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

#1.)so tan θ = 1 / (-2
`√(2)`) = -
`√(2)`/2

and cos θ = - 2
`√(2)`

#2.)

sin θ = -root(2)/root(3) = - root(6) /3 = -
`√(6)`/3

cos θ = -
`√(3)`/3

tan θ =
`√(2)`

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
`√(2)`

so tan θ = 1 / (-2
`√(2)`) = -
`√(2)`/2

and cos θ = - 2
`√(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 = -
`√(6)`/3

cos θ = -
`√(3)`/3

tan θ =
`√(2)`

sin