Задача 66111 добрый вечер,горит дедлайн,не успеваю...
Условие
Решение
30cos 2a = 30(2cos^2 a - 1) = 30(2*0,16 - 1) = -30*0,68 = -20,4
29) sin a = -0,7
-19cos 2a = -19(1 - 2sin^2 a) = -19(1 - 2*0,49) = -19*0,02 = -0,38
30) sin 2a = 0,8
[m]\frac{3sin(4a)}{5cos(2a)} = \frac{3*2sin(2a)cos(2a)}{5cos(2a)} = \frac{6sin(2a)}{5} = \frac{6*0,8}{5} = 0,96[/m]
31) sin 3a = -0,8
[m]\frac{2sin(6a)}{5cos(3a)} = \frac{2*2sin(3a)cos(3a)}{5cos(3a)} = \frac{4sin(3a)}{5} = \frac{4(-0,8)}{5} = -0,64[/m]
32) [m]\frac{7sin(a)+5cos(a)}{3sin(a)-3cos(a)}=1[/m]
7sin a + 5cos a = 3sin a - 3cos a
4sin a = -8cos a
sin a/cos a = -8/4
tg a = -2
33) tg a = -4
[m]\frac{8cos(a) + 2sin(a) + 6}{sin(a) + 4cos(a) + 3} = \frac{2(4cos(a) + sin(a) + 3)}{4cos(a) + sin(a) + 3} = 2[/m]
34) cos a = -sqrt(26)/26 = -1/sqrt(26); a ∈ [3π/2; 2π]
Так как a ∈ [3π/2; 2π], то sin a < 0, cos a > 0.
У вас cos a < 0, это ошибка. Видимо, должно быть:
sin a = -sqrt(26)/26 = -1/sqrt(26), тогда
cos^2 a = 1 - sin^2 a = 1 - 1/26 = 25/26
cos a = 5/sqrt(26) > 0
tg a = sin a / cos a = (-1/sqrt(26)) : (5/sqrt(26)) = -1/5
Или другой вариант:
cos a = sqrt(26)/26 = 1/sqrt(26), тогда
sin^2 a = 1 - cos^2 a = 1 - 1/26 = 25/26
sin a = -5/sqrt(26) < 0
tg a = sin a / cos a = (-5/sqrt(26)) : (1/sqrt(26)) = -5
35) tg a = 0,25 = 1/4
tg(a + 3π/2) = -ctg a = -1/tg a = -4
36) 5sin^2 a + 11cos^2 a = 7
5sin^2 a + 11cos^2 a = 7sin^2 a + 7cos^2 a
4cos^2 a = 2sin^2 a
tg^2 a = sin^2 a / cos^2 a = 4/2 = 2
37) tg y = 2
-4tg(-2π-y) + tg y = 4tg(2π+y) + tg y = 4tg y + tg y = 6tg y = 12