, Mathworld, Integer sequence DB
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srn437
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Re: sin cos and cos sin
« Reply #12 on: Sep 17th, 2007, 5:59pm » |
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The new question was relevant. What is radiens anyway?
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ThudnBlunder
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Re: sin cos and cos sin
« Reply #13 on: Sep 17th, 2007, 7:23pm » |
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on Sep 17th, 2007, 5:59pm, srn347 wrote:
All you paid was the looking price. Lessons are extra. - Edward G. Robinson
One minute you are telling professional mathematicians, engineers, and computer scientists what's what, and the next minute you are asking what 'radiens' are.
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| « Last Edit: Sep 18th, 2007, 9:13am by ThudnBlunder » |
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Sameer
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Re: sin cos and cos sin
« Reply #14 on: Sep 17th, 2007, 8:08pm » |
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Apparently a trigonometry class is in order!!  Whatever happened to ignoring?
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Aryabhatta
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Re: sin cos and cos sin
« Reply #15 on: Sep 17th, 2007, 8:17pm » |
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on Sep 17th, 2007, 8:08pm, Sameer wrote:| Whatever happened to ignoring? |
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I tried, but as there always seemed to be someone who responded, so I gave up.
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Aryabhatta
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Re: sin cos and cos sin
« Reply #16 on: Sep 17th, 2007, 8:19pm » |
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on Sep 17th, 2007, 5:59pm, srn347 wrote:| The new question was relevant. |
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How so?
If your answer is "trignometry" and "sin and cos", it is a bunch of bullsh*t.
The question you asked was not relevant and trivial to boot. You just polluted this thread. Sorry for the harsh words, but that is the truth.
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ThudnBlunder
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Re: sin cos and cos sin
« Reply #17 on: Sep 17th, 2007, 8:23pm » |
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on Sep 17th, 2007, 8:08pm, Sameer wrote:| Whatever happened to ignoring? |
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Ignore it is then. 
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Aryabhatta
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Re: sin cos and cos sin
« Reply #18 on: Sep 17th, 2007, 8:30pm » |
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on Sep 17th, 2007, 8:23pm, ThudanBlunder wrote:
Ignore it is then.
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Sorry. I will start ignoring again.
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srn437
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Re: sin cos and cos sin
« Reply #19 on: Sep 17th, 2007, 8:40pm » |
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Can any of you give me a straight answer?!
// resized fonts.
Oversized fonts will not make people pay more attention to you. It is simply annoying. --towr
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| « Last Edit: Sep 18th, 2007, 12:21am by towr » |
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Sameer
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Re: sin cos and cos sin
« Reply #20 on: Sep 17th, 2007, 8:45pm » |
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Come to think of this topic, Where is Nick nowadays? He is missing like Icarus...
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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srn437
Newbie
the dark lord rises again....
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Re: sin cos and cos sin
« Reply #21 on: Sep 17th, 2007, 9:01pm » |
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I wonder...
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ima1trkpny
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Re: sin cos and cos sin
« Reply #22 on: Sep 17th, 2007, 9:15pm » |
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on Sep 17th, 2007, 8:45pm, Sameer wrote:| Come to think of this topic, Where is Nick nowadays? He is missing like Icarus... |
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Nick's last post was in Jan... he's been gone even longer than Icarus (maybe he just decided 4600 was a nice even number of posts? I did actually check cause I was worried and there was no sign of him in the Wichita obits... so I assume he is fine and possibly either just busy or as I said before really likes that number)
Raven too has been missing for awhile...
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| « Last Edit: Sep 17th, 2007, 9:16pm by ima1trkpny » |
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mikedagr8
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Re: sin cos and cos sin
« Reply #23 on: Sep 18th, 2007, 12:30am » |
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Quote:| Apparently a trigonometry class is in order!! |
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Hehe, just had a triple today.
A radian is a measurement around a circle. Now for someone to have known calculus, YOU NEED TO KNOW THIS BACK TO FRONT!!! Luckily for me, I have just completed the topic, so I am well aware of what is required, you on the other hand, need to learn a few things srn347. The faster the better.
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SMQ
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Re: sin cos and cos sin
« Reply #24 on: Sep 18th, 2007, 6:10am » |
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on Sep 17th, 2007, 8:40pm, srn347 wrote:| Can any of you give me a straight answer?! |
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Sorry, we keep assuming you know how to use Wikipedia, Mathworld, or at least Google to answer simple questions for yourself. A radian is a measure of angle equal to the arc length of a corresponding segment of a circle with unit radius. Since the perimeter of a unit circle is 2 , 360o = 2 radians.
--SMQ
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--SMQ
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