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April 3

Stupid question

Hi there, kindly Mathematics help desk editors. I'm in the middle of studying and I can't figure out what the other two solutions to this equation are; I *have* checked my textbook and looked it up online, but I'm really confused :/ It'd be great if someone could explain if they had the time! :D

$sin2x={\frac {1}{2}}$ for $0\leq x\leq 2\pi$

I did the inverse sine of 1/2 (in rads) and got pi/6 and from there used the quadrant diagram (and divided by 2) to get x = pi/12 and x=5pi/12. However my textbook says that 13pi/12 and 17pi/12 are also solutions and I can't for the life of me figure out why. Any pointers in the right direction would be appreciated. Cheers, -- Naerii 00:57, 3 April 2008 (UTC)[reply]

The key thing to remember is that when x ranges from 0 to 2*pi, 2x ranges from 0 to 4*pi. When you use what you call the "quadrant diagram", you need to find all the angles up to 4*pi which have sine 1/2, and then half them, which will give you angles up to 2*pi, as required. (And remember: If you ask the question, you're stupid for a minute, if you don't ask the question, you're stupid for life!) --Tango (talk) 01:07, 3 April 2008 (UTC)[reply]

If only I could understand that article :( But thanks for both of your help, I can't believe I was so dense as to miss the 4pi thing haha. Thanks, -- Naerii 01:32, 3 April 2008 (UTC)[reply]

Your answer should be right...128.163.224.222 (talk) 22:58, 6 April 2008 (UTC)[reply]

Accidental variance discovery

While considering generalizations of the Moore-Penrose pseudoinverse, in particular matrices of the form $C=(AB)^{-1}A$ for non-square A and B, I happened across a peculiar result. Generating matrices of a fixed size by choosing their elements uniformly from $[-{\frac {1}{2}},{\frac {1}{2}})$ , I fixed B and generated two long lists of A matrices (300k each). I then evaluated the C matrices and took the sample variance of each element of C (treated as a random variable with each instantiation of C being a trial; clearly they are not independent). The result was two matrices of variances, one for each set; the oddity is that the two matrices satisfy $V'=DV$ for a diagonal matrix D.

Put simply, what does this mean about the dependence of C on A? The sample size is irrelevant; I observe the same effect generating just 3 Cs and taking the (somewhat silly) variances of $\{C_{1},C_{2}\}$ and $\{C_{2},C_{3}\}$ . My initial suspicion was that the various C matrices themselves differed only in the application of a diagonal matrix, but this is not the case: neither the matrices that satisfy $C'=MC$ nor those that then satisfy $M'=NM$ are diagonal or even symmetric. Any thoughts? --Tardis (talk) 03:58, 3 April 2008 (UTC)[reply]

question about rational equations

When solving a rational equation, why is it necessay to perform a check? —Preceding unsigned comment added by 72.94.241.26 (talk) 15:07, 3 April 2008 (UTC)[reply]

You're talking about solving an expression of the form

{\frac {P(x)}{Q(x)}}=0

, right?

For

x^{*}

to be a solution to this equation, it must be the case that

P(x^{*})=0

. But it's also necessary that

Q(x^{*})\neq 0

, for if that fails, then the expression

{\frac {P(x)}{Q(x)}}

is undefined (it is not equal to zero in this context).

Hence, solving this equation amounts to picking the values of x that solve

P(x)=0

while at the same time make

Q(x)\neq 0

. For example,

{\frac {x^{2}-1}{x-2}}

has

\pm 1

as solutions, but

{\frac {x^{2}-1}{x-1}}

is only solved by -1, since Q(1)=0. Is this what you were referring to? Pallida Mors 15:41, 3 April 2008 (UTC)[reply]

well kinda but i just wanted to know why it is necessary to do the check after you do the equation —Preceding unsigned comment added by 71.185.111.206 (talk) 16:40, 3 April 2008 (UTC)[reply]

Could you give example of a question and what you've been told to check? --Tango (talk) 16:55, 3 April 2008 (UTC)[reply]

It's good to check the answer to any problem, because different solution methods can introduce spurious answers (for example, if you square an expression, you get a positive and a negative answer in the end, but only the positive one may be an actual solution). Of course, you could also lose solutions in the process, and that's harder to detect unless you have some knowledge of how many solutions the problem should have (e.g. if you graph it). Confusing Manifestation(Say hi!) 22:17, 3 April 2008 (UTC)[reply]

Have a problem setting up equation to these questions

1. The cost, in millions of dollars, to remove x % of pollution in a lake modeled by C=6000/200-2x

 a. What is the cost to remove 75% of the pollutant?
 b. What is the cost to remove 90% of the pollutant?
 c. What is the cost to remove 99% of the poolutant?
 d. For what value is this equation underfined?

I can't figure it out how to solve this. please help —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 15:20, 3 April 2008 (UTC)[reply]

You have the cost function in terms of the percentage of the pollution you want to remove. C is the cost. x is the percent of the pollution you're removing. If you have x, how would you get C? –King Bee ^{(τ • γ)} 15:45, 3 April 2008 (UTC)[reply]

but how do i set uo the equations? —Preceding unsigned comment added by 71.185.111.206 (talk) 16:38, 3 April 2008 (UTC)[reply]

You don't need to set up any equations - you're given the equation in the question. --Tango (talk) 16:54, 3 April 2008 (UTC)[reply]

You have the equation — though surely it should be

C={\frac {6000}{200-2x}}

which is written as "C=6000/(200-2x)". You need merely identify which quantit(y/ies) in the problem are to be used for x. (You are always asked for the cost, so you can't be being given C.) For (d), you might graph the function

C(x)

, or else note that it is a rational function, to find points where it is undefined. As an aside, you have been given a silly model for the cost:

C(0)\neq 0

, so it costs money to do nothing at all! I suppose that the everpresent cost might be the cost of bringing the equipment and people to the lake in question, but it'd still be more realistic (and a lot cheaper for thorough jobs) to use

D(x)=30\left(1-\operatorname {lg} (1-x\%)\right)

(see binary logarithm), chosen to match the given formula at

x=0,x=50\%

. --Tardis (talk) 17:08, 3 April 2008 (UTC)[reply]

You are given the following:

The cost to remove x% is equal to 6000 / (200 − 2·x).

This is supposed to hold for all possible values of the variable x. So, for example:

The cost to remove 54% is equal to 6000 / (200 − 2·54).

The cost to remove 68% is equal to 6000 / (200 − 2·68).

The cost to remove 93% is equal to 6000 / (200 − 2·93).

Do you see the pattern? You can do the problems a, b and c in the same way. For d, remember that you are not allowed to divide by 0. --Lambiam 23:16, 3 April 2008 (UTC)[reply]

Now do i take 2*54 then subtract from 200 then divide it by 6000?Should there be a decimal point for the percentages when times them? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 15:34, 4 April 2008 (UTC)[reply]

By the look of it, x is the percentage, not the fraction, so you need to use x=75, not x=0.75. Then just substitute that value into the equation and calculate the answer. --Tango (talk) 16:09, 4 April 2008 (UTC)[reply]

And you would divide it into 6000 rather than by 6000.

Problem.

Theres a quadrilateral with corners A, B, C, and D. AB and CD are the long sides, and they are parallel. The length of BC is 8. The measure of angle C is 30, and the measure of angle A is 45. Find the perimeter of the object. I got $44+4{\sqrt {3}}+4{\sqrt {6}}$ , but just about everyone else I have asked to solve this problem has gotten different answers. If someone could please draw this figure using the <math> parameters, I think that ppl would be able to solve it easier-ly. So: what is the real perimeter? flaming lawyer^c 17:34, 3 April 2008 (UTC)[reply]

Unless I'm missing something, those details don't determine a single shape - there's nothing stopping you stretching AB and CD, as long as you add the same amount to each. Am I missing something, or have you missed something out? --Tango (talk) 17:49, 3 April 2008 (UTC)[reply]

Agreed with Tango. Also, I don't think the math markup can draw figures. If you tell us what you did to reach your conclusion, and hopefully supply the missing information, we can probably help further. --LarryMac | Talk 18:07, 3 April 2008 (UTC)[reply]

OK: I'll try to draw it, but I'll have to rewrite the info, too. Here it is:

   A__________________________B
   /                         /
  /                         /
 /                         /
/_________________________/D
C

Line AB is parallel to line CD. Lines BD and AC are not parallel. The length of line BD is 8. Angle D is 30 degrees. Angle A is 45 degrees. From some law of mathemeatics, in a right triangle with angles measuring 45-45-90, the measure of the hypotenuse is ${l}{\sqrt {2}}$ , where l is the length of either leg (not the hypotenuse). In a right triangle with angles measuring 30-60-90, the length of the longer leg is ${s}{\sqrt {3}}$ , where s is the length of the shorter leg. The length of the shorter leg is half of the length of the hypotenuse. You solve this equation by dividing the figure into a rectangle and two right triangles. flaming lawyer^c 19:29, 3 April 2008 (UTC)[reply]

It might help to draw it with A and D being acute angles, since you've given that they are, and I'll add the numbers in too:

   A___________________B
    \45 |             |\
     \  |             | \ 8
      \ |             |30\
      C\|_____________|___\D

Have I drawn in the triangles in the places you meant? Using the triangle with the 30 deg angle, you can get that the height is ${\frac {\sqrt {3}}{4}}$ , then using that and the other triangle, you can get that AC is ${\frac {\sqrt {6}}{4}}$ (this is all assuming I can do right angled trig in my head - the numbers may be wrong, but it's the method that's important). However, that's as far as you can go. There is no way to determine the lengths of AB or CD without more information, since I can draw another shape which still satisfies the information you've given but has different AB and CD:

   A_______________________________________B
    \45 |                                 |\
     \  |                                 | \ 8
      \ |                                 |30\
      C\|_________________________________|___\D

See? So, unless there is another piece of information you haven't given us, the perimeter is undetermined. --Tango (talk) 19:50, 3 April 2008 (UTC)[reply]

Flaming lawyer, are you sure you got the diagram right? I mean, a quadrilateral ABDC doesn't make as much sense as ABCD

   A___________________L___________B
    \45 |                  _______/|\
     \  |         ____8___/        | \ 
      \ | _______/                 |30\
      D\|/_________________________|___\C

and using this, we could probably work something out using BD = 8. -mattbuck (Talk) 19:52, 3 April 2008 (UTC)[reply]

In fact, using this, let us label the height of the shape as H, and the length of the top between the two triangles as L. Then

L^{2}+H^{2}=8^{2}

, the length of AD is

H{\sqrt {2}}

, AB = L+H, BC = 2H,

CD=L+{\frac {\sqrt {3}}{2}}H

. You will still get quite a few possibilities, but you are limited as to your choices. If nothing else, you'll get it as an equation in 1 variable, and can find the range of possibles. -mattbuck (Talk) 20:07, 3 April 2008 (UTC)[reply]

You get that L can range between 0 and 8, and everything is determined from there. That's only marginally better than the previous version - it's just the addition of an upper bound. There's still a continuum of possibilities. --Tango (talk) 20:13, 3 April 2008 (UTC)[reply]

All of the above diagrams are wrong! We are given BC = 8, not BD = 8. Since the quadrilateral has two parallel sides and angles A and C are both acute, they cannot be adjacent; therefore the four points must be in the order ABCD and not ABDC. So the correct diagram is

   A_____P______________B
    \45  |              |\
     \   |              | \ 8
      \  |              |  \
       \ |              |   \
       D\|_____________Q|__30\C

(This is equivalent to Tango's diagram except for the labeling.)

Now we do know from the given angles that BQ = PD = AP = 4, CQ = $4{\sqrt {3}}$ , and AD = $4{\sqrt {2}}$ , but distance PB is unrestricted except that it is equal to QD and (because AB was one of "the long sides") greater than 4. If the problem was reproduced correctly and I haven't goofed, the perimeter could be any number greater than $20+4({\sqrt {2}}+{\sqrt {3}})$ .

--Anonymous, edited 00:56 UTC, April 5, 2008.

The OP said BC the first time and BD the second. I based my diagram off the 2nd writing of the information. It makes little difference to the answer, though. --Tango (talk) 01:00, 5 April 2008 (UTC)[reply]

If BD=8 we can solve this.
a is the angle CDB
b is the angle DBC
H = 8 sin(a) = 8cos(b) and a+b=pi/2
so a = b = pi/4
and H = AP = PB = DQ = 4(2^(1/2)) = 32^(1/2)

AD = 8
AB = 2(32^(1/2))

Consider the triangle QBC:
QB = H = BC cos(pi/3) = 32^(1/2)
BC = 2(32^(1/2))/(3^(1/2))
and QC = BC sin(pi/3) = (32^(1/2))/(3^(1/2))

perimeter = AB + BC + CQ + QD + AD = 2(32^(1/2)) + 2(32^(1/2))/(3^(1/2)) + (32^(1/2))/(3^(1/2)) + (32^(1/2)) + 8

simplifying: perimeter = 8 + 3(32^(1/2))(1+1/(3^(1/2)))

--129.184.84.11 (talk) 12:50, 7 April 2008 (UTC)[reply]

? (Puzzle)

I was wondering: you know that puzzle, with the canoe and the river and the wolf, chicken, and chicken food (or some variation of it)? Is there any math behind that puzzle at all? Just wondering... flaming lawyer^c 19:40, 3 April 2008 (UTC)[reply]

Well, it's more a problem of logic, but since logic is the foundation of maths, I suppose you could claim there was something in it. -mattbuck (Talk) 19:48, 3 April 2008 (UTC)[reply]

I've seen the problem re-written in terms of graph theory, so in that sense there is maths behind it. Confusing Manifestation(Say hi!) 22:13, 3 April 2008 (UTC)[reply]

Someone asked for the solution to that problem, and a generalization if possible a few weeks ago on this desk, I'm too lazy to track it down, but it's in the archives somewhere. A math-wiki (talk) 22:15, 3 April 2008 (UTC)[reply]

It's here: Wikipedia:Reference desk/Archives/Mathematics/2008 March 22#Chicken, fox, and grain problem. --Lambiam 22:57, 3 April 2008 (UTC)[reply]

The solution is: Take the chicken across first, go back and get the chicken food and take it across and bring the chicken back to the beginning with you, and then take the wolf across (coming back for the chicken last). 86.149.97.133 (talk) 22:34, 3 April 2008 (UTC)[reply]

See fox, goose and bag of beans puzzle. Gandalf61 (talk) 22:48, 3 April 2008 (UTC)[reply]

I've seen it used as a very simple example of operations research. A similar riddle asks how you can toast three slices of bread in the shortest time, assuming your toaster holds two slices but only does one side of each. It made more sense with older toasters. Black Carrot (talk) 09:02, 4 April 2008 (UTC)[reply]

I've heard that puzzle phrased in terms of grilling meat, where it certainly happens one side at a time. Or, somewhat relatedly, how three surgeons can work on one patient with only two sets of gloves and no cross-contamination (even between the surgeons' hands, or it would be trivial). --Tardis (talk) 15:33, 4 April 2008 (UTC)[reply]

Trigonometric equations

Hi there, any help solving this would be appreciated.

$2sin^{2}(x)-2sin(x)=cos^{2}(x)$

The only two identities we've been taught are:

$tan(x)={\frac {sin(x)}{cos(x)}}$ and $sin^{2}(x)+cos^{2}(x)=1$

I've tried rearranging any number of different ways but trig identities get me every time x.x How do I get the x by itself? Thanks in advance! —Preceding unsigned comment added by 86.149.97.133 (talk) 22:33, 3 April 2008 (UTC)[reply]

Express the

\cos ^{2}x

in terms of

\sin ^{2}x

and then rearrange. Then let, say,

y=\sin x

, and then you might see something interesting, unrelated to trigonometric functions. x42bn6 Talk Mess 22:36, 3 April 2008 (UTC)[reply]

Thanks!!! —Preceding unsigned comment added by 86.149.97.133 (talk) 22:40, 3 April 2008 (UTC) oh quadratics :D lot easier than it looks, thank you so much —Preceding unsigned comment added by 86.149.97.133 (talk) 22:42, 3 April 2008 (UTC)[reply]