解三角函數方程

x = cos² (α + β) + cos² β - 2 cosα cosβ cos (α + β)x = cos² (α + β) + cos² β - 2 (1/2)[cos(α + β) + cos(α - β)] cos (α + β)x = cos² β - cos(α - β) cos (α + β)x = (1/2)(1 + cos 2β) - (1/2)(cos2α + cos2β)x = (1/2)(1 - cos2α)x = sin² α;2x - sin 2α = y2sin² α - sin 2α = y2sin² α - 2 sinα cosα = y(2sin α) (sin α - cos α) = y(2sin α) (- √2 sin(π/4 - α)) = y- 2√2 sin α sin(π/4 - α) = y


2011-02-06 15:44:05 補充：
- 2√2 (-1/2)[cos π/4 - cos (2α - π/4)] = y

- 2√2 (-1/2)[√2/2 - cos (2α - π/4)] = y

1 - √2 cos (2α - π/4) = y

2011-02-06 15:53:44 補充：
cos (2α - π/4) = (1 - y) / √2

2α - π/4 = 2kπ ± arccos [(1 - y)/√2]

2α = (2k + 1/4)π ± arccos [(1 - y)/√2]

α = (k + 1/8)π ± (1/2) arccos [(1 - y)/√2]