Question

Cos0 + cos p/7 + cos 2p/7 +...+ cos 6p/7

Answer 1

Cos 6pi/7 = cos (pi - pi/7) = -cos pi/7
cos 5pi/7 = cos (pi - 2pi/7) = -cos 2pi/7
cos 4pi/7 = cos (pi - 3pi/7) = -cos 3pi/7

cos 0 + cos pi/7 + cos 2pi/7 + cos 3pi/7 - cos 3pi/7 - cos 2pi/7 - cos pi/7 =
= cos 0 = 1

Answer 2

To simplify the expression, let's consider the sum:

cos(0) + cos(p/7) + cos(2p/7) + ... + cos(6p/7)

First, let's notice a pattern:

cos(0) = cos(2π * 0) = cos(0) = 1 cos(p/7) = cos(2π/7) (since p/7 = 2π/7) cos(2p/7) = cos(4π/7) (since 2p/7 = 4π/7) ... cos(6p/7) = cos(12π/7) (since 6p/7 = 12π/7)

In general, for any integer n between 0 and 6 (inclusive), we have:

cos(np/7) = cos(2nπ/7)

Now, we can rewrite the sum as:

cos(0) + cos(p/7) + cos(2p/7) + ... + cos(6p/7) = 1 + cos(2π/7) + cos(4π/7) + ... + cos(12π/7)

Since cos(2π) = 1, we can rewrite the last term:

cos(12π/7) = cos(2π + 5π/7) = cos(5π/7)

Now, the sum becomes:

1 + cos(2π/7) + cos(4π/7) + ... + cos(12π/7) = 1 + cos(2π/7) + cos(4π/7) + ... + cos(5π/7)

The above expression is now a sum of cosines of consecutive angles in arithmetic progression with a common difference of 2π/7.

To find the sum of cosines in an arithmetic progression, we can use the following formula:

Sum of cosines in an arithmetic progression = n * cos((first term + last term) / 2)

where n is the number of terms.

In this case, the number of terms is 6 (from 0 to 5), and the first term is cos(2π/7) and the last term is cos(5π/7).

Now, we can calculate the sum:

Sum = 6 * cos((2π/7 + 5π/7) / 2) = 6 * cos(7π/14) = 6 * cos(π/2) = 6 * 0 = 0

Therefore, the value of the expression cos(0) + cos(p/7) + cos(2p/7) + ... + cos(6p/7) is 0.