Hey! I'm stuck on this problem :( could you help me?

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1 Answer
Apr 19, 2018

The first thing to recognize is that we can find cosine via
#sin^2(theta) + cos^2(theta) = 1 implies costheta = pm sqrt(1 - sin^2theta) = pm 6/7#

The problem gives us that cosine is positive, so #costheta = 6/7#.

We will have some ambiguous signs so we want to figure out what quadrant this angle is in. Since the sine is negative and the cosine is positive, this angle is in the bottom righthand quadrant i.e. #(3pi)/2 < theta < 2pi #.

From this we can use a bunch of identities to find the values they want:

#sin(2theta) = 2sinthetacostheta = 2 * - sqrt13/7 * 6/7 = -(12sqrt13)/49#

#cos(2theta) = cos^2theta - sin^2theta = 36/49 - 13/49 = 23/49#

Because #theta# is in the 4th quadrant, #theta/2# is in the second quadrant, hence
#sin(theta/2) = pm sqrt((1 - costheta)/2) = \pm sqrt(1/14) = sqrt(1/14)#

#cos(theta/2) = pm sqrt((1+costheta)/2) = pm sqrt(13/14) = - sqrt(13/14) #

#tan(theta/2) = sin(theta/2)/cos(theta/2) = -sqrt(1/14) / sqrt(13/14) = - sqrt(1/13) #