Yes, since is in the third quadrant both sine and cosine are negative and, of course, [tex](-1/3)^2+ (-2\sqrt{2}/3)^2= 1/9+ 8/9= 1.

Yes, is in the fourth quadrant so cosine is positive and sine is negative. Also .

For α:

cosα = 3/5, sinα = -4/5

Yes, that is correct.

The next part of the problem is to find and simplify the following.

a)tanθ

tanθ = ((-2√2)/3)/(-1/3) = ((-6√2)/-3) = 2√2

Okay, since , , in the second quadrant, cosine is negative and you use "-".

b)cos θ/2

Using the half angle formula for cos θ/2 which is +/- (√(1+cosθ)/2), I did this:

-(√(1+(-1/3))/2) = -(√(2/3)/2) = -√(1/3) = -1/√3

You mean

c)cos2α

I used the formula for lowering powers for cos2α which is 1-sin^2α and then tried to solve it:

I would, rather, use [tex]cos(2\alpha)= cos^2(\alpha)- sin^2(\alpha)= (3/5)^2- (-4/5)^2= (9- 16)/25= -7/25

Yes, that is correct.1-2(-4/5)^2 = -7/25

d)cos(θ−α)

I used the addition angle identity for cosine:

cos(θ−α) = (cosθ)(cosα) + (sinθ)(sinα)

= (-1/3)(3/5)+((-2√2)/3)(-4/5)

= (-3/15)+((8√2)/15)

= (5√2)/15

e)sinα/2

I used the half angle formula:

sinα/2 = +/- (√(1-cosα)/2)

= (√(1-(3/5))/2)

= (√(2/5)/2)

= √(1/5)

= 1/(√5)

f)sin(θ+α)

I used the additional angle identity for sine:

sin(θ+α) = (sinθ)(cosα)+(cosθ)(sinα)

= ((-2√2)/3)(3/5) + (-1/3)(-4/5)

= ((-6√2)/15) + (4/15)

= -(2√2)/15