Let theta be an angle between pi and 3pi/2 such that cos(theta) = -1/9

so we know that θ is between π and 3π/2, which is another of saying that θ is in the III Quadrant, and obviously its cosine will be negative.

we also know that cos(θ/2) is negative, well, that rules out the I and IV Quadrants, so most likely θ/2 is on the II Quadrant, since is smaller than θ anyway, and on the II Quadrant as we know, the sine or "y" value is positive.

  (-2/3, (√5)/3)

Step-by-step explanation:

We can use the half-angle identities to find the values of the sine and cosine of the desired angle. The desired angle is a 2nd-quadrant angle, so the sine is positive, and the cosine is negative.

The coordinates of the terminal point are ...

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)