2个回答
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OA
=OC+CA
=(0,2)+(√3cosα,√3sinα)
=(√3cosα,2+√3sinα)
|OA|
=√[(√3cosα)²+(2+√3sinα)²]
=√(3cos²α+4+4√3sinα+3sin²α)
=√(7+4√3sinα)
OA·OB
=(√3cosα,2+√3sinα)·(2,0)
=2√3cosα
OA·OB
=|OA|·|OB|·cos∠AOB
=√(7+4√3sinα) * 2 * cos∠AOB
所以:
√(7+4√3sinα) * 2 * cos∠AOB = 2√3cosα
cos∠AOB = √3cosα / √(7+4√3sinα)
令√(7+4√3sinα)=t,
则sinα=(t²-7)/(4√3)
cosα=√(1-sin²α)=√[1-(t²-7)²/48]=(1/12)* √[√3 *(14t²-t^4-1)]
cos∠AOB
= √3*cosα / √(7+4√3sinα)
=(√3/12)* √[√3 *(14t²-t^4-1)] / t
--------------------------
晕,太难算了
=OC+CA
=(0,2)+(√3cosα,√3sinα)
=(√3cosα,2+√3sinα)
|OA|
=√[(√3cosα)²+(2+√3sinα)²]
=√(3cos²α+4+4√3sinα+3sin²α)
=√(7+4√3sinα)
OA·OB
=(√3cosα,2+√3sinα)·(2,0)
=2√3cosα
OA·OB
=|OA|·|OB|·cos∠AOB
=√(7+4√3sinα) * 2 * cos∠AOB
所以:
√(7+4√3sinα) * 2 * cos∠AOB = 2√3cosα
cos∠AOB = √3cosα / √(7+4√3sinα)
令√(7+4√3sinα)=t,
则sinα=(t²-7)/(4√3)
cosα=√(1-sin²α)=√[1-(t²-7)²/48]=(1/12)* √[√3 *(14t²-t^4-1)]
cos∠AOB
= √3*cosα / √(7+4√3sinα)
=(√3/12)* √[√3 *(14t²-t^4-1)] / t
--------------------------
晕,太难算了
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