maths~~~~~~

a)
cos 2A = 2cos^2 A - 1
令 A = θ/2 ,
cos θ = 2cos^2 θ/2 - 1
cos θ/2 = +/- √ [ (1 + cos θ)/2 ]
當 0° <= θ <= 180° 時 ,
0° <= θ/2 <= 90° , 在第一象限中 cos θ/2 >= 0 ,
得 cos θ/2 = √ [ (1 + cos θ)/2 ]
b)
令 θ = 135° ,
cos (135/2)° = √ [ (1 + cos 135°)/2 ]
cos 67.5° = √ [ (1 + cos (180° - 45°))/2 ]
cos 67.5° = √ [ (1 - cos 45°)/2 ]
cos 67.5° = √ [ (1 - √2/2)/2 ]
cos 67.5° = √ [ (2 - √2)/4 ]
cos 67.5° = (1/2) √ (2 - √2)

(a) By double angle formula,
cos A = 2cos^2 (A/2) - 1
cos (A/2) = sqrt [(cos A + 1) / 2]

(b) cos 67.5 = cos (135 / 2)
= sqrt[(cos135 + 1) / 2]
= sqrt[(sqrt 2 / 2 +1)/2]
= sqrt (2 - sqrt 2) / 2