Have you heard of the inverse sine function? It does what you think it'd do. That is,

sin^{-1}(1/3)=x

2 isn't "invalid". What you are trying to say is that there are no real values of x for which sin(x) is 2.

If sin(x) = 1/3 then x is the number whose sin() is 1/3

Just like sqrt(5) is the number whose square is 5 (sqrt(x) being the inverse function of x^2)

The standard notation for the inverse function of sin(x) is sin^(-1)(y). This inverse function notation shouldn't be confused with power notation where the exponent applies to the whole expression.

The function sin^(-1)(x) is sometimes called arcsin(x).

Thanks Cueballmania, and Chasqui. I'm just learning about inverse functions now. I think I'm starting to understand. So, where sin(x) = 1/3, x = arcsin(1/3)

But, how do I convert arcsin(1/3) to a multiple of pi?

For example, I have another question phrased exactly like this:

Solve the following equation in the interval [0, 2pi].

Note: Give the answer as a multiple of pi. Do not use decimal numbers. If there is more than one answer enter them separated by commas.

sin^2(t) = 3/4

t = [fill in the blank] * pi

So,
sin(t) = sqrt(3) / sqrt(4)
t = arcsin[sqrt(3) / sqrt(4)]
How do I solve this in terms of pi?

Thanks.

For this particular problem you can check the unit circle and see that sin t = sqrt(3) / 2 when t=pi/3. For "simple" points on the unit circle you can solve these equations immediately as some numbers times pi as you want to do.

The solution to sin(t)=1/3 does not correspond to a point usually taught/included with the unit circle so you would need to use a calculator. So the question you posed earlier is a bit different.

Also remember that the solutions "usually" come in pairs (except at the extreme points on the unit circle). If t is a solution to your equation sin(t)=sqrt(3)/2 then so is pi - t = 2pi/3. A slightly different rule applies to solving equations involving cos.

Thus sin(t)=sqrt(3)/2 has solutions t =pi/3 and t=2pi/3

Last point is that you are inadvertently dropping some solutions from your original problem. Notice that you took only the positive square root when you went from sin^2(t)=3/4 to sin(t)=sqrt(3)/2. You also need to solve the negative square root equation sin(t) = - sqrt(3)/2. This will give you 2 more solutions to your original problem.

The angle whose sine is 1/3 isn't a well known fraction of pi, so you leave it as arcsin(1/3). Just like sqrt(5) isn't rational so we leave it like that.

Regarding the angles whose sin(x)=+- sqrt(3)/2 , you mentioned it on your original post . Make sure to specify all the solutions in the given interval.

Thanks!

Always amazes me how much friendlier/more helpful this thread is than most of 2p2, heh.

Get a life!

If you were specifically asked for it to be t = [something]*pi, simply solve for t, and then say

t = w * pi
w = t/pi
t = (t/pi)*pi

Where you would have explicitly calculated (t/pi). For example, if been given

sin(t) = 0.25882

Then

t = arcsin(0.25882) = 0.267719

Then

t = w * pi
w = (0.266719)/pi = 1/12

Then

t = pi/12

What's the Excel formula to calculate the chances that an event with a probability p of occuring will occur at least x times out of n trials? I remember it had something to do with combine.

While good enough for government work, what you typed there is an approximation (that requires the use of a calculator).

Of course, I'm fully aware of that. Sure it requires an approximation - I didn't pick a particularly good example, and because I'd already started typing it in text, rather than tex, I couldn't throw in an approximate sign. But it still shows the method of how to get something in the form of

t = c*pi

which the initial poster seemingly didn't know how to do. For that poster, the technique is the main point, and the technique works, don't over complicate it with talk about levels of precision and approximations.

This is an example of a binomial distribution, or to be more precise, it's a cumulative binomial function, as you're working with the probability of at least x occurances.

P(X = x) = (n choose k)p^k * (1-p)^(n-k)

Is the probability of the event occurring exactly x times. How would you modify that for it to occur at least x times?

When a solution is requested as a fraction of pi it's ONLY because an exact solution is what's acceptable. Approximating to arbitrary precision and writing this wrong answer as a fraction of pi is equivalent to putting lipstick on a pi(g)

Precision in math is important. Requiring it is not an over complication, it's what it's all about.

Since you get to define what precision you want, why not use the approximation arcsin(1/3) =~ pi/9.24 or the simpler pi/1?

I just did a quick check in my Excel because I was curious. The function you are looking for is COMBIN.

Your example would be given by COMBIN(n,x). The first slot is the total and the second slot is the number chosen.

I'm sorry, did my previous post not make it perfectly clear that I understood all of that, but considered it immaterial to the question at hand?

The specific question was how to formulate something as a fraction of pi. You can keep on waffling on, but the main point is the technique the OP chose. Congratulations, you've picked out that the example I used was a bit rubbish. Anything else is immaterial. Do you have any other silly examples, how about using the simpler pi^2? Or pi^3. Just like pi/1, your precision is to 0 decimal places. So it's not an approximation, as compared to a solution with limited precision.

For example, if the question was given as arcsin of 0.333, then pi/9.27 would be a perfectly reasonable solution, for example. It's not correct to an arbitrary precision, which is something we can do with arcsins, but when asked to provide a numerical solution to a trigonometric problem, or any function that's going to give a solution to a large number of sig figs, then truncating is pretty much acceptable. And, since the question implied that we would be asked to formulate the solution as t = c * pi, where c is not necessarily an integer, some truncation is required. So since we are here to help the OP with the technique, answering the question that he is asking, rather than leaving it in the correct form of arcsin(1/3), is probably appropriate.

If we were to go to sin(t) = 1/3, and wanted to formulate it as t = c*pi, then yes, we could certainly write it as

t = arcsin(1/3)
t = (arsin(1/3)/pi) * pi

but that would be silly, yes?

Afaik the OP never said that a fraction of pi was a requirement for that first problem. He was asked for a fraction in the case where the exact solution is a well known fraction of pi.

Can you explain why pi/7 isn't a "reasonable solution" to arcsin(1/3) if we have not been given a max error?. Reasonable to whom?

And I responded under the specific caveat that this was for any arbitrary arcsin solution, and the process would see if there was an integer value that would satisfy t = (1/c) * pi. Or, to even provide a truncated non-integer solution that could be used to approximate it to a level of precision.

To quote the OP

"But, how do I convert arcsin(1/3) to a multiple of pi?"

and

"sin(t) = sqrt(3) / sqrt(4)
t = arcsin[sqrt(3) / sqrt(4)]
How do I solve this in terms of pi?"

This is what I was responding to. When you are asked for it to be in terms of pi, then you cant just keep on saying t = arcsin(sqrt(3/4)). Could you please stop trying to tell me something I'm clearly already aware of, but have brushed over in aide of helping out the OP. Which is the main point here.

Do you see the point I've been trying to make here?

To what level of precision is pi/7 the same as arcsin(1/3)? This is not a matter of error - though, if we are talking about reasonableness, typically engineers work to 10% and students solving problems when they've only just been introduced to inverse functions generally solve to 2-3 significant figures - but a matter of precision. pi/7 is not comprably to arcsin(1/3) in any way, they are not connected.

nvm i'm moving this to another thread xD

It does, but there's not just some simple formula.

Say you pick a random card from a deck 10 times (putting it back each time), and you want to know the probability you will get a club at least 7 times. Well, there are 4 ways that can happen: we can get exactly 7 clubs, or exactly 8 clubs, or exactly 9 clubs, or exactly 10 clubs. If we can find the probability that each case happens, we'll add those together to get our answer.

Now, if you want to know the probability you get exactly 7 clubs, you can do as follows: the probability of getting a club if you pick one card is 0.25, and the probability of getting a non-club is 0.75. So the probability of a particular sequence of seven clubs and three non-clubs (say, the probability your first 7 picks are clubs and the last 3 are not) is just the product of your individual probabilities: (0.25)^7 * (0.75)^3. But there are a lot of different sequences where we can get 7 clubs and 3 not. In effect we need to choose which 7 of the 10 cards will be clubs, which can be done in 10C7 (or in the notation used above, COMBINE(10, 7) ) ways. The notation nCk just means 'the number of ways to choose a set of k things from a set of n things'. In general, nCk is equal to n!/[ k! (n-k)! ] , so 10C7 is equal to 10!/ [ 7! * 3! ]. Finally if we multiply the probability we get a particular sequence containing 7 clubs by the number of all such sequences, we find the probability of getting exactly 7 clubs:

10C3 * (0.25)^7 * (0.75)^3

= 10! /[ (3!)(7!) ] * (0.25)^7 * (0.75)^3

Similarly you can do that for exactly 8, 9 and 10 clubs and add to get your answer:

10C3 * (0.25)^7 * (0.75)^3 + 10C2 * (0.25)^8 * (0.75)^2 + 10C1 * (0.25)^9 * (0.75)^1 + 10C10 * (0.25)^10


Not quite as simple as you were probably hoping for! This whole topic goes by the name 'binomial probability' if you want to read up on it somewhere.

I am a little unsure if I set this problem up correctly.

Find the volume of the solid that lies within both the cylinder x^2+y^2 = 1 and the sphere x^2+y^2+z^2 = 4.

Looks like the solid will be a cylinder and the base will be a circle w/ radius 1 where the two intersect. so I plug in x^2+y^2= 1 into the sphere equation and get z= sqrt3. Since I am trying to find the volume I go from z=0 to z=sqrt3 and multiply the whole thing by 2 then I do a standard circle w radius one for the base and 1 for the integrand.

Is this correct?

You can do it that way, but, and this may be the early morning speaking, but you wouldn't get quite the right answer, because it would be assuming the intersection of the cylinder and the sphere on the surface of there sphere would be flat, which it isn't. From memory your class is calculus based, so you might be expected to do it in a more integrally way - and the best way to do that is to convert into cylindrical polar co-ordinates, so defining your space in terms of z, r, theta.

So if you say x^2 + y^2 = r^2, then

z^2 + r^2 = 4
z = sqrt(4 - r^2)

Since you're taking the integral of the inside of the cylinder, the z co-ordinate you would need to integrate over would be 0<z<sqrt(4-r^2)

From your cylinder formula, you have x^2 + y^2 = 1 = r^2

Then r = 1 is the outer limit, so your integral will be across 0 < r < 1

When we do integrals in cylindrical polar form, theta will range from 0 to 2*pi.

To solve, you then take the integral of

r dz dr dtheta

across those limits, and you'll get the area on the inside of a cylinder within a sphere.

This requires the use of a calculator, and depending on the fraction c you decide on the error can be arbitrarily big. I understand it gives you a ballpark idea but why use pi/9 and not pi.100000/925203?.

Why insist on a fraction for an approximation with undefined precision?. The only benefit I see is to get a ballpark idea: it's between a ninth and a tenth.

This is precisely why the use of a calculator is bad. We know the exact fraction of pi arcsin(sqrt(3/4)) is.

I say "we" because you claim to be clearly aware of it.

Your contribution is understood: given a decimal d, if you need to know what fraction of m it is then divide m/d and you get an estimate by rounding.

In this particular case though, the result the calculator gives us for arcsin(1/3) is more precise than the fraction, so why not use it instead?

The fact that trying to justify pi/7 as incorrect you must use undefined terms (like reasonabless, typical, and comparable) highlights the point I'm trying to make: if we are not given a max error that is accepted in the approximation then we can claim any number or fraction as a solution.

If you want some nice terms with a concrete definitions, then by how many significant figures of alignment do your hypothetical approximations and the exact solution provide? None.

But this is the point I stop engaging with you, thanks for the condescension.

Isn't arcsin(sqrt(3/4)) exactly pi/3 (in the first quadrant)?.

Just a note on this, it should be -sqrt(4-r^2) < z < sqrt(4-r^2)