The math help thread

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.

Google Calc tells me one too. o.0

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.

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?

http://www.google.com/search?hl=en&safe=...f&oq=&aqi=

learn2google pls. :[
Since 10^-8 is tiny, and Google rounds, I'd say yea, it's 1.

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.

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) = -1

Uhhh... where'd I go wrong o.O

http://www.google.com/search?hl=en&safe=...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.

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)?

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.

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.

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.

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)

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) - 1

Seems 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 =(

This is a rather nasty thing.

First off, use the half-angle formula to prove this:


Then, we're off!


Looks almost like we hit a dead-end, but thankfully, the additive arctangent formula comes to use:


Now, we're basically done:


Noah

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..

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.

Remember harder. Pi isn't exactly 3.

I'm sure you can do better than that, being in this thread and all.

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.

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.

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.

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 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