【答案】
分析:(Ⅰ)由sinα的值,根據(jù)α的范圍,利用同角三角函數(shù)間的基本關(guān)系求出cosα的值即可;
(Ⅱ)由第一問求出的cosα的值,以及sinα的值,利用同角三角函數(shù)間的基本關(guān)系求出tanα的值,再利用二倍角的正切函數(shù)公式表示出tanα,把tanα的值代入可列出關(guān)于tan
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的方程,求出方程的解可得出tan
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,所求式子的第二項(xiàng)先利用誘導(dǎo)公式表示,再利用二倍角的余弦函數(shù)公式變形后,把sinα的值代入即可求出cos2α的值,進(jìn)而求出所求式子的值.
解答:解:(Ⅰ)∵
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,
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,
∴cosα=-
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=-
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;
(Ⅱ)∵tanα=
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=-
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,又tanα=
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,
∴
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=-
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,即(5tan
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+1)(tan
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-5)=0,
解得:tan
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=-
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,或tan
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=5,
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182603678540909/SYS201310241826036785409025_DA/16.png">,所以
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∈(
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,
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),
所以tan
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>0,故tan
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=5,
又cos(π-2α)=-cos2α=-2cos
2α+1=-2×
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+1=-
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,
則
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=5+
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=5
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.
點(diǎn)評(píng):此題考查了同角三角函數(shù)間的基本關(guān)系,二倍角的余弦、正切函數(shù)公式,以及誘導(dǎo)公式,熟練掌握公式是解本題的關(guān)鍵,學(xué)生在求值時(shí)注意角度的范圍.