Question

Matematica - trigonometrie
Aaaaaaaaaaa​

Answer 1

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: