sin0 = 4/5 where 90< 0 < 180, find cos 0/2

Answer 1

cosx = -3/5 the adjacent is negative in quadrant II. a^2 = h^2 -o^2 = 25-16 = 9 = (-3)^2, a =-3

cosx = adjacent side over hypotenuse = a/h = -3/5

sinx = 4/5

use half angle formula

cos(x/2) = + or - sqr((1+cosx)/2) =sqr((1-3/5)/2) = sqr2/5 =

= sqr(2/5)

x was in quadrant II

x/2 is in quadrant I so cos(x/2) > 0 (ignore the negative square root)