Предмет: Алгебра,
автор: fklm88
Помогите, срочно!! Любые номера
Приложения:
Ответы
Автор ответа:
0
1)
sqrt(3) * sin(60) + cos(60) * sin(30) - tg(45) * ctg(135) + ctg(90) = 3 / 2 + 1 / 4 - (1 * -1) + 0 = 6 / 4 + 1 / 4 + 1 = 11/4
2)
(1-cosA)(1+cosA)/sinA = 1 - cos^2A / sinA = sin^2(A) / sin(A) = sin(A)
sin(2pi + a) + cos(pi + a) + sin(-a) + cos(-a) = sin(a) - cos(a) - sin(a) + cos(a) = 0
3)
(sin(a)+cos(a))^2 - 2sin(a)cos(a) = sin^2(a) + cos^2(a) + 2sin(a)cos(a) - 2sin(a)cos(a) = sin^2(a) + cos^2(a) = 1
tg(a) + ctg(a) = (sin^2(a) + cos^2(a))/sin(a)cos(a) = 1 / 0.4 = 2.5
4)
sin(a) = sqrt(3) / 2 => a = (-1)^k * pi / 3 + pi * k, k ∈ Z
cos(a) = sqrt(2) / 2 => a = +- 2pi / 3 + 2pi*k, k ∈ Z
tg(a) = sqrt(3) => a = pi / 3 + pi * k, k ∈ Z
ctg(a) = sqrt(2) / 2 => a = -pi / 4 + pi * k, k ∈ Z
sqrt(3) * sin(60) + cos(60) * sin(30) - tg(45) * ctg(135) + ctg(90) = 3 / 2 + 1 / 4 - (1 * -1) + 0 = 6 / 4 + 1 / 4 + 1 = 11/4
2)
(1-cosA)(1+cosA)/sinA = 1 - cos^2A / sinA = sin^2(A) / sin(A) = sin(A)
sin(2pi + a) + cos(pi + a) + sin(-a) + cos(-a) = sin(a) - cos(a) - sin(a) + cos(a) = 0
3)
(sin(a)+cos(a))^2 - 2sin(a)cos(a) = sin^2(a) + cos^2(a) + 2sin(a)cos(a) - 2sin(a)cos(a) = sin^2(a) + cos^2(a) = 1
tg(a) + ctg(a) = (sin^2(a) + cos^2(a))/sin(a)cos(a) = 1 / 0.4 = 2.5
4)
sin(a) = sqrt(3) / 2 => a = (-1)^k * pi / 3 + pi * k, k ∈ Z
cos(a) = sqrt(2) / 2 => a = +- 2pi / 3 + 2pi*k, k ∈ Z
tg(a) = sqrt(3) => a = pi / 3 + pi * k, k ∈ Z
ctg(a) = sqrt(2) / 2 => a = -pi / 4 + pi * k, k ∈ Z
Автор ответа:
0
Добавил 1) а
Автор ответа:
0
1.б - cos(pi / 6) - sqrt(2) * sin (pi/ 4) + sqrt(3) * tg(pi / 3) = sqrt(3) / 2 - 1 + 3 = 2 + sqrt(3) / 2
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