§ 30: 30.25 (1)

30.25 Обчисліть значення виразу:

1) sin(arctg 2) = застосуємо формулу 1sin2α = 1 + ctg²α, де α = arctg2 = ± 11+ ctg2(arctg2) = оскільки arctg2 > 0 і arctgx ∈ (–π2; π2), то sin(arctg 2) > 0 = 1 :(1+ 1tg2arctg2) = 1 :(1+ 14) = 1 : 54 = 45 = 255;

2) cos(arctg12 – arcctg3) = cos(arctg12) • cos(arcctg3) + sin(arctg12) • sin(arcctg3) • cos(arctg12) : 1cos2(arctg12) = 1 + tg²(arctg12);

cos(arctg12) = ± 1+ 1tg2(arctg 12) = оскільки arctg12 ∈ 1, то cos(arctg12) > 0 = 11+ 14 = 45 = 25; sin²(arctg12) + cos²(arctg12) = 1 ⇒ sin(arctg12) = ± 1- cos2(arctg12) = оскільки arctg12 ∈ 1, то sin(arctg12) > 0 = 1-(25)2 = 15 = 15.

Знайдемо sin(arcctg3): 1sin2(arcctg3) = 1 + ctg²(arcctg 3); оскільки arctg3 ∈ 1, то sin(arcctg3) > 0 = 11+9= 110 = 110;

sin²(arcctg3) + cos²(arcctg3) = 1 ⇒ ± 1- sin2(arcctg3) =

= 25 • 310 + 15 • 110 = 150 (2 • 3 + 1) = 752 = 7210.