Hi

b) Given that -pi<theta<-1/2pi and cos theta =-1/6 use appropriate trigonometric formulas to find the exact values of the following:

- Cos (2theta)
- sin theta
- sin (2theta)

I don't have a clue where to start for part a but for part b i've worked out the following. Is any of it correct?

- Cos (2theta) = -1/6 x 2 = -1/3

no ... cos(2t) = 2cos^{2}t - 1
cos(2t) = 1 - 2sin

^{2}t

- Sin theta: Because cos
^{2}theta + sin^{2}theta = - Sin
^{2}theta = 1 + 1/6^{2} - Sin
^{2}theta = 37/36 - Sin theta = sqrt 37/36
- Sin theta = -16 sqrt 37 I've put negative answer because of the interval it lies in but not sure if that's correct

sin^{2}t = 1 - cos^{2}t

sin^{2}t = 1 - (-1/6)^2 = 35/36 - Sin (2theta) = 2 sin theta x cos theta
- = 2 x (-1/6 sqrt 37) x -1/6
- = 1/18 sqrt 37

now use what you were given and the correct value of sin(t)