三角比方程

(sinx + cosx) = 2sin[(π/4) + x]sin[(π/4) - x]
sinx+cosx+2sinxcosx=2{-1/2[cos(π/2)-cos(2x)]}
1+2sinxcosx=-cos(π/2)+cos2x
1+2sinxcosx=1-2sinx
2sinxcosx+2sinx=0
2sinx(cosx+sinx)=0
sinx=0 or tanx=-1
x=0 , (3π)/4 , π , (7π)/4 or 2π

2009-06-16 17:46:36 補充：
(sinx + cosx)^2 = 2sin[(π/4) + x]sin[(π/4) - x]
sin^2x+cos^2x+2sinxcosx=2{-1/2[cos(π/2)-cos(2x)]}
1+2sinxcosx=-cos(π/2)+cos2x
1+2sinxcosx=1-2sin^2x
2sinxcosx+2sin^2x=0
2sinx(cosx+sinx)=0
sinx=0 or tanx=-1
x=0 , (3π)/4 , π , (7π)/4 or 2π