Abstract Algebra Question dealing with Constructible Numbers

[26xc37]

Homework Statement

Let c=cos(2pi/5). It can be shown that (4c^2)+(2c)-1=0. Use this fact to prove that a 72degree angle is constructible.

Homework Equations

The Attempt at a Solution

I can see that using the equation and what c equals that you get the statement 0=0 and I know that 2pi/5 relates to 72degree angle on the unit circle but I have no idea how to formally prove it. Also can equate cos(2pi/5) but don't see how that would help either.

 

[mighty2000]

Hello!

You are on the right track with your approach. Let's break down the steps to formally prove that a 72 degree angle is constructible using the given equation.

Step 1: Start by stating the given equation, (4c^2)+(2c)-1=0, and substitute c=cos(2pi/5) into it. This will give us (4cos^2(2pi/5))+(2cos(2pi/5))-1=0.

Step 2: Use the double angle formula for cosine, cos(2x)=2cos^2(x)-1, to simplify the equation. This will give us (4cos^2(2pi/5))+(2cos(2pi/5))-1=0 becomes (2cos(4pi/5)+1)(2cos(2pi/5)+1)=0.

Step 3: Notice that we have two factors in the equation and both of them are equal to 0. This means that either (2cos(4pi/5)+1)=0 or (2cos(2pi/5)+1)=0.

Step 4: Now we can solve each of these equations separately. For (2cos(4pi/5)+1)=0, we can solve for cos(4pi/5) by subtracting 1 from both sides and then dividing by 2. This gives us cos(4pi/5)=-1/2.

Step 5: We know that cos(4pi/5)=cos(2pi-2pi/5) using the angle sum formula for cosine. This means that cos(2pi-2pi/5)=-1/2.

Step 6: We can now use the inverse cosine function to find the angle that has a cosine of -1/2. This angle is 2pi-2pi/5=4pi/5.

Step 7: Since we know that cos(2pi/5)=cos(36 degrees), we can conclude that cos(4pi/5)=cos(72 degrees). Therefore, we have proven that a 72 degree angle is constructible.

I hope this helps! Let me know if you have any further questions.