Prove that cos 20 is irrational

[xax]

I've started to work on the it, just tell me if I'm on the right track.
cos (45-30) = ($$\sqrt{3}$$ + 1) / 2$$\sqrt{2}$$ so cos 15 is irrational.
cos3x = 4cos^3x - 3cos x $$\Rightarrow$$ 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

 

[robert Ihnot]

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=$$\frac{\pm1}{2^k}$$. We see that none work,and so there is no rational solution.

 

[xax]

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?

 

[robert Ihnot]

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?

 

[xax]

Thanks again robert.