2024-2025學(xué)年新教材高中數(shù)學(xué)第10章三角恒等變換10.2二倍角的三角函數(shù)課時(shí)素養(yǎng)評(píng)價(jià)含解析蘇教版必修第二冊(cè)

PAGE課時(shí)素養(yǎng)評(píng)價(jià)十三二倍角的三角函數(shù)(20分鐘35分)1.sin22°30'·cos22°30'的值為 ()A.QUOTE B.QUOTE C.-QUOTE D.QUOTE【解析】選B.原式=QUOTEsin45°=QUOTE.2.若sinQUOTE=QUOTE,則cosα等于 ()A.-QUOTE B.-QUOTE C.QUOTE D.QUOTE【解析】選C.因?yàn)閟inQUOTE=QUOTE,所以cosα=1-2sin2QUOTE=1-2×QUOTE=QUOTE.3.若QUOTE=QUOTE,則tan2α等于 ()A.-QUOTE B.QUOTE C.-QUOTE D.QUOTE【解析】選B.因?yàn)镼UOTE=QUOTE,所以QUOTE=QUOTE,故tanα=-3,所以依據(jù)倍角公式,得tan2α=QUOTE.【補(bǔ)償訓(xùn)練】設(shè)f(tanx)=tan2x,則f(2)的值為 ()A.QUOTEB.-QUOTEC.-QUOTED.4【解析】選B.因?yàn)閒(tanx)=QUOTE,所以f(2)=QUOTE=-QUOTE.4.化簡(jiǎn)QUOTE-2QUOTE= ()A.2sin4 B.-2sin4C.2cos4 D.-2cos4【解析】選A.原式=QUOTE-2QUOTE=2|cos4|-2|sin4+cos4|,因?yàn)棣?lt;4<QUOTE,所以cos4<0,sin4+cos4<0.所以原式=-2cos4+2(sin4+cos4)=2sin4.5.已知tanα=-QUOTE,則QUOTE=________.

【解析】QUOTE=QUOTE=QUOTE=tanα-QUOTE=-QUOTE.答案:-QUOTE6.已知函數(shù)f(x)=cosQUOTE+sin2x-cos2x+2QUOTE·sinxcosx.(1)化簡(jiǎn)f(x);(2)若f(α)=QUOTE,2α是第一象限角,求sin2α.【解析】(1)f(x)=QUOTEcos2x-QUOTEsin2x-cos2x+QUOTEsin2x=QUOTEsin2x-QUOTEcos2x=sinQUOTE.(2)f(α)=sinQUOTE=QUOTE,2α是第一象限角,即2kπ<2α<QUOTE+2kπ(k∈Z),所以2kπ-QUOTE<2α-QUOTE<QUOTE+2kπ,所以cosQUOTE=QUOTE,所以sin2α=sinQUOTE=sinQUOTEcosQUOTE+cosQUOTEsinQUOTE=QUOTE×QUOTE+QUOTE×QUOTE=QUOTE.(30分鐘60分)一、單選題(每小題5分,共20分)1.若tanα=3,則QUOTE= ()A.2 B.3 C.4 D.6【解析】選D.QUOTE=QUOTE=QUOTE=2tanα=6.2.QUOTE-QUOTE等于 ()A.-2cos5° B.2cos5°C.-2sin5° D.2sin5°【解析】選C.原式=QUOTE-QUOTE=QUOTE(cos50°-sin50°)=2QUOTE=2sin(45°-50°)=-2sin5°.3.已知等腰三角形底角的正弦值為QUOTE,則頂角的正弦值是 ()A.QUOTE B.QUOTEC.-QUOTE D.-QUOTE【解析】選A.設(shè)底角為θ,則θ∈QUOTE,頂角為180°-2θ.因?yàn)閟inθ=QUOTE,所以cosθ=QUOTE=QUOTE.所以sin(180°-2θ)=sin2θ=2sinθcosθ=2×QUOTE×QUOTE=QUOTE.4.下列關(guān)于函數(shù)f(x)=1-2sin2QUOTE的說(shuō)法錯(cuò)誤的是 ()A.最小正周期為πB.最大值為1,最小值為-1C.函數(shù)圖象關(guān)于直線x=0對(duì)稱D.函數(shù)圖象關(guān)于點(diǎn)QUOTE對(duì)稱【解析】選C.函數(shù)f(x)=1-2sin2QUOTE=cosQUOTE=sin2x,函數(shù)的最小正周期T=π,A正確.最大值為1,最小值為-1,B正確.由2x=kπ+QUOTE?x=QUOTE+QUOTE,k∈Z,得函數(shù)圖象關(guān)于直線x=QUOTE+QUOTE,k∈Z對(duì)稱,C不正確.由2x=kπ?x=QUOTE,k∈Z,得函數(shù)圖象關(guān)于點(diǎn)QUOTE,k∈Z對(duì)稱,D正確.二、多選題(每小題5分,共10分,全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)5.下列選項(xiàng)中,值為QUOTE的是 ()A.cos72°cos36° B.sinQUOTEsinQUOTEC.QUOTE+QUOTE D.QUOTE-QUOTEcos215°【解析】選AB.對(duì)于A,cos36°cos72°=QUOTE=QUOTE=QUOTE=QUOTE,故A正確;對(duì)于B,sinQUOTEsinQUOTE=sinQUOTEcosQUOTE=QUOTE·2sinQUOTEcosQUOTE=QUOTEsinQUOTE=QUOTE,故B正確;對(duì)于C,原式=QUOTE=QUOTE=QUOTE=QUOTE=4,故C錯(cuò)誤;對(duì)于D,QUOTE-QUOTEcos215°=-QUOTE(2cos215°-1)=-QUOTEcos30°=-QUOTE,故D錯(cuò)誤.6.若2cos2α=sinQUOTE,則sin2α的值為 ()A.-QUOTE B.QUOTE C.1 D.QUOTE【解析】選AC.若2cos2α=sinQUOTE,即2(cos2α-sin2α)=QUOTEcosα-QUOTEsinα,當(dāng)cosα=sinα?xí)r,滿意條件,此時(shí),tanα=1,sin2α=1.當(dāng)cosα≠sinα?xí)r,則2(cosα+sinα)=QUOTE,即cosα+sinα=QUOTE,所以1+2sinαcosα=QUOTE,即sin2α=2sinαcosα=-QUOTE.綜上可得,sin2α=1或-QUOTE.【補(bǔ)償訓(xùn)練】(多選題)若函數(shù)f(x)=(1+QUOTEtanx)cosx,則f(x)的()A.最小正周期為π B.最大值是2C.最小正周期為2π D.最大值是1【解析】選BC.f(x)=(1+QUOTEtanx)cosx=QUOTEcosx=QUOTEsinx+cosx=2sinQUOTE.所以f(x)的最小正周期為2π,當(dāng)x+QUOTE=2kπ+QUOTE,k∈Z時(shí),f(x)取到最大值2.三、填空題(每小題5分,共10分)7.函數(shù)f(x)=2cos2QUOTE-1的最小正周期為_(kāi)_______.

【解析】f(x)=cosQUOTE=sin2x,故f(x)的最小正周期為π.答案:π【補(bǔ)償訓(xùn)練】函數(shù)f(x)=cos2x+4sinx的值域是________.

【解析】f(x)=cos2x+4sinx=1-2sin2x+4sinx=-2sin2x+4sinx+1=-2(sinx-1)2+3.當(dāng)sinx=1時(shí),f(x)max=3;當(dāng)sinx=-1時(shí),f(x)min=-5.答案:[-5,3]8.若sinα+2cosα=0(0<α<π),則tanα=________,cosQUOTE=________.

【解析】因?yàn)閟inα+2cosα=0(0<α<π),所以sinα=-2cosα,即tanα=-2.所以cos(2α+QUOTE)=QUOTEcos2α-QUOTEsin2α=QUOTE·QUOTE-QUOTE·QUOTE=QUOTE·QUOTE-QUOTE·QUOTE=QUOTE×QUOTE-QUOTE×QUOTE=QUOTE.答案:-2QUOTE四、解答題(每小題10分,共20分)9.已知α為其次象限角,且sinα=QUOTE,求QUOTE的值.【解析】原式=QUOTE=QUOTE.因?yàn)棣翞槠浯蜗笙藿?且sinα=QUOTE,所以sinα+cosα≠0,cosα=-QUOTE,所以原式=QUOTE=-QUOTE.10.設(shè)函數(shù)f(x)=2cos2ωx+sinQUOTE+a(其中ω>0,a∈R),且f(x)的圖象在y軸右側(cè)的第一個(gè)最高點(diǎn)的橫坐標(biāo)為QUOTE.(1)求ω的值;(2)設(shè)f(x)在區(qū)間QUOTE上的最小值為QUOTE,求a的值.【解析】f(x)=1+cos2ωx+QUOTEsin2ωx-QUOTEcos2ωx+a=sinQUOTE+a+1.(1)由2ωx+QUOTE=2kπ+QUOTE(k∈Z),得ωx=kπ+QUOTE(k∈Z).又ω>0,所以當(dāng)k=0時(shí),f(x)的圖象在y軸右側(cè)的第一個(gè)最高點(diǎn)的橫坐標(biāo)為x=QUO