# Thread: Finding an angle in radians and finding exact values

1. ## Finding an angle in radians and finding exact values

Hi

I'm really stuck with these two questions would anyone be able to help?

a) Given that sec theta=-sqrt2, tan theta=-1 and -pi<theta<pi, find the exact value of the angle theta in radians. Justify your answer.

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

• Sin theta: Because cos2theta + sin2theta =
• Sin2theta = 1 + 1/62
• Sin2theta = 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 (2theta) = 2 sin theta x cos theta
• = 2 x (-1/6 sqrt 37) x -1/6
• = 1/18 sqrt 37

I would be incredibly grateful if someone could help me out i've been looking at these and trying to work them out for hours

2. ## Re: Finding an angle in radians and finding exact values

(a) $\displaystyle \sec{t} = -\sqrt{2} \implies \cos{t} = -\frac{\sqrt{2}}{2}$

from this piece of info, we can surmise that $\displaystyle t = \frac{3\pi}{4} \, or \, -\frac{3\pi}{4}$, a quad III and quad III angle, respectively.

now ... in what quadrant is $\displaystyle \tan{t} = -1$ ?

3. ## Re: Finding an angle in radians and finding exact values

Originally Posted by soapqueen2011
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) = 2cos2t - 1

cos(2t) = 1 - 2sin2t

• Sin theta: Because cos2theta + sin2theta =
• Sin2theta = 1 + 1/62
• Sin2theta = 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

sin2t = 1 - cos2t

sin2t = 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)

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