Question

Sendo sen x = 21/29 e x € ao 2° Q. Calcule Cos 2x, Sen 2x e Tg 2x.

Answer 1

Resposta:

Explicação :

Sabemos que:

sen²x+cos²x=1

cos²x=1-sen²x

cos²x=1-(21/29)²

cos²x=1-441/841

cos²x=(841-441):841

cos²x=400/841

cosx=20/29

Mas o arco está no 2º quadrante, logo:

cosx=-20/29

cos2x=cos²x-sen²x

cos2x=(-20/29)²-(21/29)²

cos2x=400/841-441/841

cos2x=-41/841

sen2x=2senxcosx

sen2x=2.21/29(-20/29)

sen2x=42/29(-20/29)

sen2x=-840/841

sen2x=-0,9988109

tgx=senx/cosx

tgx=[21/29]:[-20/29]

tgx=[21/29].[-29/20]

tgx=-21/20

tg2x=[2tgx]:[1-tg²x]

tg2x=[2(-21/20)]:[1-(-21/20)²]

tg2x=[-42/20]:[1-441/400]

tg2x=[-21/10]:[(400-441):400]

tg2x=[-21/10]:[-41/400]

tg2x=[-21/10].[-400/41]

tg2x=840/41

tg2x=20,487805