Задача 65847 N° 52 , 53 , 56 . help я тупая ...
Условие
Решение
sin^2 a + cos^2 a = 1
sin^2 a + 1/4 = 1
sin^2 a = 3/4
sin a = sqrt(3)/2
tg a = sin a / cos a = (sqrt(3)/2) : (1/2) = sqrt(3)
б) cos a = 0,2
sin^2 a + cos^2 a = 1
sin^2 a + 0,04 = 1
sin^2 a = 0,96
sin a = sqrt(0,96) = sqrt(0,16*6) = 0,4*sqrt(6)
tg a = sin a / cos a = (0,4*sqrt(6)) : (0,2) = 2sqrt(6)
в) cos a = -0,5
sin^2 a + cos^2 a = 1
sin^2 a + 0,25 = 1
sin^2 a = 0,75 = 3/4
sin a = sqrt(3)/2
tg a = sin a / cos a = (sqrt(3)/2) : (-0,5) = -sqrt(3)
53) а) sin a = 1/4
sin^2 a + cos^2 a = 1
1/16 + cos^2 a = 1
cos^2 a = 15/16
cos a = sqrt(15)/4
tg a = sin a / cos a = (1/4) : (sqrt(15)/4) = 1/sqrt(15) = sqrt(15)/15
б) sin a = sqrt(3)/2
sin^2 a + cos^2 a = 1
3/4 + cos^2 a = 1
cos^2 a = 1/4
cos a = 1/2
tg a = sin a / cos a = (sqrt(3)/2) : (1/2) = sqrt(3)
в) sin a = 0,72
sin^2 a + cos^2 a = 1
0,5184 + cos^2 a = 1
cos^2 a = 0,4816
cos a = sqrt(0,4816) = sqrt(0,16*3,01) = 0,4*sqrt(3,01)
tg a = sin a / cos a = (0,72) : (0,4*sqrt(3,01)) = 1,8/sqrt(3,01) = 1,8*sqrt(3,01)/3,01
56) а) sin 117° = sin (180° - 63°) = sin 63°
б) cos 142° = cos (90° + 52°) = -sin 52°
в) tg 118° = tg (180° - 62°) = -tg 62°
г) sin 136° = sin (180° - 44°) = sin 44°