Prove that cos(2∏/7), cos(4∏/7), cos(6∏/7) are the roots of this equation.

Michael_Light · Dec 19, 2012

1. The problem statement, all variables and given/known data

When cos(4θ)=cos(3θ). prove that θ=0, 2∏/7, 4∏/7, 6∏/7

Hence prove that cos(2∏/7), cos(4∏/7), cos(6∏/7) are the roots of 8x³+4x²-4x-1

2. Relevant equations

3. The attempt at a solution

I can do the first part, but i have some difficulty in solving the second part. For the second part, I start by letting x= cosθ and try to solve the equation, however, i notice that i couldn't simplify the equation... so it is correct to let x=cosθ? How is the second part related to the first part? Thanks in advance.

Dick · Dec 19, 2012

Michael_Light said: ↑

1. The problem statement, all variables and given/known data

When cos(4θ)=cos(3θ). prove that θ=0, 2∏/7, 4∏/7, 6∏/7

Hence prove that cos(2∏/7), cos(4∏/7), cos(6∏/7) are the roots of 8x³+4x²-4x-1

2. Relevant equations

3. The attempt at a solution

I can do the first part, but i have some difficulty in solving the second part. For the second part, I start by letting x= cosθ and try to solve the equation, however, i notice that i couldn't simplify the equation... so it is correct to let x=cosθ? How is the second part related to the first part? Thanks in advance.

Look at the equation cos(4θ)-cos(3θ)=0 and expand the multiple angles in terms of cos(θ). Yes, put x=cos(θ). Then try to factor it.

haruspex · Dec 19, 2012

Michael_Light said: ↑

so it is correct to let x=cosθ?

Yes. Do you know how to expand cos(a+b)?

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Prove that cos(2∏/7), cos(4∏/7), cos(6∏/7) are the roots of this equation.

Michael_Light

Dick

haruspex

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