Proving the Roots of the Polynomial x^3 - 3x - 1 0 Using Trigonometric Identities

To prove that the roots of the polynomial equation

x^3 - 3x - 1 0

are

2cos2π/9 2cos8π/9 2cos14π/9

We can follow these steps:

Step 1: Use the Roots Formulation

We can express the roots of the polynomial in terms of cosine. The roots can be expressed as:

xk 2cos

for k 0, 1, 2.

- For k 0: x0 2cos - For k 1: x1 2cos - For k 2: x2 2cos

Step 2: Verify the Roots

We need to show that these values satisfy the equation x^3 - 3x - 1 0.

First, calculate x_k^3: Using the triple angle formula, we have:

x_k^3 2cos(theta_k)^3 8cos(theta_k)^3

where theta_k frac{(2k-1)pi}{9}

Next, use the triple angle identity: cos(3theta) 4cos(theta)^3 - 3cos(theta) Rearranging gives:

4cos(theta_k)^3 cos(3theta_k) - 3cos(theta_k)

Therefore: cos(theta_k)^3 frac{cos(3theta_k) - 3cos(theta_k)}{4} Substituting this into the expression for x_k^3:

x_k^3 8left(frac{cos(3theta_k) - 3cos(theta_k)}{4}right) 2cos(3theta_k) - 6cos(theta_k) Calculate 3x_k: 3x_k 3(2cos(theta_k)) 6cos(theta_k) Substituting into the polynomial:

x_k^3 - 3x_k - 1 2cos(3theta_k) - 6cos(theta_k) - 6cos(theta_k) - 1

This simplifies to: x_k^3 - 3x_k - 1 2cos(3theta_k) - 1 Evaluate cos(3theta_k): - For theta_0 frac{2pi}{9}, we have 3theta_0 frac{6pi}{9} frac{2pi}{3}, thus:

cos(3theta_0) -frac{1}{2}

- For theta_1 frac{8pi}{9}, we have 3theta_1 frac{24pi}{9} frac{8pi}{3} equiv frac{2pi}{3}, thus:

cos(3theta_1) -frac{1}{2}

- For theta_2 frac{14pi}{9}, we have 3theta_2 frac{42pi}{9} frac{14pi}{3} equiv frac{2pi}{3}, thus:

cos(3theta_2) -frac{1}{2}

Therefore, for all k:

2cos(3theta_k) - 1 2left(-frac{1}{2}right) - 1 -1 - 1 0

Conclusion

Since x_k^3 - 3x_k - 1 0 for k 0, 1, 2, we conclude that the values

2cos2π/9 2cos8π/9 2cos14π/9

are indeed the roots of the polynomial x^3 - 3x - 1 0.