trig problem?

Question

find the exact value of:

a) sin 2theta
b)cos 2 theta
c)sin theta/2
d) cos theta/2

all that is given is sin theta = (-sqroot3)/3

please help and explain this, I'm having a hard time!!! thanks

Answer 1

I'll show how to do a).
a) sin 2theta = 2sin theta cos theta = ±2(√3/3)(√(2/3) = ±2√2/3, since cos theta = ±√(2/3)
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Ideas: Use sin^2(theta) + cos^2(theta) = 1 or the right triangle with three sides of 1, √2, and √3. Pay attention to the two different signs for the answer.
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To the answerer below me,
sin theta < 0 can be in either the third or the fourth quadrant.

Answer 2

Let's use t for theta. Since sin t < 0, we know t is a fourth quadrant angle. Sketch a ray from the origin into the 4th quadrant. Select a point (x,y) on this ray, and from that point drop a perpendicular to the x-axis.

From sin t = -sqrt(3)/3, we know that y = sqrt(3), and the hypotenuse of the triangle is 3. Since x^2 + y^2 = 3^2, we have x^2 = 6, and x = sqrt(6), so cos t = sqrt(6)/3.

Now you know sin t and cos t. Use the double angle formulas and the half angle formulas to answer your questions.