Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
Aaaboy97
|
  |
| Joined: 05 Apr 2009 |
| Total Posts: 6612 |
|
|
| 26 Dec 2011 08:28 PM |
oysi... confused with math?
*world ends* |
|
|
| Report Abuse |
|
|
shayan414
|
  |
| Joined: 11 Aug 2008 |
| Total Posts: 8090 |
|
|
| 26 Dec 2011 08:28 PM |
| I can see how they are equivalent, yet the ^2 threw me off of my argument. |
|
|
| Report Abuse |
|
|
|
| 26 Dec 2011 08:33 PM |
They aren't equal for me!
When x=0.5, the first is exactly 1, however, the second is 0.8535... |
|
|
| Report Abuse |
|
|
|
| 26 Dec 2011 08:35 PM |
| ...that is, when I calculate it. When I plug it in as a graph function, they both come out the same >-> |
|
|
| Report Abuse |
|
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
|
| 26 Dec 2011 08:50 PM |
"You're doing something wrong, poke"
I'm not, my calculator is. Damn you TI, your 84-Plus Silver Edition can't agree with itself! I got two separate answers at first, then I graphed it, and they became the same, and then I highlighted each and hit enter to recalculate them, and then they came out the same. Bipolar calculation much? This calculator really angers me sometimes, it can't even do symbolic differentiation ;-; |
|
|
| Report Abuse |
|
|
shayan414
|
  |
| Joined: 11 Aug 2008 |
| Total Posts: 8090 |
|
|
| 26 Dec 2011 08:51 PM |
| Rofl, bipolar calculation. That must trend. |
|
|
| Report Abuse |
|
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
stravant
|
  |
 |
| Joined: 22 Oct 2007 |
| Total Posts: 2893 |
|
|
| 26 Dec 2011 08:57 PM |
You're not stupid, it's a bit tricky of a problem. How I eventually got to the solution:
Start with: cos(2x) = cos^2(x) - sin^2(x) cos(2x) = (1 - sin^2(x)) - sin^2(x) [By cos^2 + sin^2 = 1] cos(2x) = 1 - 2sin^2(x) sin(2x + pi/2) = 1 - 2sin^2(x) [By cos = sin+pi/2] sin(2(x-pi/4) + pi/2) = 1 - 2sin^2(x - pi/4) [By replacing x with x-pi/4] sin(2x) = 1 - 2sin^2(x - pi/4) sin(x) = 1 - 2sin^2(x/2 - pi/4) [By replacing x with x/2] -sin(-x) = 1 - 2sin^2(x/2 - pi/4) [By -sin(x) = sin(-x)] sin(-x) = 2sin^2(x/2 - pi/4) - 1 1/2(sin(-x) + 1) = sin^2(x/2 - pi/4) 1/2(sin(x) + 1) = sin^2(-(x/2 + pi/4)) 1/2(sin(x) + 1) = sin^2(x/2 + pi/4) [By sin(-x) = -sin(x) => sin^2(-x) = sin^2(x)]
Which is obviously equal to what you wanted. QED.
Edit: Bit late, but this shows all the steps in an obvious way, so it should be clear to anyone reading it.
|
|
|
| Report Abuse |
|
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
stravant
|
  |
 |
| Joined: 22 Oct 2007 |
| Total Posts: 2893 |
|
|
| 26 Dec 2011 09:04 PM |
What? All I used is sin(-x) = -sin(x) [This is obvious from the fact that sin is symmetrical to the origin], and translations from sin to cos [Which is obvious from the fact that the graph of sin is just a translation of the graph of cos].
The rest is just plain algebra. |
|
|
| Report Abuse |
|
|
|
| 26 Dec 2011 09:06 PM |
"I need to learn more trig stuff..."
Learn basic Calculus first, it's so much fun and it makes a lot of math make sense <3 I got bored and started teaching myself some of it last week, and understanding limits, derivatives, and integrals have explained everything I've learned in Algebra II so far, and it explains a lot of the odd trigonometric identities I encountered in my trig unit in Geometry last year :V |
|
|
| Report Abuse |
|
|
stravant
|
  |
 |
| Joined: 22 Oct 2007 |
| Total Posts: 2893 |
|
|
| 26 Dec 2011 09:32 PM |
@pokelover
Better yet learn real analysis, so that you can actually prove more complicated things. It's quite interesting, for example here is the proof that proof by induction works using the least element axiom of natural numbers:
Note that proof by induction states that if I have some function "P(n)" where n is a natural number, then if I know that P(1) is true, and that if P(n) is true then P(n+1 ) is also true for all n, then P(n) must be true for all n. For example: If it is raining today, and I know that it raining on one day means it will rain on the next as well, it must continue to rain forever.
But how can you prove that that works?
-Start by assuming that induction does not work, that is, you know that P(1) is true and that (if P(n) then P(n+1)), but P(n) fails for some number.
-Since P(n) fails for some n, we can construct a non-empty list, let's call it "A", of all the numbers "n" that P(n) fails for.
-Now since A is non-empty, let's take the least element in that list A, and call it "k". P(1) is not in A since we assumed that P(1) is true to start with, that means that if we take k-1, it must also be a natural number.
P(k-1) must be true, since by definition P(k) was the first index that P is false for. By by out original assumption, "if P(n) then P(n+1)", P(k-1) is true means that P(k) must also be true. But by definition P(k) was in the set of things P fails for, so it must be false.
That is, P(k) must be both true and false, a contradiction. As a result, we can deduce that our original assumption was false, and since our original assumption was that induction is NOT true, and that assumption is false, via double-negative we know that induction must be true. QED.
As you can see it's quite interesting to construct arguments like that. Most stuff turns out to be considerably more lengthy than that. That's the shortest interesting proof that I can think of at the moment that would fit in a forum post easily, and is still easy to understand.
If you want to read some on it this is the online textbook that I've been using for it: http://www.math.ualberta.ca/~bowman/m117/m117.pdf |
|
|
| Report Abuse |
|
|
|
| 27 Dec 2011 05:00 PM |
| @Stravant: Thanks for that link, I took a quick look through it to see what it contains and it's quite interesting, I'm going to use it to try and learn some more stuff. I'm going to go back to my Algebra II class and everyone is going to hate me more than they already do because I'll know even more math stuff than they do. It's an Honors class though, they should all be math nerds like me, trying just as hard to go above the expectations! Of course, honors is just a silly name slapped on the class now, only AP seem to really make a huge difference at this point. AP Calculus as a Senior for me though :D |
|
|
| Report Abuse |
|
|
stravant
|
  |
 |
| Joined: 22 Oct 2007 |
| Total Posts: 2893 |
|
|
| 27 Dec 2011 05:09 PM |
| For future reference, that proof of proof by induction is a particularly good one to show to people, because they'll think it's really complicated and awesome, but they can still understand it completely. |
|
|
| Report Abuse |
|
|
NXTBoy
|
  |
| Joined: 25 Aug 2008 |
| Total Posts: 4533 |
|
|
| 28 Dec 2011 03:30 AM |
| FYI, google now graphs stuff for you: http://goo.gl/JuCOu |
|
|
| Report Abuse |
|
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
NXTBoy
|
  |
| Joined: 25 Aug 2008 |
| Total Posts: 4533 |
|
|
| 28 Dec 2011 10:20 AM |
| Whoops. Google Instant tricked me. http://goo.gl/dj8Do. |
|
|
| Report Abuse |
|
|
Oysi
|
  |
| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
|
| |
|
Catblox
|
  |
| Joined: 23 Apr 2008 |
| Total Posts: 2223 |
|
|
| 28 Dec 2011 11:50 AM |
| Wolfram Alpha does some pretty nice graphing. |
|
|
| Report Abuse |
|
|
shayan414
|
  |
| Joined: 11 Aug 2008 |
| Total Posts: 8090 |
|
|
| 29 Dec 2011 01:21 PM |
| I'll check out both of those graphs. |
|
|
| Report Abuse |
|
|