Question

`cos  (2π)/7 + cos  (4π)/7 + cos  (6π)/7`

विकल्प

• is equal to zero

• lies between 0 and 3

• is a negative number

• lies between 3 and 6

Answer 1

is a negative number

Explanation:

Let `(2πr)/7` = θ

`\implies` 2πr = 3θ + 4θ

`\implies` 4θ = 2πr – 3θ

`\implies` sin 4θ = sin (2πr – 3θ)

`\implies` sin 4θ = – sin 3θ

`\implies` 2 sin 2θ cos 2θ = – [3 sin θ – 4 sin³ θ]

`\implies` 2 × 2 sin θ cos θ (2 cos² θ – 1) = – 3 sin θ + 4 sin³ θ

`\implies` sin θ [8 cos³ θ – 4 cos θ + 3 – 4 (1 – cos² θ)] = 0

`\implies` 8 cos³ θ + 4 cos² θ – 4 cos θ – 1 = 0  ...(i)

Thus, `cos  (2π)/7, cos  (4π)/7` and `cos  (6π)/7` are the roots of equation (i).

∴ `cos  (2π)/7 + cos  (4π)/7 + cos  (6π)/7 = - 1/2`