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The math help thread - Corn - 2009-11-09
More trig. I have no frigging clue why I can't understand trig as well as geometry and Algebra II.
Okay. So I'm using De Moivres Theorem to solve (- square root of 3 + i ) ^ 3
divided by (1 - i) ^ 6. I keep on getting one, but I'm 95% sure the answer is supposed to be i.
The math help thread - Hazzy - 2009-11-09
Google Calc tells me one too. o.0
The math help thread - Nikkey - 2009-11-09
ClawofBeta Wrote:More trig. I have no frigging clue why I can't understand trig as well as geometry and Algebra II.
Okay. So I'm using De Moivres Theorem to solve (- square root of 3 + i ) ^ 3
divided by (1 - i) ^ 6. I keep on getting one, but I'm 95% sure the answer is supposed to be i.
It is one.
The math help thread - Corn - 2009-11-09
Great. Either my teacher is an idiot or I copied wrong. To check...
(this is in degrees)
(cos 36 + i sin 36) ^ 100 ....does this equal 1?
The math help thread - Hazzy - 2009-11-09
http://www.google.com/search?hl=en&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=NVM&q=%28cos+0.628318531++%2B+i+sin+0.628318531+%29+^+100&aq=f&oq=&aqi=
learn2google pls. :[
Since 10^-8 is tiny, and Google rounds, I'd say yea, it's 1.
The math help thread - KajitiSouls - 2009-11-09
ClawofBeta Wrote:More trig. I have no frigging clue why I can't understand trig as well as geometry and Algebra II.
Okay. So I'm using De Moivres Theorem to solve (- square root of 3 + i ) ^ 3
divided by (1 - i) ^ 6. I keep on getting one, but I'm 95% sure the answer is supposed to be i.
I have never heard of De Moirvres before o.O *googles and wikis*
The rest of you are lazy cheaters.
![[Image: b29923f7db2847c9f284814eca4e78a9.png]](http://upload.wikimedia.org/math/b/2/9/b29923f7db2847c9f284814eca4e78a9.png)
![[Image: 265px-Imaginarynumber2.svg.png]](http://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Imaginarynumber2.svg/265px-Imaginarynumber2.svg.png)
Code:
(-sqrt(3) + [I]i[/I])^3
arctan(-1/sqrt(3)) = θ = -30 degrees
r = sqrt(sqrt(3)^2 + 1^2) = 2
z = 2*[cos(-30) + [I]i[/I]*sin(-30)]
[2*(cos(-30) + [I]i[/I]*sin(-30))]^3 = 8*[cos(-90) + [I]i[/I]*sin(-90)] = 8*[I]i[/I]*-1
(1 - [I]i[/I])^6
arctan(-1/1) = -45 degrees
r = sqrt(2*1^2) = sqrt(2)
z = sqrt(2)*[cos(-45) + [I]i[/I]*sin(-45)]
[sqrt(2)*[cos(-45) + [I]i[/I]*sin(-45)]]^6 = 8*[cos(-270) + [I]i[/I]*sin(-270)] = 8*[I]i[/I]*1
(8*[I]i[/I]*-1) / (8*[I]i[/I]*1) = -1Uhhh... where'd I go wrong o.O
The math help thread - Corn - 2009-11-09
Hazzy Wrote:http://www.google.com/search?hl=en&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=NVM&q=%28cos+0.628318531++%2B+i+sin+0.628318531+%29+^+100&aq=f&oq=&aqi=
learn2google pls. :[
Since 10^-8 is tiny, and Google rounds, I'd say yea, it's 1.
I'm not sure if that's actually 1, since the i screws up everything. Darn imaginary numbers.
KajitiSouls Wrote:I have never heard of De Moirvres before o.O *googles and wikis*
The rest of you are lazy cheaters.
Code:(-sqrt(3) + [I]i[/I])^3
[B][I][U]arctan(-1/sqrt(3)) = θ = -30 degrees[/U][/I][/B]
r = sqrt(sqrt(3)^2 + 1^2) = 2
z = 2*[cos(-30) + [I]i[/I]*sin(-30)]
[2*(cos(-30) + [I]i[/I]*sin(-30))]^3 = 8*[cos(-90) + [I]i[/I]*sin(-90)] = 8*[I]i[/I]*-1
(1 - [I]i[/I])^6
arctan(-1/1) = -45 degrees
r = sqrt(2*1^2) = sqrt(2)
z = sqrt(2)*[cos(-45) + [I]i[/I]*sin(-45)]
[sqrt(2)*[cos(-45) + [I]i[/I]*sin(-45)]]^6 = 8*[cos(-270) + [I]i[/I]*sin(-270)] = 8*[I]i[/I]*1
(8*[I]i[/I]*-1) / (8*[I]i[/I]*1) = -1
Uhhh... where'd I go wrong o.O
Isn't the bolded/unerlined part supposed to be arctan 1/-sqrt(3)?
The math help thread - KajitiSouls - 2009-11-09
ClawofBeta Wrote:Isn't the bolded/unerlined part supposed to be arctan 1/-sqrt(3)?
Yes, but what's inside the arctan function is still negative. It's -30 degrees either way, right?
And (cos(36) + i*sin(36))^100 is indeed 1.
The math help thread - Corn - 2009-11-09
KajitiSouls Wrote:Yes, but what's inside the arctan function is still negative. It's -30 degrees either way, right?
And (cos(36) + i*sin(36))^100 is indeed 1.
But my thingy turns it into a Quadrant II angle (while yours is a QIV), so it'll be 150 degrees, not 330 =/.
And good riddance. I'm going to rage at my teacher, since I've been stuck on these two problems for the past 2 hours when I think it's actually correct.
The math help thread - KajitiSouls - 2009-11-09
ClawofBeta Wrote:But my thingy turns it into a Quadrant II angle (while yours is a QIV), so it'll be 150 degrees, not 330 =/.
And good riddance. I'm going to rage at my teacher, since I've been stuck on these two problems for the past 2 hours when I think it's actually correct.
Oh poo you're right (wtp 330? lol). They're both in the same line though xD Stupid graphing calculator D=
Yeah I got 1 for that problem then.
The math help thread - Horusmaster - 2009-11-12
Need help on my math assignment due friday morning.
I don't get why they throw in a trig identity question in the assignment when we're not even learning it...
prove:
arcsin ((x-1)/(x+1))=2arctan(sqrt(x))-(pi/2)
The math help thread - KajitiSouls - 2009-11-12
Horusmaster Wrote:Need help on my math assignment due friday morning.
I don't get why they throw in a trig identity question in the assignment when we're not even learning it...
prove:
arcsin ((x-1)/(x+1))=2arctan(sqrt(x))-(pi/2)
I haven't quite gotten there, but this is what I've managed...
Code:
Reference:
Hypotenuse = x + 1
Opposite = x - 1
Adjacent = 2*sqrt(x)
ᴨ = PI
arcsin((x - 1) / (x + 1)) = 2*arctan(sqrt(x)) - (ᴨ / 2)
1/2*arcsin((x - 1) / (x + 1)) = arctan(sqrt(x)) - (ᴨ / 4)
tan(1/2 * arcsin((x - 1) / (x + 1))) = sqrt(x) - 1
(x - 1) / (4*sqrt(x)) = sqrt(x) - 1Seems to me that trying to do this problem without trigonometric identities or formulas isn't possible, given the nature of tangent.
Okay I've gotten closer: x = sqrt(x)*sqrt(x + 1). Sorry, seems like I can't help after all =(
The math help thread - Noah - 2009-11-12
This is a rather nasty thing.
First off, use the half-angle formula to prove this:
![[Image: yzdwuk2.png]](http://mathurl.com/yzdwuk2.png)
Then, we're off!
![[Image: ycr3ycs.png]](http://mathurl.com/ycr3ycs.png)
Looks almost like we hit a dead-end, but thankfully, the additive arctangent formula comes to use:
![[Image: yhgwmsw.png]](http://mathurl.com/yhgwmsw.png)
Now, we're basically done:
![[Image: yblxo4n.png]](http://mathurl.com/yblxo4n.png)
Noah
The math help thread - ZottenKerel - 2009-11-12
Pi is exactly 3.
Hey just thought I'd shake things up a bit, nerd style. Btw I haven't had maths in 4 years and it wasn't super advanced, the stuff I see here makes me a little jealous that I can't remember all of it anymore. XD
make that any of it.. altho the arctan frmules did ring a bell.. u+v/1-uv xD man ages ago..
The math help thread - KajitiSouls - 2009-11-12
ZottenKerel Wrote:Pi is exactly 3.
Hey just thought I'd shake things up a bit, nerd style. Btw I haven't had maths in 4 years and it wasn't super advanced, the stuff I see here makes me a little jealous that I can't remember all of it anymore. XD
make that any of it.. altho the arctan frmules did ring a bell.. u+v/1-uv xD man ages ago..
Remember harder. Pi isn't exactly 3.
The math help thread - ZottenKerel - 2009-11-12
KajitiSouls Wrote:Remember harder. Pi isn't exactly 3.
I'm sure you can do better than that, being in this thread and all.
The math help thread - Horusmaster - 2009-11-15
So, I finally understood how to do that question (though i messed that up for the assignment)
I'm suppose to rearrange the equation into
arcsin ((x-1)/(x+1))-2arctan(sqrt(x))-(pi/2)=0
then let f(x)=arcsin ((x-1)/(x+1))-2arctan(sqrt(x))-(pi/2)
take the derivate, and f '(x)=0
since f '(x)=0 that means f(x) is a constant for all value of x, plug in any number for f(x) and it =0.
The math help thread - KajitiSouls - 2009-11-15
ZottenKerel Wrote:I'm sure you can do better than that, being in this thread and all.
I can show you the series to derive all the digits of Pi if you want.
Horusmaster Wrote:So, I finally understood how to do that question (though i messed that up for the assignment)
I'm suppose to rearrange the equation into
arcsin ((x-1)/(x+1))-2arctan(sqrt(x))-(pi/2)=0
then let f(x)=arcsin ((x-1)/(x+1))-2arctan(sqrt(x))-(pi/2)
take the derivate, and f '(x)=0
since f '(x)=0 that means f(x) is a constant for all value of x, plug in any number for f(x) and it =0.
Oh schnapples, that is GENIUS!
Okay not exactly, but not something you'd think up all the time.
The math help thread - Noah - 2009-11-15
Horusmaster Wrote:So, I finally understood how to do that question (though i messed that up for the assignment)
I'm suppose to rearrange the equation into
arcsin ((x-1)/(x+1))-2arctan(sqrt(x))-(pi/2)=0
then let f(x)=arcsin ((x-1)/(x+1))-2arctan(sqrt(x))-(pi/2)
take the derivate, and f '(x)=0
since f '(x)=0 that means f(x) is a constant for all value of x, plug in any number for f(x) and it =0.
Not bad, not bad at all.
(However, you cannot plug in all values of x. x = -1 makes bad stuff here. Though, it's maybe obvious that it's in the interval (-1, 1))
Noah
The math help thread - Veneni - 2009-11-15
Quote:The exact chance? Do ...
And yeah, 2/7 is a lower chance than 1/3, so it follows that the slates with the highest probability of working, assuming you have infinite of them, are the 1, 3 and 5% slates.
Noah
Oh lol how could I've been so stupid, it was alot simpler than I thought it would be
ty