№1
sin(1260°) + tg(-2460°) = ?
sin(1260°) = sin(180° • 7) = sin(360° • 3 + 180°) = sin(180°) = 0
tg(-2460°) = -tg(180° • 13 + 120°) = -tg(120°) = -√3
sin(1260°) + tg(-2460°) = -√3
№2
sin α = -√3/3
3π/2 < α < 2π
Найти:
cos α
tg α
ctg α
• cos α = ± √(1 - sin²α) = ± √(1 - ⅓) = ± √⅔
Так как 3π/2 < α < 2π, значит α ∈ IV четверти, ⇒ cos α > 0
⇒ cos α = √⅔
• tg α = sin α / cos α = -√3/3 : √⅔ = - 3/3√2 = -3√2/6 = - √2/3
• ctg α = 1/tg α = 1 : - √2/3 = - 3√2/2
№3
(1 + ctg²α) • sin²α - 1 = 1/sin²α • sin²α - 1 = 1 - 1 = 0
