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А) cos(π/19) + сos(3π/19) + cos(5π/19) + ... + cos(17π/19) = 1/2

Пусть φ = π/19, тогда f(φ) = cos(φ) + cos(3φ) + cos(5φ) + ... + cos(17φ) = 1/2

Домножим обе части на 2sin(φ) и применим известную формулу:

2cos(α)•sin(β) = sin(α + β) - sin(α - β)

f(φ)•2sin(φ) = 2cos(φ)•sin(φ) + 2cos(3φ)•sin(φ) + 2cos(5φ)•sin(φ) + ... + 2cos(15φ)•sin(φ) + 2cos(17φ)•sin(φ) = sin(2φ) + sin(4φ) - sin(2φ) + sin(6φ) - sin(4φ) + ... + sin(16φ) - sin(14φ) + sin(18φ) - sin(16φ) =  sin(18φ)

f(φ)•2sin(φ) = sin(18φ)  ;  sin(18φ) = sin(18π/19) = sin(π - (π/19)) = sin(π/19)

f(φ)•2sin(φ) = sin(φ)  ⇒  f(φ) = 1/2, ч.т.д.

Б) cos(6π/7) + cos(4π/7) + sin(11π/14) = - 1/2

sin(11π/14) = sin((π/2) + (2π/7)) = cos(2π/7) ⇒ cos(2π/7) + cos(4π/7) + cos(6π/7) = - 1/2

Пусть φ = π/7, тогда f(φ) = cos(2φ) + cos(4φ) + cos(6φ) = - 1/2

Таким ж образом, домножим обе части на 2sin(φ) и применим ту ж формулу:

f(φ)•2sin(φ) = 2cos(2φ)•sin(φ) + 2cos(4φ)•sin(φ) + 2cos(6φ)•sin(φ) = sin(3φ) - sin(φ) + sin(5φ) - sin(3φ) + sin(7φ) - sin(5φ) = sin(7φ) - sin(φ) = sin(7π/7) - sin(φ) = - sin(φ)

f(φ)•2sin(φ) = - sin(φ)  ⇒ f(φ) = - 1/2 , ч.т.д.