Răspuns:
1, sinx=12/13 x∈(π/2,π)
Aplici formula sin²x+cos²x=1 s-o retii ca e importanta in trigonometrie
1] (12/13)²+cos²x=1
144/169+cos²x=1
cos²x=1-144/169
cos²x=25/169
cosx=√25/169
cosx=±5/13
deoarece esti in cadranul 2 se ia valoarea negariva
cosx= -5/13
ctgx cosx/sinx=(-5/13)/(12/13)= -5/12
sin2x=2sinx*cosx=2*12/13*(-5/13)= -120/169
cos2x=cos²x-sin²x=(-5/13)²-(12/13)²=25/169-144/169=
3.]a)
aplici formula
sinxcosy+cosxsiny=sin(x+y)
x=136°, y=44°
sin136cos44+cos44*sin136=sin(136+44)=sin180°=0
b) aplici formula cosxcos y+sinxsiny=cos(x-y)
x=52°, y=7°
cos52*cos7+sin52*sin7=cos(52-7)=cos45=√2/2
5.tgx+ctg x=5/2
ctgx=1/tgx=>
tgx+1/tgx=5/2
2tg²x+2=5tgx
2tg²x-5tgx+2=0
tgx=y
2y²-5g+2=0
y1=2 y2=1/2=>
tgx1=2
tgx2=1/2
Explicație pas cu pas:
