Problem 5 - Algebra and Geometry

Question

Show that \(\cos (\pi / 5)=\lambda / 2\), where \(\lambda=(1+\sqrt{5}) / 2\) (the Golden Ratio). [Hint: As \(\cos 5 \theta=1\), where \(\theta=2 \pi / 5\), we see from De Moivre's theorem that \(P(\cos \theta)=0\) for some polynomial \(P\) of degree five. Now observe that \(P(z)=(1-z) Q(z)^{2}\) for some quadratic polynomial \(\left.Q .\right]\)

Step-by-Step Solution

Answer

The polynomial roots confirm that \( \cos(\pi/5) = \frac{1+\sqrt{5}}{4} = \frac{\lambda}{2} \).

Step 1: Set Up the Problem

We need to show that \( \cos(\pi / 5) = \frac{\lambda}{2} \), where \( \lambda = \frac{1 + \sqrt{5}}{2} \). The key hint is that \( \cos 5\theta = 1 \) gives \( \cos 2\theta = \cos(\pi - 2\theta) \) when \( \theta = \frac{2\pi}{5} \), thus you need to find the equation that \( \cos \theta \) satisfies.

Step 2: Identify the Polynomial

Using De Moivre's theorem, \( \cos(5\theta) = 1 \) implies a polynomial \( P(\cos \theta) = 0 \). If we let \( z = \cos \theta \), observe that \( P(z) = 0 \) for the degree 5 polynomial \( P(z) = 16z^5 - 20z^3 + 5z - 1 \) by expanding \( \cos(5\theta) \) using \( \cos \theta \).

Step 3: Factor the Polynomial

Given that \( P(z) = (1-z)Q(z)^2 \) for some quadratic polynomial \( Q(z) \), we can find \( Q(z) = 4z^2 - 2z - 1 \). This implies \( (1-z)(4z^2 - 2z - 1)^2 \) should yield \( P(z) \) when expanded.

Step 4: Solve the Quadratic Equation

To find specific roots, solve \( 4z^2 - 2z - 1 = 0 \) using the quadratic formula, \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 4, b = -2, c = -1 \) leading to: \[ z = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{2 \pm \sqrt{20}}{8} = \frac{2 \pm 2\sqrt{5}}{8} = \frac{1 \pm \sqrt{5}}{4} \].

Step 5: Select the Correct Root

Since \( \theta = \frac{\pi}{5} \) corresponds to a positive angle in \( (0,\pi/2) \), use the positive root \( z = \frac{1 + \sqrt{5}}{4} \). This gives \( \cos \left(\frac{\pi}{5}\right) = \frac{\lambda}{2} \), confirming that \( \lambda = \frac{1 + \sqrt{5}}{2} \).

Key Concepts

De Moivre's Theorem

De Moivre's theorem is a powerful tool in complex number arithmetic, particularly useful in finding powers and roots of complex numbers expressed in polar form. It states that for any complex number in polar coordinates, if \[ z = r (\cos \theta + i \sin \theta), \] then its nth power is given by:\[ z^n = r^n (\cos(n\theta) + i \sin(n\theta)). \]
This theorem is instrumental when dealing with polynomial roots, as it allows us to express powers of cosine and sine in terms of the angles. In our exercise, it's used to expand \( \cos(5\theta) \), helping us to form a polynomial equation.

• Relates to any angle \( \theta \), with n being the power.
• Links trigonometry with complex numbers.

Understanding De Moivre's theorem is crucial for solving problems involving trigonometric identities and polynomials, as it simplifies the process of raising complex numbers to powers.

Cosine of Angle

The cosine of an angle is a fundamental trigonometric function that describes the adjacent side over the hypotenuse in a right triangle. In the context of this exercise, calculating the cosine of an angle related to the golden ratio links trigonometry with algebra.
Given \( \theta = \frac{\pi}{5} \), our goal was to demonstrate that \( \cos(\pi/5) = \frac{\lambda}{2} \). This involves analyzing the angle through the lens of polynomial equations derived from De Moivre's theorem.

• \( \cos \theta \) is given by the formula \( \cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta}). \)
• Assists in connecting angles to polynomial roots by the derived expression.

Thus, understanding how to manipulate the cosine function and express it through equations is key to linking angles with solutions of quadratic equations in such exercises.

Polynomial Roots

Polynomials play an essential role in algebra, and their roots are the solutions to the equation set equal to zero. In our exercise, we identified a polynomial of degree 5 to determine the roots through which the cosine of a particular angle can be expressed.
Finding the roots involves using De Moivre's theorem, with the polynomial expressed as \[ P(z) = 16z^5 - 20z^3 + 5z - 1. \]This polynomial emerges from expanding the cosine of angles.

• The roots are found by solving polynomial equations derived from the cosine identity.
• Each root corresponds to specific angles when considered in trigonometric contexts.

Connecting polynomials and trigonometric identities, like in this example, highlights the intersection of different mathematical fields, showcasing how algebra and trigonometry complement each other.

Quadratic Equations

Quadratic equations are polynomials of degree two, typically having the form \( ax^2 + bx + c = 0 \). In our analysis, the quadratic equation in question arises from factoring a higher-degree polynomial and finding roots that correspond to trigonometric values.
The quadratic equation derived from the polynomial factor is given by \[ 4z^2 - 2z - 1 = 0. \]It is solved using the quadratic formula:\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

• Solving this quadratic yields values directly connected to cosine angles.
• This step involves determining which root aligns with specific trigonometric conditions, such as angle signs.

Ultimately, these calculations demonstrate how quadratic equations provide a simpler means to solve problems initially given in more complex polynomial forms.