Вычислите значение выражения.
1)arcsin 0
2)arccos 1
3)arcsin√2/2
4)arccos 3
5)arcsin (-1)
6)arccos(-√3/2)
7)arctg 0
8)arctg 1
9)arctg(-√3)
10)arcctg(-√3/3)
11)arcsin(-1/2)+arccos 1
12)arcsin -1/2+ arccos 1
13)cos ( arccos 1)
14)sin(arcsin√2/2)
15)arcsin (sin пи/4)
16)arccos ( cos(-пи/4))
17)cos (arcsin(-1/3))
18)tg(arccos(-1/4))
19)sin(arcctg(-2))
20) arcsin(cos пи/9)

Ответ

Проверено экспертом

Автор - oganesbagoyan

1)arcsin 0 =0
2)arccos 1= 0 ;
3)arcsin√2/2 =π/4 ;
4)arccos 3 не существует угол косинус которой =3 ;
5)arcsin (-1) = -π/2 ;
6)arccos(-√3/2) = π -π/6 = 5π/6 ;
7)arctg 0 = 0 ;
8)arctg 1 =π/4 ;
9)arctg(-√3) = - π/3 ;
10)arcctg(-√3/3) = π -π/3= 2π/3 ;
11)arcsin(-1/2)+arccos 1 = -π/6 +0 = -π/6 ;
12) (arcsin -1)/2+ arccos 1 = -π/4+0= -π/4;
13)cos ( arccos 1) =1;
14)sin(arcsin√2/2) =√2/2 ;
15)arcsin (sin π/4) =arcsin(√2/2) =π/4 ;
16)arccos ( cos(-π/4))=arccos ( cos(π/4))=arccos (√2/2))=π/4 ;
17)cos (arcsin(-1/3))=cos(arccos(√8/3)= √8/3 =2√2/3 ;
18)tg(arccos(-1/4)) =tq(arctq(-√15) = - √15; 1+tq²α= 1/cos²α
19)sin(arcctg(-2)) =sin(arcsin(1/√5)=1/√5 ;
20) arcsin(cos π/9) =arcsin(sin(π/2 - π/9))=arcsin(sin7π/18) =7π/18 .