Question

gente preciso de ajuda!!!!

1)Dado seno x = 3/4 no I quadrante. Calcule

1) Tgx:
2) Cotgx:
3) Sec X:
4) Cossec x:
5) y= sen2x . cosx + tg2x
_______________
cos2x . sen2x

Answer 1

sen²x+cos²x=1

(3/4)²+cos²x=1

9/16+cos²x=1

cos²x=1-9/16

cos²x=7/16

cosx=+√7/4(cosseno no primeiro quadrante é positivo)

1)Tgx=senx/cosx=3/4/√7/4=3/√7=3√7/7

2)Cotgx=1/tgx=7/3√7=7√7/21=√7/3

3)Secx=1/cosx=4/√7=4√7/7

4)Cossecx=1/senx=3/4=4/3

5)sen2x=2.senx.cosx

sen2x=2.3/4.√7/4=6√7/16= 3√7/8

tg2x=2tgx/1-tg²x

tg2x=2.3√7/7/1-(3√7/7)²

tg2x=6√7/7/1-(9.7/49)

tg2x=6√7/7/1-(9/7)

tg2x=6√7/7/1/7

tg2x=6√7/7.7/1

tg2x=6√7

cos2x= cos²x-sen²x

cos2x=(√7/4)²-(3/4)²

cos2x=7/16-9/16

cos2x=-2/16

cos2x=-1/8

Substituindo:

3√7/8.√7/4+6√7 / -1/8.3√7/8

168/32+6√7 / -3√7/64

21/4+6√7 /-3√7/64

21+24√7/4 / -3√7/64

3(7+8√7)/4 / -3√7/64

3(7+8√7)/4.64/-3√7

7+8√7/-16√7