一條maths

2009-02-02 3:34 am
Solve 2 sin A = tanA

回答 (2)

2009-02-03 3:40 am
2 sin A = tanA

2 sin A = sin A / cos A

2 sin A cos A = sin A

2 sin A cos A - sin A = 0

sin A ( 2 cos A - 1 ) = 0

sin A = 0 or 2 cos A - 1 = 0

sin A = 0 or cos A = 1 / 2

A = 180 n + [(-1)^(n)](0) or A = 360 n (+/-) 60 ( where n is an integer )

A = 180 n or A = 360 n ± 60

2009-02-02 19:41:18 補充:
A = 180 n or A = 360 n (+/-) 60
參考: me
2009-02-02 3:40 am
2 sin A = tanA
2 sin A = sin A/cos A
2 sin A cos A=sin A
sinA (2cosA-1)=0
sin A = 0 or cos A = 1/2
A= 0, 180, 360 , 60, 300

2009-02-01 19:42:00 補充:
要計A的general rule代入相應公式便成


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