If sin x = - cos x and x is in 4th quadrant, find sin (x/2) and cos (x/2)
x is actually one of the "nice" angles
since sinx = -cosx, and x is in IV, x = 315° , from (360° - 45°)
and sinx = -√2/2, and cosx = √2/2
from cos (2A) = 1 - 2sin^2 A = 2cos^2 A - 1
case1
cos x = 1 - 2sin^2 (x/2)
√2/2 = 1 - 2sin^2 (x/2)
2sin^2 (x/2) = 1 - √2/2 = (2 - √2)/2
sin^2 (x/2) = (2 - √2)/4
sin x/2 = ± √(2-√2)/2 , but x/2 is in quad II
sin x/2 = √(2-√2)/2
since cosx = 2cos^2 (x/2) - 1
2cos^2 (x/2) = cosx + 1
cos^2 (x/2) = √2/2 + 1 = (2 + √2)/4
cos x/2 = -√(2 + √2)/2 , since the cosine is negative in quad II