Найти область определения функции а) у=3х-2/5х+3 б)у=5+6х/2х-4

Question

Алгебра: Найти область определения функции а) у=3х-2/5х+3 б)у=5+6х/2х-4

artyom11111112 · Accepted Answer

а)
cos105° =cos(60° +45°) = cos60°*cos45° -sin60°*sin45° =
(1/2)*(√2)/2 - (√3)/2 *(√2)/2= (√2)/4 (1-√3).
или cos105° =cos(90°+15°) = - sin15° = -√(1-cos30°)/2 = -√(1-√3/2)/2 =
-(1/2)√(2-√3) = -(1/2)√((1-√3)²/2) =- (1/2√2)(1-√3)=(√2)/4 (1-√3) .
б)
sin75° =sin(45° +30°) =sin45°cos30° +cos45°sin30°=
1/√2 *(√3)/2 +(1/√2)*(1/2) = (√3+1)/2√2 =√2(√3+1)/4.
или sin75° =sin(90° -15°) =cos15° =√ ((1+cos30°)/2) =√((1+(√3)/2)/2) =
(1/2)√(2+√3) =(1/2)√( (√3+1)²*1/2) = (√3+1)/2√2 =√2(√3+1)/4.
в)
ctq15° =ctq(45° - 30°) =(ctq45°ctq30° +1)/(ctq30° -ctq45°)=(√3+1)/(√3 -1)=
((√3+1)²/2 =2+ √3 .
или ctq15° =√((1+cos30°)/(1-cos30°) ) =√(2+√3)/(2-√3) =2+√3.

MOGZ ответил