2024-2025學(xué)年高中數(shù)學(xué)第三章三角恒等變換3.3二倍角的三角函數(shù)二課時(shí)素養(yǎng)評(píng)價(jià)含解析北師大版必修4

PAGE課時(shí)素養(yǎng)評(píng)價(jià)二十七二倍角的三角函數(shù)(二)(20分鐘35分)1.已知sinα-cosα=QUOTE,則sin2α= ()A.-QUOTE B.-QUOTE C.QUOTE D.QUOTE【解析】選A.sin2α=2sinαcosα=QUOTE=-QUOTE.【補(bǔ)償訓(xùn)練】已知cosα-sinα=QUOTE,則cosQUOTE= ()A.-QUOTE B.-QUOTE C.QUOTE D.QUOTE【解析】選C.因?yàn)閏osα-sinα=QUOTE,所以cos2α-2sinαcosα+sin2α=1-sin2α=QUOTE,所以sin2α=QUOTE,所以cosQUOTE=sin2α=QUOTE.2.若sinQUOTE=QUOTE,則cosQUOTE= ()A.-QUOTE B.-QUOTE C.QUOTE D.QUOTE【解析】選A.由題意,可得cosQUOTE=-cosQUOTE=-cosQUOTE=-cosQUOTE=-QUOTE=-QUOTE.3.若tanθ=QUOTE,則cos2θ= ()A.-QUOTE B.-QUOTE C.QUOTE D.QUOTE【解析】選D.cos2θ=cos2θ-sin2θ=QUOTE.分子分母同時(shí)除以cos2θ,得:cos2θ=QUOTE=QUOTE=QUOTE.4.已知α∈QUOTE,sin2α=QUOTE,則sinQUOTE=.

【解析】因?yàn)?-2sin2QUOTE=cosQUOTE=-sin2α,所以sin2QUOTE=QUOTE,因?yàn)棣痢蔘UOTE,所以α+QUOTE∈QUOTE,所以sinQUOTE=QUOTE.答案:QUOTE5.已知α是其次象限角,且sinQUOTE=-QUOTE,則QUOTE=.

【解析】由sinQUOTE=-QUOTE,得cosα=-QUOTE,又因?yàn)棣潦瞧浯蜗笙藿?所以tanα=-2,所以QUOTE=QUOTE=QUOTE·QUOTE=QUOTE.答案:QUOTE6.若θ∈QUOTE,sin2θ=QUOTE,求sinθ.【解析】因?yàn)棣取蔘UOTE,所以2θ∈QUOTE,所以cos2θ≤0,所以cos2θ=-QUOTE=-QUOTE=-QUOTE.又cos2θ=1-2sin2θ,所以sin2θ=QUOTE=QUOTE=QUOTE,因?yàn)棣取蔘UOTE,所以sinθ>0,所以sinθ=QUOTE.(30分鐘60分)一、選擇題(每小題5分,共25分)1.QUOTE= ()A.-QUOTE B.-1 C.QUOTE D.1【解析】選D.QUOTE=QUOTE=2×QUOTE=2sin30°=1.2.設(shè)α∈QUOTE,β∈QUOTE,且QUOTE=QUOTE,則 ()A.2α+β=QUOTE B.2α-β=QUOTEC.α+2β=QUOTE D.α-2β=QUOTE【解析】選B.由QUOTE=QUOTE,可得:sinα-sinαsinβ=cosαcosβ.所以sinα=cosαcosβ+sinαsinβ=cos(α-β),因?yàn)棣痢蔘UOTE,β∈QUOTE,所以cos(α-β)>0,所以α+α-β=QUOTE,即2α-β=QUOTE.3.若sin(α+β)cosβ-cos(α+β)sinβ=0,則sin(α+2β)+sin(α-2β)等于()A.1 B.-1 C.0 D.±1【解析】選C.因?yàn)閟in(α+β)cosβ-cos(α+β)sinβ=sin(α+β-β)=sinα=0,所以sin(α+2β)+sin(α-2β)=2sinαcos2β=0.4.計(jì)算:4cos50°-tan40°= ()A.QUOTE B.QUOTE C.QUOTE D.2QUOTE【解析】選A.4cos50°-tan40°=4cos50°-QUOTE=QUOTE=QUOTE=QUOTE=QUOTE=QUOTE=QUOTE.【誤區(qū)警示】由于50°+40°=90°,故想用其中一個(gè)表示另外一個(gè),沒有考慮到其他特別角,從而思路斷掉.5.已知sinα+cosα=QUOTE,則2cos2QUOTE-1= ()A.QUOTE B.QUOTE C.-QUOTE D.-QUOTE【解析】選C.由sinα+cosα=QUOTE平方得,1+sin2α=QUOTE,故sin2α=-QUOTE,故2cos2QUOTE-1=cosQUOTE=sin2α=-QUOTE.二、填空題(每小題5分,共15分)6.函數(shù)f(x)=sinQUOTE-3cosx的最小值為.

【解題指南】首先應(yīng)用誘導(dǎo)公式,轉(zhuǎn)化得到二倍角的余弦,進(jìn)一步應(yīng)用二倍角的余弦公式,得到關(guān)于cosx的二次函數(shù),從而得解.【解析】f(x)=sinQUOTE-3cosx=-cos2x-3cosx=-2cos2x-3cosx+1=-2QUOTE+QUOTE,因?yàn)?1≤cosx≤1,所以當(dāng)cosx=1時(shí),f(x)min=-4.答案:-47.已知θ∈QUOTE,且sinQUOTE=QUOTE,則tan2θ=.

【解析】由sinQUOTE=QUOTE得,QUOTE=QUOTE?sinθ-cosθ=QUOTE,解方程組:QUOTE得QUOTE或QUOTE因?yàn)棣取蔘UOTE,所以sinθ>0,所以QUOTE不合題意,舍去.所以tanθ=QUOTE,所以tan2θ=QUOTE=QUOTE=-QUOTE.答案:-QUOTE8.QUOTE·cos10°+QUOTEsin10°tan70°-2cos40°=.

【解析】原式=QUOTE+QUOTE-2cos40°=QUOTE+QUOTE-2cos40°=QUOTE-2cos40°=QUOTE-2cos40°=QUOTE-2cos40°=4cos220°-2(2cos220°-1)=2.答案:2三、解答題(每小題10分,共20分)9.已知2sinθ-cosθ=1,求QUOTE的值.【解析】已知等式變形得:2sinθ=1+cosθ,即4sinQUOTEcosQUOTE=2cos2QUOTE,即2sinQUOTE=cosQUOTE或cosQUOTE=0,當(dāng)2sinQUOTE=cosQUOTE時(shí),原式=QUOTE=QUOTE==QUOTE=2.當(dāng)cosQUOTE=0時(shí),原式=0,綜上所述,原式的值為0或2.10.設(shè)A,B,C是△ABC的三個(gè)內(nèi)角,求證:sin2A+sin2B+sin2C=4sinAsinBsinC.【證明】左邊=(sin2A+sin2B)+sin2[π-(A+B)]=2sin(A+B)cos(A-B)-sin2(A+B)=2sin(A+B)cos(A-B)-2sin(A+B)cos(A+B)=2sin(A+B)[cos(A-B)-cos(A+B)]=2sin(π-C)·(cosAcosB+sinAsinB-cosAcosB+sinAsinB)=2sinC·2sinA·sinB=4sinAsinBsinC=右邊.所以原式得證.1.函數(shù)f(x)=sin2x+QUOTEsinxcosx在區(qū)間QUOTE上的最小值為.

【解析】f(x)=QUOTE+QUOTEsin2x=QUOTE+QUOTE=sinQUOTE+QUOTE,又x∈QUOTE,所以2x-QUOTE∈QUOTE,所以sinQUOTE∈QUOTE,故f(x)min=QUOTE+QUOTE=1.答案:12.求證:QUOTE=QUOTE.【證明】原式等價(jià)于1+sin