Prove that cos 20 is irrational

In summary, the conversation is discussing whether or not there is a rational solution to the equation 8x^3 - 6x - 1. However, when looked at from a Euclidean perspective, there is no such solution.
  • #1
xax
26
0
I've started to work on the it, just tell me if I'm on the right track.
cos (45-30) = ([tex]\sqrt{3}[/tex] + 1) / 2[tex]\sqrt{2}[/tex] so cos 15 is irrational.
cos3x = 4cos^3x - 3cos x [tex]\Rightarrow[/tex] cos 5 is irrational
cos 4x... cos 20
If this is a bad way, maybe someone knows a better one.

Here is what I think a better way:

cos3x = 4cos^3x - 3cosx so cos20 is the solution for 8x^3 - 6x - 1 and this one has no rational solutions so cos20 is irrational
 
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  • #2
xax: this one has no rational solutions so cos20 is irrational

You are correct. This can be looked at from the standpoint of Euclidean construction. And since, as you may have heard, you can not trisect an angle using the method of the Greeks, so you can not construct a 20 degree angle by trisecting a 60 degree angle. Since you can construct any fraction, this is saying there is no rational solution.

On the other hand, set r=p/q, which is in lowest terms, and consider 8p^3/q^3 -6p/q-1 = 0. Multiplying by q we get (8qp^3)/q^2 -6p-q= 0. This indicates first that q divides 8 and second that p divides q. Since p and q are in lowest terms, we find that p must be plus or minus 1, and that q is a divisor of 8.

So that trying the various cases, k=0,1,2,3 we have r=[tex] \frac{\pm1}{2^k}[/tex]. We see that none work,and so there is no rational solution.
 
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  • #3
Hi robert. I was thinking instead of working on 8x^3 - 6x - 1 to substitute 2x with y and so the equivalent equation y^3 - 3y - 1 will be easier to prove it has no rational solutions( r can be only +/- 1). What do you think?
 
  • #4
xax said:
Hi robert. I was thinking instead of working on 8x^3 - 6x - 1 to substitute 2x with y and so the equivalent equation y^3 - 3y - 1 will be easier to prove it has no rational solutions( r can be only +/- 1). What do you think?

If it helps to make make things clearer, why not?
 
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  • #5
Thanks again robert.
 

1. What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a ratio of two integers (whole numbers), or in other words, if it cannot be written as a fraction.

2. How can you prove that cos 20 is irrational?

To prove that cos 20 is irrational, we need to assume the opposite, which is that cos 20 can be expressed as a fraction, and then show that this leads to a contradiction.

3. Can you provide an example of a contradiction when assuming cos 20 is rational?

Yes, if we assume that cos 20 can be expressed as a fraction a/b, where a and b are integers, we can use the double angle identity for cosine to get cos 20 = (2cos^2 10 - 1)/2. This means that 2a^2 = 2b^2 + b, which leads to a contradiction since the left side is even and the right side is odd.

4. Is there a simpler way to prove that cos 20 is irrational?

Yes, we can also use the fact that cos 20 is a zero of the polynomial 8x^3 - 6x - 1, which has integer coefficients. This means that if cos 20 were rational, it would be a root of a polynomial with integer coefficients, which is a known theorem in number theory. However, since cos 20 is not a rational number, this polynomial cannot have a rational root.

5. Why is it important to prove that cos 20 is irrational?

Proving that cos 20 (and other trigonometric functions) is irrational is important in mathematics as it helps us understand the nature of real numbers and their properties. It also has applications in other fields, such as physics and engineering, where irrational numbers are commonly used in calculations.

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