HaloMods "Classic" Read-only Archive

Posted: **Sun Nov 16, 2008 7:19 pm**

In response to [cc]z@nd! 's request, this is the Math Help/Tutoring Thread. Ask about problems you are having or a concept you are having trouble with.

I am taking Calculus BC AP and I'd be happy to help as well.

Posted: **Sun Nov 16, 2008 7:28 pm**

Vector calculus, multi-dimensional calculus (partial differentials and the like), differential equations and linear algebra. Been a year since my last class of math but still use a lot of it.

Posted: **Sun Nov 16, 2008 7:31 pm**

I'll help as well. I'm taking Calculus AB AP, but I should have taken BC, the class is SOOOO SLOW.

Posted: **Sun Nov 16, 2008 7:52 pm**

I need to come up with a list of 10 points that the line y=x passes through, and I'm lost. Can you help me?!

Haha, just kidding. Really though, I hope this thread proves to be helpful.

Posted: **Sun Nov 16, 2008 7:56 pm**

http://www.stupidhomework.com

Posted: **Sun Nov 16, 2008 8:57 pm**

I'm in PreCal so I can help people I guess. Might need help too.

Posted: **Sun Nov 16, 2008 10:07 pm**

Yeah, uh, whats uh grahm's number as a regular number?
lol, this thread may come in handy.
Actually i do have a question is there a way to translate coordinates into matrices?

Posted: **Mon Nov 17, 2008 2:02 pm**

Matrices are devices for storing numbers.

So instead of coordinates in (x,y) form, you could just put then in a 1x2 matrix [[x,y]].

Posted: **Mon Nov 17, 2008 3:22 pm**

DrXThirst wrote:http://www.stupidhomework.com

i went there and signed up, its actually a good website haha

Posted: **Mon Nov 17, 2008 6:40 pm**

i didn't know there'd be so many calc guys in here. i'm just in precal myself at college, so i guess i'm getting some good fundamental work.

anyways, i'm taking calculus next semester, so i figure i should get prepared ahead of time. anything i should look into to help smooth the transition? practice problems i can do?

and for people wanting to just come in here to do work, take a look at these problems. average precal stuff, but who knows, maybe it'll remind you of something you should brush up on, or give you a chance to get some nerd cred

oh, and i don't care if the answer is in degrees or radians, although radians are much better than degrees. (why?). and yes, i do have the answers.

Code: Select all

1) evaluate sin(pi/12) using an appropriate half-angle formula. leave the answer exact.

2) find all possible values of x where sin(x) + cos(x) = 1

as you can see, we're in the trig chapter right now. we have been for the past month, and it gets kind of boring after a while D:

Posted: **Tue Nov 18, 2008 2:06 pm**

[cc]z@nd! wrote:
Code: Select all
1) evaluate sin(pi/12) using an appropriate half-angle formula. leave the answer exact.

2) find all possible values of x where sin(x) + cos(x) = 1
as you can see, we're in the trig chapter right now. we have been for the past month, and it gets kind of boring after a while D:

Come nerd-boy to the geek-mobile!

1) first change it into a X/2 fraction so:

Code: Select all

 pi/12 = 144pi/12 -> ((144/6)*pi)/(12/6) --> (7pi/3)/(2) now we have an X/2

then:

sin(x/2)= sqrt( 1 - cos (x) 
                      ------------
                              2             )
so subsitute:

sin((7pi/3)/(2))= sqrt( 1 - cos (7pi/3) 
                                 ------------
                                        2             )

so cos(420) = cos(360+60) = cos(360)*cos(60) - sin(360)*sin(60)
                                                  1           1/2            0      whatever
                                                         1/2          -            0
                                                                          1/2
back agian to the equation

1-1/2
-------
2

1/2
----
2


1/4

sqrt(1/4)= 1/2

and finally because X/2 (where X = 7*pi/3 in this case) lies in the bottom quadrant the fraction is negative so:

THE ANSWER IS:
-(1/2)

shit uh did the fraction wrong in the first step it should be: (pi/6)/2

the steps are similar with different numbers though

Posted: **Tue Nov 18, 2008 3:59 pm**

CompKronos wrote: 1) first change it into a X/2 fraction so:
Code: Select all
lots 'o math. well, mostly math.
THE ANSWER IS:
-(1/2)

**** uh did the fraction wrong in the first step it should be: (pi/6)/2

the steps are similar with different numbers though

yeah, i was going to point out the fraction thing. i was wondering how you turned pi/12 into 144pi/12. as for starting out that problem, (pi/6)/2 is what you get, then it's the steps you wrote down from there, just with cos(pi/6). also, starting your half-angle formula, it looks like you missed the +/-, although it didn't end up changing anything since pi/6 is in the first quadrant meaning it's sin() and cos() would be positive.

ANSWER wrote:+ sqrt(1/2 - sqrt(3)/4)

now for the next two problems. these are on the homework i'm doing right now, so you can be doing them alongside me pretty much. i've already done the first work. currently working on the second.

Code: Select all

solve
cos(arccos(5/4))

cos[ arcsin(1/3) + arccos(1/2) ]

Posted: **Tue Nov 18, 2008 5:25 pm**

yeah sorry about the fraction, every now and again my algebra with factions/exponents can become very nonsensical. I'll look at the other see if i can find it.

wait is the cos(arccos(5/4)) part of your solution for values of sin(x) + cos(x) = 1 ?

Posted: **Tue Nov 18, 2008 5:57 pm**

No those are two new problems I'm fairly sure. I don't see anyone actually getting cos(arccos(x)) in calculation.

Sorry I won't be much help on these. Cosine and sine are just things I use for evaluation and modeling. Getting the values is something left to calculators and computers.

Posted: **Tue Nov 18, 2008 8:15 pm**

yes, those are two completely new problems, but i guess i'll post the solution to the sin(x) + cos(x) = 1 problem.

sin(x) + cos(x) = 1 wrote:don't mistake this for the identity, where x can be any real number.
if you think of the unit circle, and adding the x and y coordinates of each point on the circle, you can figure this one out pretty easy. when sin(x) = 1, cos(x) = 0, and when cos(x) = 1, sin(x) = 0. these are the cases where the statement would be true, so we just have to find what angles this happens at.
looking at the unit circle, sin(x) = 1 at pi/2, and cos(x) = 1 at 0. those are our two first answers, but we have to alter them because sin and cos are periodic functions, and we weren't given a specific interval our answers had to be in. since the period of sin and cos is 2pi, we simply add multiples of 2pi to each solution.
x = pi/2 + 2Kpi
x = 2Kpi
where k is an integer.

a related brainteaser is sin(x)cos(x) = 1. it's kind of tricky, although i kind of gave it away in the solution.

and yeah, i used to hate dealing with fractions, but once you work with them enough, they're sooooo much better than decimals. and while i'm on one number type vs. another rant, radians are much better than degrees, because radians are real numbers and a ratio, instead of a virtual division of a circle like degrees.

Posted: **Wed Nov 19, 2008 10:54 am**

wouldn't cos(arccos(x))=x?

Posted: **Wed Nov 19, 2008 11:07 am**

Yes it would. These questions are mostly for understanding the basic trigonometric functions right? Make sure you're familiar with them because of how prevalent they are in higher level math.

Posted: **Wed Nov 19, 2008 3:28 pm**

CompKronos wrote:wouldn't cos(arccos(x))=x?

AH! you would THINK so, but you have to be careful here. cos(arccos(x)) wouldn't always equal x, and here's why:

arccos is the inverse function of cos, but there's a little caveat here. since a function can only have an inverse function if it's one-to-one (so the inverse function is a function, as in has one output for each input in it's domain), the cos, sin and tan functions normally wouldn't have inverses, because they are periodic.

what we can do, however, is get the inverse of a restricted sin, cos, or tan function. for example, we can limit the domain of our cos to [0,pi], and it would be one-to-one, allowing it to have an inverse function. now that we've restricted it so it can have an inverse, i'll list each function's domain and range

Code: Select all

function [] domain [] range
cos         [0,pi]     [-1,1]
arccos      [-1,1]     [0,pi]

what's important here is the domain for arccos. since cosine's range was [-1,1], arccos' domain is [-1,1], so anything outside that domain isn't valid input for arccos. that's why cos(arccos(5/4)) isn't a valid statement, because (5/4) > 1, and is outside arccos' domain. i believe you can try this on your calculator and it would give you a domain error.

and yes, that means the cos(arccos(5/4)) problem i posted was a trick.

Posted: **Wed Nov 19, 2008 3:58 pm**

Yeah it does have a limited domain. Forgot about that little bit.

Posted: **Wed Nov 19, 2008 4:06 pm**

I am in Pre-calc now, would be in AP Calc AB, but due to a class mess-up I am Honor PreCalc now, I am also Taking Statistics this year, so I can lend that help too.