Question

16. (UF-AL) A +1sen300°/tg540+ cos(-120°) é igual a:
( ) - √3/3
( )√3/4
( )2-√3/4
( )2+√3
( ) -2+√3

Answer 1

sen(300)=sen(360-60)

sen(300)=sen(360)*cos(60)- sen(60)*cos(360)

sen(300)= 0 *cos(60) -sen(60) * 1  

sen(300) =-sen(60)=-√3/2

540-360 = 180

tg(540)=tg(180)=sen(180)/cos(180) =0/(-1)=0

cos(-120)=cos(120)

cos(120)=cos(60+60)=2cos²(60)-1 = 2*(1/2)²-1 =2*1/4-1=1/2-1=-1/2

[1+sen(300)]/[tg(540)+cos(-120)]

=

[1-√3/2]/[0-1/2]

[1-√3/2]/[-1/2]

= -2+√3

Letra E