【答案】
分析:(1)先根據(jù)α,β的范圍求得sinα和sin(α+β)進(jìn)而根據(jù)兩角和公式求得答案.
(2)先求得sin(x+
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),進(jìn)而求得tan(x+
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),根據(jù)正切的兩角和公式求得tanx,進(jìn)而根據(jù)萬(wàn)能公式求得sin2x和cos2x,代入
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中即可.
(3)先根據(jù)α,β的范圍求得sin(α-
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)和cos(
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-β),進(jìn)而根據(jù)兩角和公式求得cos
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,進(jìn)而根據(jù)倍角公式求得cos(α+β).
解答:解:(1)∵
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∴sinα=
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=
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,sin(α+β)=
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=
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∴sinβ=sin(α+β-α)=sin(α+β)cosα-cos(α+β)sinα=
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×
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+
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×
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=
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(2)∵
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∴
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<x+
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<2π
∴sin(x+
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)=-
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=-
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∴tan(x+
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)=
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=
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∴tanx=7
∴sin2x=
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=
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,cos2x=
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=-
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∴
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=
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=-
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(3)∵
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∴sin(α-
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)=
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=
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,cos(
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-β)=
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=
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∴cos
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=cos(α-
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-
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+β)=cos(α-
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)cos(
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-β)+sin(α-
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)sin(
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-β)=-
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×
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+
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×
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=
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∴cos(α+β)=2cos
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-1=-
點(diǎn)評(píng):本題主要考查了利用三角函數(shù)的基本公式化簡(jiǎn)求值.解題的時(shí)候要特別注意三角函數(shù)值的正負(fù)號(hào).