已知在平面直角坐標(biāo)系xOy中,△AOB三個頂點的直角坐標(biāo)分別為A(4,3),O(0,0),B(b,0).
(1)若b=5,求cos2A的值;
(2)若△AOB為銳角三角形,求b的取值范圍.
【答案】
分析:(1)法一:由題意義可得,要求cos2A,可先求 cosA,而A可以看成
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與
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的夾角,代入向量夾角公式
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,然后利用二倍角公式可求cos2A
(法二)由題可得A=B,cos2A=cos(∠A+∠B)=cos(π-∠AOB),利用誘導(dǎo)公式進行化簡可求
(2)由△AOB為銳角三角形可得A,B,O都為銳角,由∠A為銳角可得
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,
由∠B為銳角可得,
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由∠O為銳角可得,
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,代入整理即可求
解答:解:(1)
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,
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,
若b=5,則
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所以,
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所以,
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(法二)cos2A=cos(∠A+∠B)=cos(π-∠AOB)=
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(2)若∠A為銳角,則
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,即-4b+16+9>0,得
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若∠B為銳角,則
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,即-b(4-b)>0,得b<0或b>4
若∠O為銳角,則
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,即4b>0,得b>0綜上所述,
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【解二】用平面幾何或解析幾何的方法同樣給分.
點評:本題主要考查了向量夾角公式的應(yīng)用,二倍角公式的運用,向量的數(shù)量積的符號在判斷角的范圍中的應(yīng)用.