Задача 46285 [r][b][i][red]Решите...

apricot

2020-04-07 11:32:25

[r][b][i][red]Решите пожалуйста[/red][/i] [/b]
[blue]2 sin(- 60°)+ соs(- 45° )+tg 30°
cos 13 ° - cos 75°
cos 105° + cos 135° [/r]
[/blue]

sova

2020-04-07 11:43:36

★

1)
sin(-60 °) =-sin60 ° =-sqrt(3)/2
cos(-45 ° )=cos45 ° =sqrt(2)/2
tg30 ° =sqrt(3)/3

[b]2 sin(– 60°)+ соs(– 45° )+tg 30°=[/b] -sqrt(3)+sqrt(2)/2+(sqrt(3)/3)=

=(3sqrt(2)-4sqrt(3))/6

2) cos135 ° - cos75= -2 sin(1/2)*(135 ° +75 ° )* sin(1/2)*(135 ° -75 ° )

=-2 sin (105 ° )*sin45 °

sin105 ° =sin(60 ° +45 ° )=sin60 ° *cos45 ° +cos60 ° *sin45 ° =

так как sin45 ° =cos45 ° =sqrt(2)/2

=sqrt(2)/2 * (sqrt(3)+1)/2

3)
[b]cos105 ° +cos135 °[/b] =2 cos(1/2)*(105 ° +135 ° )*cos(1/2)(105 ° -135 ° )=

=2cos(120 ° )*cos(-15 ° )=-sqrt(2+sqrt(3))/2



cos120 ° =cos(180- ° 60 ° )=-cos60 ° =-1/2

cos(-15 ° )=cos15 °

cos^215 ° =(1+cos30 ° )/2=(2+sqrt(3))/4

cos15 ° =2)=sqrt(2+sqrt(3))/2