Answer:
#1.)so tan θ = 1 / (-2[tex]\sqrt{2}[/tex]) = - [tex]\sqrt{2}[/tex]/2
and cos θ = - 2 [tex]\sqrt{2}[/tex]
#2.)
sin θ = -root(2)/root(3) = - root(6) /3 = - [tex]\sqrt{6}[/tex]/3
cos θ = -[tex]\sqrt{3}[/tex]/3
tan θ = [tex]\sqrt{2}[/tex]
Step-by-step explanation:
1.) we know sin(θ) = 1/3
and tan θ < 0
so to find cosθ and tan θ
if tan θ < 0, then θ is is either quadrant 2 or 4. since sine is positive, θ must be in quadrant 2
so ...opposite side length= 1 , hypotenuse = 3, adjacent leg = root(3^2 - 1)
adjacent leg = - 2 [tex]\sqrt{2}[/tex]
so tan θ = 1 / (-2[tex]\sqrt{2}[/tex]) = - [tex]\sqrt{2}[/tex]/2
and cos θ = - 2 [tex]\sqrt{2}[/tex]
2.) (cos θ)^2 = 1/3
cos θ = - 1/root(3)
but θ is in quadrant 4,
triangle here hypotenuse = root(3) ,
adjacent leg= -1,
opposite leg =root( (root(3))^2 - 1 ) = root(3 - 1) = -root(2)
so
sin θ = -root(2)/root(3) = - root(6) /3 = - [tex]\sqrt{6}[/tex]/3
cos θ = -[tex]\sqrt{3}[/tex]/3
tan θ = [tex]\sqrt{2}[/tex]
sin
