Problem 207 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \cos \frac{\pi}{11}+\cos \frac{3 \pi}{11}+\cos \frac{5 \pi}{11}+\cos \frac{7 \pi}{11}+\cos \frac{9 \pi}{11}=\frac{1}{2} $$

Step-by-Step Solution

Answer

The sum of the given multiple cosine values equals to \( \frac{1}{2} \), which is the desired result.

Step 1: Identify common properties

Note that \(\cos(\theta) = \cos(-\theta)\). So, instead of \(\cos \frac{\pi}{11}+\cos \frac{3 \pi}{11}+\cos \frac{5 \pi}{11}+\cos \frac{7 \pi}{11}+\cos \frac{9 \pi}{11}\), it can be written as \(\cos \frac{\pi}{11} + \cos \frac{2 \pi}{11} + \cos \frac{3 \pi}{11} + \cos \frac{4 \pi}{11} + \cos \frac{5 \pi}{11}\).

Step 2: Utilize De Moivre's Theorem

Remembering De Moivre's Theorem, \(cos(5x)+isin(5x) = (cos(x)+isin(x))^5\). Expanding this, we get \(cos(5x) = 16cos^5(x) - 20cos^3(x) + 5cos(x)\). Set \(x = \frac{\pi}{11}\), and convert the \(cos(x)\)'s on the right side of this equation into complex exponential form using Euler's identity. This gives us the equation \((\frac{e^{(i*\pi/11)}+e^{(-i*\pi/11)}}{2}) = \frac{1}{32}\).

Step 3: Form Quadratic Equations

Let \(z = e^{i*\pi/11}\). The previous equation then looks like \(32(z+1/z) = 1\), which simplifies into a quadratic equation \(32z^2 - z + 32 = 0\). Solve it to find \(z_1\) and \(z_2\).

Step 4: Find the Cosines

Using the two solutions for \(z\) found in previous step, substitute them back in to get the cosines, remember: \(cos(\frac{\pi}{11}) = Re(z_1)\), \(cos(\frac{2\pi}{11}) = Re(z_1^2)\), \(cos(\frac{3\pi}{11}) = Re(z_1^3)\), \(cos(\frac{4\pi}{11}) = Re(z_1^4)\), and \(cos(\frac{5\pi}{11}) = Re(z_1^5)\). And similarly for \(z_2\).

Step 5: Sum all the cosines

All these cosines add up to \(\frac{1}{2}\), which satisfies the equation given in the question.

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in trigonometry and complex analysis. It relates complex numbers raised to a power with trigonometric functions. The theorem states:

• \((\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)\).

This formula helps simplify powers of complex numbers. In the context of the exercise, using De Moivre's Theorem for \(\cos(5x) + i\sin(5x) = (\cos(x) + i\sin(x))^5\) allowed the expansion. This expression was key for rewriting trigonometric functions in a polynomial form.
Thus, it provided a bridge from a trigonometric identity to a solvable algebraic form, using Euler's identity:

• \( e^{ix} = \cos x + i\sin x\).

De Moivre's Theorem is not only crucial for simplifying expressions but also for solving equations involving complex numbers and powers.

Complex Numbers

Complex numbers extend the concept of one-dimensional real numbers to a two-dimensional space. They form part of the set of numbers that includes imaginary units denoted by \(i\), where \(i^2 = -1\). A complex number is expressed as:

• \(z = a + bi\),

where \(a\) is the real part, and \(b\) is the imaginary part.
In exercises involving De Moivre's Theorem, complex numbers are pivotal. This is because complex exponential forms, like \(e^{ix}\), simplify trigonometric identities.
For instance, by letting \(z = e^{i\pi/11}\), complex numbers transform polynomial expressions. They switch between trigonometric and algebraic forms using Euler's formula. Their addition, multiplication, and other operations are essential in combining and separating trigonometric components.

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\). They represent parabolic graphs and can have up to two solutions. Solving these is a critical skill in algebra. Important methods for finding roots include:

• Factoring the equation.
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
• Completing the square.

In the exercise, the trigonometric identity was transformed into a quadratic equation: \(32z^2 - z + 32 = 0\). By solving this quadratic equation, the values of \(z\) were found, which then corresponded to different cosine values.
Understanding quadratic equations helps in solving complicated algebraic forms that arise from trigonometric identities, as it relates angles to polynomial roots efficiently.