Question

If cos B = -0.8 and 180°<=B<=270°, find cos (0.5B).

How would I go about solving this problem?

Answer 1

B is between 180 and 270

cut those values in half to get 180/2 = 90 and 270/2 = 135

so B/2 is between 90 and 135 placing the angle B/2 in quadrant 2. Therefore, cos(B/2) is negative

Answer 2

you will use the identity

\[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1+\cos(x)}{2}}\]

see top of page 2 in the link below

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

since we know cos(B/2) is negative, we can change the "plus/minus" to just "minus"

\[\Large \cos\left(\frac{B}{2}\right) =-\sqrt{\frac{1+\cos(B)}{2}}\]

\[\Large \cos\left(\frac{B}{2}\right) =??\]

Answer 3

Sorry, I lost connection there. Thanks for the help!

Answer 4

no problem