Trigonometry1題 (急)

sin^2 x-2sin x cos x-cos^2 x=?
-cos(2x)-sin(2x)=?
cos(2x)+sin(2x)=-?
cos(2x)+cos(90-2x)=-?
2cos[(2x+(90-2x))/2]sin[(2x-(90-2x))/2]=-?
2cos(90/2)sin[(4x-90)/2]=-?
2(√2 / 2)sin(2x-45)=-?
√2sin(45-2x)=?
sin(45-2x)=? / √2
If ?=0
sin(45-2x)=0
sin(45-2x)=sin 0 or sin 180
45-2x=0
2x=45
x=22.5----(1)
OR
45-2x=180
2x=-135
x=-67.5----(2)
From (1)
sin(22.5)=sin(180-22.5)
=sin(157.5)
From(2)
sin(-67.5)=sin(360+(-67.5))
=sin(292.5)
sin(292.5)=sin[180+(360-292.5)]
=sin(247.5)
∴x=22.5 or 157.5 or 247.5 or 292.5
By substituting x into the equation
x=22.5 or 292.5 are rejected
計係計到不過我5識點表達...應該會pp-1

sin²x-2sinxcosx-cos²x=0
-cos2x-sin2x=0
tan2x=0
x=90,180 or 270

2012-05-06 21:05:51 補充：
修正
-cos2x-sin2x=0
tan2x=-1
x=67.5,157.5,247.5 or 337.5

2012-05-07 07:13:54 補充：
sin²x-2sinxcosx-cos²x
2sinxcosx=sin2x
cos²x-sin²x=cos2x