Question

If the value of $\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} + \cos\frac{7\pi}{7} = \frac{l}{2}$. Find the value of $l$.

Answer 1

-3

Answer 2

The problem asks for the value of l given the equation: cos(2π/7) + cos(4π/7) + cos(6π/7) + cos(7π/7) = l/2.

First, let's evaluate the last term: cos(7π/7) = cos(π) = -1.

Now, consider the sum of the first three terms: S' = cos(2π/7) + cos(4π/7) + cos(6π/7). This sum can be evaluated using the properties of roots of unity. Consider the equation z^7 - 1 = 0. The roots are z_k = e^(i2kπ/7) for k = 0, 1, ..., 6. The sum of these roots is zero: Σ_(k=0)^6 z_k = 0 e^(i0) + e^(i2π/7) + e^(i4π/7) + e^(i6π/7) + e^(i8π/7) + e^(i10π/7) + e^(i12π/7) = 0.

Using Euler's formula e^(ix) = cos(x) + i sin(x), and taking the real part of the sum: 1 + cos(2π/7) + cos(4π/7) + cos(6π/7) + cos(8π/7) + cos(10π/7) + cos(12π/7) = 0.

We use the property cos(2π - x) = cos(x) and cos(π + x) = -cos(x): cos(8π/7) = cos(π + π/7) = -cos(π/7) cos(10π/7) = cos(π + 3π/7) = -cos(3π/7) cos(12π/7) = cos(2π - 2π/7) = cos(2π/7)

However, a simpler symmetry can be used: cos(2(n-k)π/n) = cos(2kπ/n). For n=7: cos(12π/7) = cos(2π(6/7)) = cos(2π(1-1/7)) = cos(2π/7). cos(10π/7) = cos(2π(5/7)) = cos(2π(1-2/7)) = cos(4π/7). cos(8π/7) = cos(2π(4/7)) = cos(2π(1-3/7)) = cos(6π/7).

Substituting these into the sum of cosines: 1 + cos(2π/7) + cos(4π/7) + cos(6π/7) + cos(6π/7) + cos(4π/7) + cos(2π/7) = 0. 1 + 2[cos(2π/7) + cos(4π/7) + cos(6π/7)] = 0. 2[cos(2π/7) + cos(4π/7) + cos(6π/7)] = -1. cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2.

This matches the result from the similar question.

Now, substitute this value back into the original equation: (-1/2) + cos(7π/7) = l/2. (-1/2) + (-1) = l/2. -1/2 - 2/2 = l/2. -3/2 = l/2. Therefore, l = -3.