Question

объясните пожалуйста, как они пришли к таким формулам приведения

Answer 1

1) х = π/12 + πn, n ∈Z

Sin(3x + π/4) = Sin(3*(π/12 + πn) + π/4) = Sin(π/4 + 3πn + π/4 ) =

=Sin(π/2 + 3πn) = Sin(π/2 + πn + 2πn) = Sin(π/2 + πn) = Cosπn = (-1)^n ≥ 0, ⇒

⇒ n = 2k.

2) x = 5π/12 + πn

Sin(3x +π/4) = Sin(3* (5π/12 + πn) + π/4) = Sin(5π/4 + 3πn + π/4) =

= Sin(3π/2 + 3πn) = Sin(3π/2 + π + 2πn) = Sin(3π/2 + πn) =

= -Cosπn = -(-1) ^k ≥ 0, ⇒ k = 2m+1