考點(diǎn)規(guī)范練23三角恒等變換基礎(chǔ)鞏固1.2sin47°-3sin17°A.-3 B.-1 C.3 D.答案:D解析:原式=2×sin47°-sin17°cos30°cos17°=2×sin(2.已知2sin2α=1+cos2α,則tan2α=()A.43 B.-43 C.43或0 答案:C解析:因?yàn)?sin2α=1+cos2α,所以2sin2α=2cos2α.所以2cosα(2sinα-cosα)=0,解得cosα=0或tanα=1若cosα=0,則α=kπ+π2,k∈Z,2α=2kπ+π,k∈Z,所以tan2α=0若tanα=12,則tan2α=綜上所述,故選C.3.已知f(x)=sin2x+sinxcosx,則f(x)的最小正周期和一個單調(diào)遞增區(qū)間分別為()A.π,[0,π] B.2π,-C.π,-π8,3π8 答案:C解析:由f(x)=sin2x+sinxcosx=1-cos2x2=1=12+2則T=2π2=又2kπ-π2≤2x-π4≤2kπ+π2∴kπ-π8≤x≤kπ+3π8(k∈Z)為函數(shù)的單調(diào)遞增區(qū)間4.已知5sin2α=6cosα,α∈0,π2,則tanA.-23 B.13 C.答案:B解析:由題意,知10sinαcosα=6cosα,又α∈0,π2,∴sinα=35∴tanα5.(2021全國Ⅱ,理9)若α∈0,π2,tan2α=cosα2-A.1515 B.55 C.53 答案:A解析:由題意可知sin2αcos2α=因?yàn)棣痢?,π2,所以cosα>0,所以2sinα1-2sin2α=12-sinα,解得sinα=6.為了得到函數(shù)y=sin2x+cos2x的圖象,可以將函數(shù)y=cos2x-sin2x的圖象()A.向右平移π4個單位長度 B.向左平移πC.向右平移π2個單位長度 D.向左平移π答案:A解析:∵y=sin2x+cos2x=2=2cos2x-π8,y=cos2x-sin2x=2=2cos2=2cos2x∴只需將函數(shù)y=cos2x-sin2x的圖象向右平移π4個單位長度可得函數(shù)y=sin2x+cos2x的圖象7.已知函數(shù)f(x)=cos4x-π3+2cos22x,將函數(shù)y=f(x)的圖象上所有點(diǎn)的橫坐標(biāo)伸長為原來的2倍,縱坐標(biāo)不變,再將所得函數(shù)圖象向右平移π6個單位,得到函數(shù)y=g(x)的圖象,則函數(shù)y=gA.-π3,π6 B.答案:B解析:∵函數(shù)f(x)=cos4x-π3+2cos22x=cos4x-π3+1+cos4x=12cos4x+32sin4x+1+cos4x=32cos4x+3∴函數(shù)y=f(x)的圖象伸縮后的圖象對應(yīng)的解析式為y=3sin2x+π3+1,再平移后得y=g(x)=3sin由2kπ-π2≤2x≤2kπ+π2,k得kπ-π4≤x≤kπ+π4,k當(dāng)k=0時,得-π4≤x≤π48.已知2cos2x+sin2x=Asin(ωx+φ)+b(A>0),則A=,b=.

答案:21解析:因?yàn)?cos2x+sin2x=1+cos2x+sin2x=2sin2x+π4+1,所以A=2,b=1.9.設(shè)f(x)=1+cos2x2sinπ2-x+sinx+a2sinx+π答案:±3解析:f(x)=1+2cos2x-12cosx=cosx+sinx+a2sinx=2sinx+π4+a=(2+a2)sinx依題意有2+a2=2+3,則a=±310.已知點(diǎn)π4,1在函數(shù)f(x)=2asinxcosx+cos2(1)求a的值和f(x)的最小正周期;(2)求函數(shù)f(x)在區(qū)間(0,π)內(nèi)的單調(diào)遞減區(qū)間.解:(1)函數(shù)f(x)=2asinxcosx+cos2x=asin2x+cos2x.∵f(x)的圖象過點(diǎn)π4∴1=asinπ2+cosπ2,可得a=∴f(x)=sin2x+cos2x=2sin2∴函數(shù)的最小正周期T=2π2=(2)由2kπ+π2≤2x+π4≤3π2+2可得kπ+π8≤x≤5π8+kπ∴函數(shù)f(x)的單調(diào)遞減區(qū)間為kπ+π8,∵x∈(0,π),∴當(dāng)k=0時,可得單調(diào)遞減區(qū)間為π11.函數(shù)f(x)=cos-x2+sinπ-x2(1)求f(x)的最小正周期;(2)若f(α)=2105,α∈0解:(1)f(x)=cos-x2+sinπ-x2=sinx2+故f(x)的最小正周期T=2π12=(2)由f(α)=2105,得sinα2+cosα2即1+sinα=85,解得sinα=3又α∈0,π2,則cos故tanα=sinα所以tanα+π4能力提升12.已知函數(shù)f(x)=cosωx(sinωx+3cosωx)(ω>0),若存在實(shí)數(shù)x0,使得對任意的實(shí)數(shù)x,都有f(x0)≤f(x)≤f(x0+2016π)成立,則ω的最小值為()A.12016π B.1答案:C解析:由題意可得,f(x0)是函數(shù)f(x)的最小值,f(x0+2016π)是函數(shù)f(x)的最大值.又f(x)=cosωx(sinωx+3cosωx)=12sin2ωx+=sin2ωx所以要使ω取最小值,只需保證區(qū)間[x0,x0+2016π]為一個完整的單調(diào)遞增區(qū)間即可.故2016π=12·2πωmin,求得ωmin=1201613.已知cosα=13,cos(α+β)=-13,且α,β∈0,π2A.-12 B.12 C.-1答案:D解析:∵α∈0,π2,∴2∵cosα=13,∴cos2α=2cos2α-1=-7∴sin2α=1-又α,β∈0,π2,∴α+∴sin(α+β)=1-∴cos(α-β)=cos[2α-(α+β)]=cos2αcos(α+β)+sin2αsin(α+β)=-=2314.已知函數(shù)f(x)=2sinx+5π24cosx+5π24-2cos2x+5π24+1,則f(x)的最小正周期為答案:πkπ-π3,解析:f(x)=2sinx+5π24cosx+5=sin2x+5=2=2sin2=2sin2∴f(x)的最小正周期T=2π2=因此f(x)=2sin2當(dāng)2kπ-π2≤2x+π6≤2kπ+π2即kπ-π3≤x≤kπ+π6(k∈Z∴函數(shù)f(x)的單調(diào)遞增區(qū)間是kπ-π3,k15.已知函數(shù)f(x)=3sinxcosx+cos2x.(1)求函數(shù)f(x)的最小正周期;(2)若-π2<α<0,f(α)=56,求sin2α解:(1)∵函數(shù)f(x)=3sinxcosx+cos2x=32sin2x+1+cos2x2=∴函數(shù)f(x)的最小正周期為2π2=(2)若-π2<α<0,則2α+∵f(α)=sin2α∴sin2α+π6=13∴cos2α∴sin2α=sin2α+π6-π6=sin2高考預(yù)測16.已知f(x)=1+1tanxsin2x-2sin(1)若tanα=2,求f(α)的值;(2)若x∈π12,π2,求解:(1)f(x)=(sin2x+sinxcosx)+2sinx+π=1-cos2x2+1=12+12(sin2x-cos2x)+=12(si