赞 怡醉芳羞 幼苗
共回答了21个问题采纳率：95.2% 举报
如图,作单位圆O,半径为1；作∠AOB=a,∠BOC=b
作CD⊥OA于D,CE⊥OB于E,BF⊥OA于F
则 cosa=OF, sina=BF; cosb=OE, sinb=CE; cos(a+b)=OD
GD∥BF => GD/BF=OD/OF => GD=OD/OF*BF=cos(a+b)/cosa*sina (1)
∠OGD=∠CGE, CD⊥OD, CE⊥OE => △OGD∽△CGE
=> GD/OD=GE/CE = tana=(OE-OG)/CE=(cosb-GD/sina)/sinb
整理,得 GD=sina*cosb-sina*sinb*tana (2)
(1)(2)式相等,得 cos(a+b)*sina/cosa=sina*cosb-sina*sinb*tana
两边同乘cosa/sina,即得 cos(a+b)=cosa*cosb-sina*sinb
作CD⊥OA于D,CE⊥OB于E,BF⊥OA于F
则 cosa=OF, sina=BF; cosb=OE, sinb=CE; cos(a+b)=OD
GD∥BF => GD/BF=OD/OF => GD=OD/OF*BF=cos(a+b)/cosa*sina (1)
∠OGD=∠CGE, CD⊥OD, CE⊥OE => △OGD∽△CGE
=> GD/OD=GE/CE = tana=(OE-OG)/CE=(cosb-GD/sina)/sinb
整理,得 GD=sina*cosb-sina*sinb*tana (2)
(1)(2)式相等,得 cos(a+b)*sina/cosa=sina*cosb-sina*sinb*tana
两边同乘cosa/sina,即得 cos(a+b)=cosa*cosb-sina*sinb
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