2024-2025學(xué)年高中數(shù)學(xué)第三章三角恒等變換單元素養(yǎng)評價含解析新人教A版必修4

PAGE單元素養(yǎng)評價(三)(第三章)(120分鐘150分)一、選擇題(每小題5分,共60分)1.函數(shù)y=2cos2QUOTE+1的最小正周期是 ()A.4π B.2π C.π D.QUOTE【解析】選B.因為y=2cos2QUOTE+1=QUOTE+2=cosx+2,所以函數(shù)的最小正周期T=2π.2.(2024·長春高一檢測)已知sinα=QUOTE,cosα=QUOTE,則tanQUOTE等于 ()A.2-QUOTE B.2+QUOTEC.QUOTE-2 D.±(QUOTE-2)【解析】選C.因為sinα=QUOTE,cosα=QUOTE,所以tanQUOTE=QUOTE=QUOTE=QUOTE-2.3.若3sinx-QUOTEcosx=2QUOTEsin(x+φ),φ∈(-π,π),則φ等于 ()A.-QUOTE B.QUOTE C.QUOTE D.-QUOTE【解析】選A.3sinx-QUOTEcosx=2QUOTE=2QUOTEsinQUOTE.又φ∈(-π,π),所以φ=-QUOTE.4.(2024·紹興高一檢測)已知角α的頂點在坐標(biāo)原點O,始邊與x軸的非負(fù)半軸重合,將α的終邊按順時針方向旋轉(zhuǎn)QUOTE后經(jīng)過點(3,4),則tanα= ()A.-7 B.-QUOTE C.QUOTE D.7【解析】選A.依據(jù)題意tanQUOTE=QUOTE,tanQUOTE=QUOTE=QUOTE,所以tanα=-7.5.(2024·全國卷Ⅲ)函數(shù)f(x)=2sinx-sin2x在[0,2π]的零點個數(shù)為 ()A.2 B.3 C.4 D.5【解析】選B.令f(x)=2sinx-sin2x=2sinx-2sinxcosx=2sinx(1-cosx)=0,則sinx=0或cosx=1,又x∈[0,2π],所以x=0,π,2π,共三個零點.6.若α∈(0,π),且cosα+sinα=-QUOTE,則cos2α= ()A.QUOTE B.-QUOTE C.-QUOTE D.QUOTE【解析】選A.因為cosα+sinα=-QUOTE,α∈(0,π),所以sin2α=-QUOTE,cosα<0,且α∈QUOTE,所以2α∈QUOTE,所以cos2α=QUOTE=QUOTE.7.QUOTE= ()A.QUOTE B.-QUOTE C.-1 D.1【解析】選B.原式=QUOTE=-QUOTE=-QUOTE=-QUOTE=-QUOTE.8.(2024·珠海高一檢測)在△ABC中,已知tanQUOTE=sinC,則△ABC的形態(tài)為 ()A.正三角形 B.等腰三角形C.直角三角形 D.等腰直角三角形【解析】選C.在△ABC中,tanQUOTE=sinC=sin(A+B)=2sinQUOTEcosQUOTE,所以2cos2QUOTE=1,所以cos(A+B)=0,從而A+B=QUOTE,即△ABC為直角三角形.9.已知0<β<α<QUOTE,點P(1,4QUOTE)為角α的終邊上一點,且sinαsinQUOTE+cosαcosQUOTE=QUOTE,則角β= ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選D.因為P(1,4QUOTE),所以|OP|=7(O為坐標(biāo)原點),所以sinα=QUOTE,cosα=QUOTE.又sinαcosβ-cosαsinβ=QUOTE,所以sin(α-β)=QUOTE.因為0<β<α<QUOTE,所以0<α-β<QUOTE,所以cos(α-β)=QUOTE,所以sinβ=sin[α-(α-β)]=sinαcos(α-β)-cosαsin(α-β)=QUOTE×QUOTE-QUOTE×QUOTE=QUOTE.因為0<β<QUOTE,所以β=QUOTE.10.已知0<α<QUOTE<β<π,又sinα=QUOTE,cos(α+β)=-QUOTE,則sinβ等于()A.0 B.0或QUOTE C.QUOTE D.±QUOTE【解析】選C.因為0<α<QUOTE<β<π且sinα=QUOTE,cos(α+β)=-QUOTE,所以cosα=QUOTE,QUOTE<α+β<QUOTEπ,所以sin(α+β)=±QUOTE,當(dāng)sin(α+β)=QUOTE時,sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα=QUOTE×QUOTE-QUOTE×QUOTE=QUOTE;當(dāng)sin(α+β)=-QUOTE時,sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα=-QUOTE×QUOTE-QUOTE×QUOTE=0.又β∈QUOTE,所以sinβ>0,故sinβ=QUOTE.11.(2024·廣州高一檢測)已知函數(shù)f(x)=sinQUOTE,若方程f(x)=QUOTE的解為x1,x2(0<x1<x2<π),則sin(x1-x2)= ()A.-QUOTE B.-QUOTE C.-QUOTE D.-QUOTE【解題指南】由已知可得x2=QUOTE-x1,結(jié)合x1<x2求出x1的范圍,再由sin(x1-x2)=sinQUOTE=-cosQUOTE求解即可.【解析】選D.因為方程f(x)=QUOTE的解為x1,x2(0<x1<x2<π),所以QUOTE=QUOTE,所以x2=QUOTE-x1,所以sin(x1-x2)=sinQUOTE=-cosQUOTE.因為x1<x2,x2=QUOTE-x1,所以0<x1<QUOTE,所以2x1-QUOTE∈QUOTE,所以由f(x1)=sinQUOTE=QUOTE,得cosQUOTE=QUOTE,所以sin(x1-x2)=-QUOTE.12.已知不等式f(x)=3QUOTEsinQUOTEcosQUOTE+QUOTEcos2QUOTE-QUOTE-m≤0對于隨意的-QUOTE≤x≤QUOTE恒成立,則實數(shù)m的取值范圍是 ()A.m≥QUOTE B.m≤QUOTEC.m≤-QUOTE D.-QUOTE≤m≤QUOTE【解析】選A.f(x)=3QUOTEsinQUOTEcosQUOTE+QUOTEcos2QUOTE-QUOTE-m=QUOTEsinQUOTE+QUOTEcosQUOTE-m,=QUOTEsinQUOTE-m≤0,所以m≥QUOTEsinQUOTE,因為-QUOTE≤x≤QUOTE,所以-QUOTE≤QUOTE+QUOTE≤QUOTE,所以-QUOTE≤QUOTEsinQUOTE≤QUOTE,所以m≥QUOTE.二、填空題(每小題5分,共20分)13.若函數(shù)f(x)=QUOTEsin2x+2cos2x+m在區(qū)間QUOTE上的最大值為6,則m=.

【解析】f(x)=QUOTEsin2x+2cos2x+m=QUOTEsin2x+1+cos2x+m=2sinQUOTE+m+1,因為0≤x≤QUOTE,所以QUOTE≤2x+QUOTE≤QUOTE.所以當(dāng)2x+QUOTE=QUOTE,即x=QUOTE時,f(x)max=2+m+1=6,所以m=3.答案:314.tanQUOTE+tanQUOTE+QUOTEtanQUOTE·tanQUOTE+θ的值是.

【解析】因為tanQUOTE=tanQUOTE=QUOTE=QUOTE,所以QUOTE=tanQUOTE+tanQUOTE+QUOTEtanQUOTE-θtanQUOTE.答案:QUOTE15.(2024·臨汾高一檢測)若sinα+2cosα=-QUOTE(0<α<π),則cosQUOTE=.

【解析】由sinα+2cosα=-QUOTE(0<α<π)可知,α為鈍角,又sin2α+cos2α=1,可得sinα=QUOTE,cosα=-QUOTE,所以sin2α=2sinαcosα=-QUOTE,cos2α=cos2α-sin2α=-QUOTE,所以cosQUOTE=cos2αcosQUOTE-sin2αsinQUOTE=QUOTE.答案:QUOTE16.關(guān)于函數(shù)f(x)=cosQUOTE+cosQUOTE,則下列命題:①y=f(x)的最大值為QUOTE;②y=f(x)的最小正周期是π;③y=f(x)在區(qū)間QUOTE上是減函數(shù);④將函數(shù)y=QUOTEcos2x的圖象向右平移QUOTE個單位后,將與已知函數(shù)的圖象重合.其中正確命題的序號是.

【解析】f(x)=cosQUOTE+cosQUOTE=cosQUOTE+sinQUOTE=cosQUOTE-sinQUOTE==QUOTEcosQUOTE=QUOTEcosQUOTE,所以y=f(x)的最大值為QUOTE,最小正周期為π,故①、②正確.又當(dāng)x∈QUOTE時,2x-QUOTE∈[0,π],所以y=f(x)在QUOTE上是減函數(shù),故③正確.由④得y=QUOTEcos2QUOTE=QUOTEcosQUOTE,故④正確.答案:①②③④三、解答題(共70分)17.(10分)(1)求值:QUOTE.(2)已知sinθ+2cosθ=0,求QUOTE的值.【解析】(1)原式=QUOTE=QUOTE=QUOTE=2+QUOTE.(2)由sinθ+2cosθ=0,得sinθ=-2cosθ,又cosθ≠0,則tanθ=-2,所以QUOTE=QUOTE=QUOTE=QUOTE=QUOTE.18.(12分)已知sinQUOTEsinQUOTE=QUOTE,且α∈QUOTE,求tan4α的值.【解析】因為sinQUOTE=sinQUOTE=cosQUOTE,則已知條件可化為sinQUOTEcosQUOTE=QUOTE,即QUOTEsinQUOTE=QUOTE,所以sinQUOTE=QUOTE,所以cos2α=QUOTE.因為α∈QUOTE,所以2α∈(π,2π),從而sin2α=-QUOTE=-QUOTE,所以tan2α=QUOTE=-2QUOTE,故tan4α=QUOTE=-QUOTE=QUOTE.19.(12分)(2024·晉中高一檢測)設(shè)向量a=(QUOTEsinx,sinx),b=(cosx,sinx),x∈QUOTE.(1)若|a|=|b|,求x的值.(2)設(shè)函數(shù)f(x)=a·b,求f(x)的最大值.【解析】(1)由|a|2=(QUOTEsinx)2+(sinx)2=4sin2x,|b|2=(cosx)2+(sinx)2=1及|a|=|b|,得4sin2x=1.又x∈QUOTE,從而sinx=QUOTE,所以x=QUOTE.(2)f(x)=a·b=QUOTEsinx·cosx+sin2x=QUOTEsin2x-QUOTEcos2x+QUOTE=sinQUOTE+QUOTE,又x∈QUOTE,所以當(dāng)x=QUOTE時,sinQUOTE取得最大值為1,所以f(x)的最大值為QUOTE.20.(12分)已知函數(shù)f(x)=cosx(sinx+cosx)-QUOTE.(1)若0<α<QUOTE,且sinα=QUOTE,求f(α)的值.(2)求函數(shù)f(x)的最小正周期及單調(diào)遞增區(qū)間.【解析】方法一:(1)因為0<α<QUOTE,sinα=QUOTE,所以cosα=QUOTE.所以f(α)=QUOTE×QUOTE-QUOTE=QUOTE.(2)因為f(x)=sinxcosx+cos2x-QUOTE=QUOTEsin2x+QUOTE-QUOTE=QUOTEsin2x+QUOTEcos2x=QUOTEsinQUOTE,所以T=QUOTE=π.由2kπ-QUOTE≤2x+QUOTE≤2kπ+QUOTE,k∈Z,得kπ-QUOTE≤x≤kπ+QUOTE,k∈Z.所以f(x)的單調(diào)遞增區(qū)間為QUOTE,k∈Z.方法二:f(x)=sinxcosx+cos2x-QUOTE=QUOTEsin2x+QUOTE-QUOTE=QUOTEsin2x+QUOTEcos2x=QUOTEsinQUOTE.(1)因為0<α<QUOTE,sinα=QUOTE,所以α=QUOTE.從而f(α)=QUOTEsinQUOTE=QUOTEsinQUOTE=QUOTE.(2)T=QUOTE=π.由2kπ-QUOTE≤2x+QUOTE≤2kπ+QUOTE,k∈Z,得kπ-QUOTE≤x≤kπ+QUOTE,k∈Z.所以f(x)的單調(diào)遞增區(qū)間為QUOTE,k∈Z.21.(12分)如圖所示,已知α的終邊所在直線上的一點P的坐標(biāo)為(-3,4),β的終邊在第一象限且與單位圓的交點Q的縱坐標(biāo)為QUOTE.(1)求tan(2α-β)的值.(2)若QUOTE<α<π,0<β<QUOTE,求α+β.【解析】(1)由三角函數(shù)的定義知tanα=-QUOTE,所以tan2α=QUOTE=QUOTE.又由三角函數(shù)線知sinβ=QUOTE.因