转化为三角函数的最值问题（2023全国乙，12）

2023-06-09 14:30 作者:数学老顽童 0人读过 | 我要投稿

（2023全国乙，12）已知 $%5Codot%20O$ 的半径为 $1$ ，直线 $PA$ 与 $%5Codot%20O$ 相切于点 $A$ ，直线 $PB$ 与 $%5Codot%20O$ 交于 $B$ 、 $C$ 两点， $D$ 为 $BC$ 的中点，若 $%5Cleft%7C%20PO%20%5Cright%7C%3D%5Csqrt%7B2%7D$ ，则 $%5Coverrightarrow%7BPA%7D%5Ccdot%20%5Coverrightarrow%7BPB%7D$ 的最大值为（）

A. $%5Cfrac%7B1%2B%5Csqrt%7B2%7D%7D%7B2%7D$

B. $%5Cfrac%7B1%2B2%5Csqrt%7B2%7D%7D%7B2%7D$

C. $1%2B%5Csqrt%7B2%7D$

D. $2%2B%5Csqrt%7B2%7D$

解：如图

连接 $OA$ 、 $OD$ ，

易知 $OA%5Cbot%20PA$ ，

又因为 $%5Cvert%20PO%20%5Cvert%20%3D%5Csqrt%7B2%7D%20$ ， $%5Cvert%20OA%20%5Cvert%20%3D1$ ，

所以 $%5Ccolor%7Bred%7D%7B%5Cleft%7C%20%5Coverrightarrow%7BPA%7D%20%5Cright%7C%3D1%7D$ ，且 $%5Cmeasuredangle%20APO%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20$ .

又易知 $OD%5Cbot%20PD$ ，

设 $%5Cmeasuredangle%20DPO%3D%5Ctheta%20$ ，其中 $%5Ctheta%20%5Cin%20%5Cleft(-%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2C%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Cright)$ ，

所以 $%5Ccolor%7Bred%7D%7B%5Cleft%7C%20%5Coverrightarrow%7BPD%7D%20%5Cright%7C%7D%3D%5Cvert%20PO%20%5Cvert%20%5Ccdot%20%5Ccos%20%5Ctheta%20%20%3D%5Ccolor%7Bred%7D%7B%5Csqrt%7B2%7D%20%5Ccos%20%5Ctheta%7D$ ，

且 $%5Ccolor%7Bred%7D%7B%5Cleft%3C%20%5Coverrightarrow%7BPA%7D%2C%5Coverrightarrow%7BPD%7D%20%5Cright%3E%20%3D%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B4%7D-%5Ctheta%20%7D$ .

所以

$%5Cbegin%7Baligned%7D%0A%09%5Coverrightarrow%7BPA%7D%5Ccdot%20%5Coverrightarrow%7BPD%7D%26%3D1%5Ctimes%20%5Csqrt%7B2%7D%5Ccos%20%5Ctheta%20%5Ccdot%20%5Ccos%20%5Cleft(%20%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B4%7D-%5Ctheta%20%5Cright)%5C%5C%0A%09%26%3D%5Csqrt%7B2%7D%5Ccos%20%5Ctheta%20%5Cleft(%20%5Ccos%20%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B4%7D%5Ccos%20%5Ctheta%20%2B%5Csin%20%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B4%7D%5Csin%20%5Ctheta%20%5Cright)%5C%5C%0A%09%26%3D%5Csqrt%7B2%7D%5Ccos%20%5Ctheta%20%5Cleft(%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Csin%20%5Ctheta%20%2B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Ccos%20%5Ctheta%20%5Cright)%5C%5C%0A%09%26%3D%5Csin%20%5Ctheta%20%5Ccos%20%5Ctheta%20%2B%5Ccos%20%5E2%5Ctheta%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Csin%202%5Ctheta%20%2B%5Cfrac%7B1%7D%7B2%7D%5Cleft(%201%2B%5Ccos%202%5Ctheta%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20%5Csin%202%5Ctheta%20%2B%5Ccos%202%5Ctheta%20%2B1%20%5Cright)%5C%5C%0A%09%26%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Csqrt%7B2%7D%5Csin%20%5Cleft(%20%5C%2C%5C%2C2%5Ctheta%20%2B%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B4%7D%20%5Cright)%20%2B1%20%5Cright%5D%7D%5C%5C%26%5Ccolor%7Bred%7D%7B%5Cleqslant%20%5Cfrac%7B1%2B%5Csqrt%7B2%7D%7D%7B2%7D%7D%5C%5C%0A%5Cend%7Baligned%7D$