解:(1)
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=
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=
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=1;
(2)令tanx=t,∵
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,∴t>1,
∴y=tan2xtan
3x=
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=
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=
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≤
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=-8
∴函數(shù)y=tan2xtan
3x的最大值為-8;
(3)解:(1)f(x)=sin(2x+θ)+2
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×
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-
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=sin(2x+θ)+
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cos(2x+θ)=2sin(2x+θ+
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);
要使f (x)為偶函數(shù),則必有f(-x)=f(x),
∴2sin(-2x+θ+
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)=2sin(2x+θ+
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),即-sin[2x-(θ+
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)]=sin(2x+θ+
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),
整理得:-sin2xcos(θ+
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)+cos2xsin(θ+
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)=sin2xcos(θ+
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)+cos2xsin(θ+
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)
即2sin2xcos(θ+
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)=0對(duì)x∈R恒成立,
∴cos(θ+
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)=0,
又0≤θ≤π,則θ=
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故答案為:1,-8,A.
分析:(1)通分母,積化和差化簡(jiǎn)sin70°sin10°,再利用誘導(dǎo)公式,約分得出結(jié)果;
(2)利用二倍角公式,轉(zhuǎn)化成關(guān)于tanx的函數(shù),將tanx看成整體,最后轉(zhuǎn)化成函數(shù)的最值問(wèn)題解決;
(3)利用二倍角的正弦函數(shù)公式、余弦函數(shù)公式化簡(jiǎn),合并整理后,再利用兩角和與差的正弦函數(shù)公式及特殊角的三角函數(shù)值化為一個(gè)角的正弦函數(shù),把函數(shù)解析式中的x化為-x,確定出f(-x)的解析式,根據(jù)偶函數(shù)的性質(zhì)f(-x)=f(x),列出關(guān)系式,利用正弦函數(shù)的奇偶性以及兩角和與差的正弦函數(shù)公式化簡(jiǎn),整理后可得cos(θ+
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)=0,根據(jù)θ的范圍,利用特殊角的三角函數(shù)值即可求出滿足題意θ的度數(shù).
點(diǎn)評(píng):本題考查三角函數(shù)知識(shí),考查函數(shù)的性質(zhì),考查學(xué)生解決問(wèn)題的能力,綜合性強(qiáng).