Question

Could someone please help with these two questions I am very confused and I want to know to know the answer and how to get there.

Answer 1

Answer:

#1.)so tan θ = 1 / (-2$\sqrt{2}$)  = - $\sqrt{2}$/2

and cos θ  = - 2 $\sqrt{2}$

#2.)

sin θ = -root(2)/root(3) =  - root(6) /3 = - $\sqrt{6}$/3

cos θ = -$\sqrt{3}$/3

tan θ = $\sqrt{2}$

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 $\sqrt{2}$

so tan θ = 1 / (-2$\sqrt{2}$)  = - $\sqrt{2}$/2

and cos θ  = - 2 $\sqrt{2}$

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 = - $\sqrt{6}$/3

cos θ = -$\sqrt{3}$/3

tan θ = $\sqrt{2}$

sin