Trying to figure out a single variable expression for Sin

In summary, the conversation discusses the search for a single variable algebraic expression for the sin wave, and various suggestions and strategies are proposed, including the use of Taylor polynomials and the unit circle. However, it is ultimately concluded that there is no exact algebraic expression for sin, and attempting to find one may be a futile effort.
  • #1
mesa
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Well, I am now trying to figure out a single variable expression for Sin. I have a couple ideas using some pieces of geometric formulas I have played with recently but this is still new to me. I'm not talking about sin x, but an algebraic expression for the sin wave.

Any thoughts?
 
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  • #2
You need to clarify what you have in mind.
single variable expression for Sin
 
  • #3
mesa said:
Well, I am now trying to figure out a single variable expression for Sin. I have a couple ideas using some pieces of geometric formulas I have played with recently but this is still new to me. I'm not talking about sin x, but an algebraic expression for the sin wave.

Any thoughts?

Do you mean that you want an expression for sin that involves only addition/subtraction, multiplication/division and exponentiation? Have you tried a Taylor polynomial?
 
  • #4
y= sin(x- ct) is a perfectly good formula for a sine wave moving with speed c. Did you have something else in mind?
 
  • #5
mathman said:
You need to clarify what you have in mind.

An algebraic expression that will compute y for a given x on a sin wave.
 
  • #6
Number Nine said:
Do you mean that you want an expression for sin that involves only addition/subtraction, multiplication/division and exponentiation? Have you tried a Taylor polynomial?

Basically I want an algebraic expression (if it can be done in one piece) that will spit out the right y values for a given x
What is a Taylor polynomial?
 
  • #7
mesa said:
Basically I want an algebraic expression (if it can be done in one piece) that will spit out the right y values for a given x
What is a Taylor polynomial?

Something that you'll learn about in calculus; it's a way of approximating certain kinds of functions using polynomials. The sin function is nice in the sense that it is actually everywhere equal to it's Taylor series.
 
  • #8
In terms of the Taylor series expression for x, about x= 0, also called the "McLaurin series", is [itex]x- x^3/3!+ x^5/5!- \cdot\cdot\cdot+ ((-1)^n/(2n+1)!)x^{2n+1}+ \cdot\cdot\cdot[/itex].
 
  • #9
Number Nine said:
Something that you'll learn about in calculus; it's a way of approximating certain kinds of functions using polynomials. The sin function is nice in the sense that it is actually everywhere equal to it's Taylor series.

Is it possible to get an exact function for sin?
 
  • #10
mesa said:
Is it possible to get an exact function for sin?

With a finite number of terms, it can only be approximated (to any conceivable degree of precision, mind you). I don't think there is any other algebraic expression.
 
  • #11
Well including the complex plane I'd reckon that [itex]\frac{e^ix-cos(x)}{i}[/itex] should be equal to sin(x), but I doubt that's what you are looking for. I don't think there is a non-infinite algebraic function that will accomplish what you are looking for.
 
  • #12
Vorde said:
Well including the complex plane I'd reckon that [itex]\frac{e^ix-cos(x)}{i}[/itex] should be equal to sin(x), but I doubt that's what you are looking for. I don't think there is a non-infinite algebraic function that will accomplish what you are looking for.

Technically correct but not quite what I'm looking for, cos put's me back to square one lol

Number Nine said:
With a finite number of terms, it can only be approximated (to any conceivable degree of precision, mind you). I don't think there is any other algebraic expression.

Interesting, well let's give it a shot.
Any thoughts on where to start? I was going to use the unit circle and partially filled areas to see what I can pull from that.
 
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  • #13
mesa said:
Is it possible to get an exact function for sin?
I still don't know what you mean by this. "y= sin(x)" is a perfectly good function and is every bit as "exact" as, say, [itex]\sqrt{x}[/itex] and [itex]e^x[/itex]. So the question is what do you mean by "exact" here?
 
  • #14
HallsofIvy said:
I still don't know what you mean by this. "y= sin(x)" is a perfectly good function and is every bit as "exact" as, say, [itex]\sqrt{x}[/itex] and [itex]e^x[/itex]. So the question is what do you mean by "exact" here?

That it is, but I want to see what sin actually looks like algebraically, and it should be a fun exercise.
 
  • #15
mesa said:
That it is, but I want to see what sin actually looks like algebraically, and it should be a fun exercise.

It doesn't look like anything, algebraically; it's a trigonometric function. The closest you can get is it's Taylor series, which has already been posted.
 
  • #16
Number Nine said:
It doesn't look like anything, algebraically; it's a trigonometric function. The closest you can get is it's Taylor series, which has already been posted.

Most likely true. Worst case this is a good exercise for an older returning student. I want to get a strong handle on this stuff and I really enjoy digging into what is taught in my classes.

I have an idea about what I want to try and would love some input from people like you. Heck, you likely will have a completely different and better strategy than mine, but most importantly I will learn from this process!

Math is a big subject and I am regularly amazed at what it can accomplish at the hands of those who know how to wield it ;)
 
  • #17
Well I'd say the only luck you are going to get is with the taylor polynomial, but that won't tell you what sine 'actually looks like'- just how to mimic it to an arbitrary precision.

I can't see a way to write an equation for sine without using other trigonometric functions, but good luck to you!
 
  • #18
Well I have to admit this one has been challenging, although I'm not throwing in the towel yet.
 
  • #19
mesa said:
Well I have to admit this one has been challenging, although I'm not throwing in the towel yet.

Well, you should. As has been clearly explained to you, sin has no expression in term of elementary algebraic operations. Trying to square the circle is not noble, just futile.
 
  • #20
Here is another way. Do you remember the Pythagorean theorem? A2 + B2 = C2, where C is the hypotenouse, and A is close to the origin?
Choose C=1, and using the angle between A and C, slowly increase the angle from 0 to 90 (or 180) degrees. The length of B is equal to the sine of the angle.
 
  • #21
Number Nine said:
Well, you should. As has been clearly explained to you, sin has no expression in term of elementary algebraic operations. Trying to square the circle is not noble, just futile.

This is supposed to be a learning exercise ;)

Bob S said:
Here is another way. Do you remember the Pythagorean theorem? A2 + B2 = C2, where C is the hypotenouse, and A is close to the origin?
Choose C=1, and using the angle between A and C, slowly increase the angle from 0 to 90 (or 180) degrees. The length of B is equal to the sine of the angle.

I've played with pythagorea's but realized that would just give me a function of y for a given x when the Sine curve is based on a value for y based on the length of the radian created on a unit circle. Could play around with deformations and such with pythagorea's but I thought I would try something else first.

The formula for calculating the area of a partially filled circle is based on a value for θ while the formula for the area of a partially filled sphere is based on a value for our y. If I can convert the first formula into 3 dimensions while preserving θ I could try to pull out what we need from the two by setting them equal to each other.

Here is what I have so far:

∏ x r^2 - ((∏ x r^2/2) + (∏ x r^2(2(θ/ 2 x Pi)) + (sin θ)(cos θ))

It calculates the area of a partially filled circle (at least one that is less than half full lol) with θ being the angle created by the dividing line of the circle to the radius to the top of the segment for the area which we are calculating for. It spits out the right number for area given the restriction above.

Next I wanted to see what took a circle to a sphere so divided (4/3)∏r^3 by ∏r^2 resulting in 4/3r. Unfortunately I can't just take 4/3r and multiply it into the formula above; that would be like taking the area of the circle segment, setting it equal to a smaller circles' area and then calculating for the volume of a sphere of that smaller circles' dimension of radius. Basically useless in most cases.

I noticed that ∏ can be looked at as a constant (2Pi really but that's besides the point). I was thinking I could use a function of 4/3r and the arc length to get my third dimension and get our θ based formula.

Any thoughts, and/or corrections?

Trig is remarkable! :)
 
  • #22
Finding a polynomial expression for sine is not really possible. All of equivalent sine definitions take sine as a transcendental function that does not have an expansion in terms of polynomials or algebraic expressions. I will still give some definitions for sine, maybe you might be interested in one of them.

[itex]\displaystyle \sin(x)=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}[/itex]
[itex]\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/itex] where i is the imaginary unit.
[itex]\displaystyle \sin(x)=\Re(e^{ix})[/itex]
[itex]\displaystyle \sin(x)=\frac{\pi}{\Gamma(x/\pi)\Gamma(1-x/\pi)}[/itex] where [itex]\Gamma(x)[/itex] denotes the Gamma function.

Sine is the only solution to the differential equation [itex]\displaystyle \frac{d^2 y}{d x^2}+y=0[/itex] satisfying [itex]y(0)=0,\,y'(0)=1[/itex].

Sine is the only solution to the integral equation [itex]\displaystyle \int_{0}^{f(a)}\frac{dx}{1-x^2}=a[/itex] for the function f.

You should be able to derive more by yourself.
 

1. How do you solve a single variable expression for Sin?

The first step is to identify the variable in the expression. In this case, it is the angle (x). Then, use the trigonometric identity sin^2(x) + cos^2(x) = 1 to eliminate the squared term. Next, use algebraic manipulations to isolate sin(x) on one side of the equation. Finally, use a calculator or table to find the value of sin(x) for the given angle.

2. Can you explain the process of finding the value of Sin for a given angle?

To find the value of Sin for a given angle, first convert the angle from degrees to radians. Then, use the unit circle to determine the coordinates of the point on the circle corresponding to the angle. The y-coordinate of this point is the value of Sin for the given angle.

3. What are some common mistakes to avoid when solving a single variable expression for Sin?

Some common mistakes include forgetting to convert the angle from degrees to radians, using the wrong trigonometric identity, and not isolating sin(x) properly. It is important to double check all steps and use a calculator or table to verify the final answer.

4. Can you provide an example of solving a single variable expression for Sin?

Sure. Let's say we have the expression sin(2x) - 1 = 0. First, we use the double angle formula sin(2x) = 2sin(x)cos(x) to expand the expression. Then, we use algebra to isolate sin(x) and get sin(x) = 1/2. Using a calculator or table, we find that the angle corresponding to this value of sin(x) is 30 degrees or π/6 radians.

5. How is solving a single variable expression for Sin useful in real world applications?

Solving single variable expressions for Sin is useful in various fields such as engineering, physics, and astronomy. It can help calculate the height of objects, the angles of trajectory, and the position of celestial bodies. It is also used in everyday tasks such as navigation and construction.

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