新教材數(shù)學(xué)高中必修一課時(shí)素養(yǎng)評價(jià)7-2-3三角函數(shù)的誘導(dǎo)公式（二）

課時(shí)素養(yǎng)評價(jià)三十八三角函數(shù)的誘導(dǎo)公式(二)(15分鐘30分)1.如果cos(π+A)=QUOTE,那么sinQUOTE= ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選B.因?yàn)閏os(π+A)=cosA=QUOTE,所以cosA=QUOTE,所以sinQUOTE=cosA=QUOTE.2.已知sinQUOTE=QUOTE,則cosQUOTE的值是()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選C.cosQUOTE=cosQUOTE=sinQUOTE=QUOTE.3.(2020·重慶高一檢測)已知角θ是第二象限角,且滿足sinQUOTE=QUOTE,則tan(π+θ)= ()A.QUOTE B.1 C.QUOTE D.QUOTE【解析】選A.因?yàn)榻铅仁堑诙笙藿?且滿足sinQUOTE=cosθ=QUOTE,可得cosθ=QUOTE,所以sinθ=QUOTE=QUOTE,所以tan(π+θ)=tanθ=QUOTE=QUOTE.4.已知cosα=QUOTE,則sinQUOTE·cosQUOTE·tan(πα)=________.

【解析】sinQUOTEcosQUOTEtan(πα)=cosαsinα·(tanα)=sin2α=1cos2α=1QUOTE=QUOTE.答案:QUOTE5.(2020·延吉高一檢測)已知α是第三象限角,且f(α)=QUOTE.(1)若cosQUOTE=QUOTE,求f(α)的值.(2)求函數(shù)y=f2(x)+sinx,x∈QUOTE的值域.【解析】(1)因?yàn)棣潦堑谌笙藿?cosQUOTE=QUOTE=sinα,所以sinα=QUOTE,所以f(α)==QUOTE=cosα=QUOTE=QUOTE.(2)因?yàn)閤∈QUOTE,所以sinx∈QUOTE,函數(shù)y=f2(x)+sinx=1sin2x+sinx=QUOTEQUOTE,故當(dāng)sinx=QUOTE時(shí),函數(shù)取得最大值為QUOTE;當(dāng)sinx=QUOTE時(shí),函數(shù)取得最小值為QUOTE,故該函數(shù)的值域?yàn)镼UOTE.(20分鐘40分)一、選擇題(每小題5分,共20分)1.如果角θ的終邊經(jīng)過點(diǎn)QUOTE,那么sinQUOTE+θ+cos(πθ)+tan(2πθ)= ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選B.易知sinθ=QUOTE,cosθ=QUOTE,tanθ=QUOTE.原式=cosθcosθtanθ=QUOTE.2.若f(sinx)=3cos2x,則f(cosx)= ()A.3cos2x B.3sin2xC.3+cos2x D.3+sin2x【解析】選C.f(cosx)=fQUOTE=3cos(π2x)=3+cos2x.3.已知f(x)=sinx,下列式子成立的是 ()A.f(x+π)=sinx B.f(2πx)=sinxC.f(πx)=f(x) D.fQUOTE=cosx【解析】選D.f(x+π)=sin(x+π)=sinx;f(2πx)=sin(2πx)=sin(x)=sinx;fQUOTE=sinQUOTE=sinQUOTE=cosx;f(πx)=sin(πx)=sinx=f(x).【補(bǔ)償訓(xùn)練】計(jì)算sin21°+sin22°+sin23°+…+sin289°= ()A.89 B.90 C.QUOTE D.45【解析】選C.原式=(sin21°+sin289°)+(sin22°+sin288°)+…+(sin244°+sin246°)+sin245°=44+QUOTE=QUOTE.4.(多選題)下列與cosQUOTE的值相等的是 ()A.sin(πθ) B.sin(π+θ)C.cosQUOTE D.cosQUOTE【解析】選BD.cosQUOTE=cosQUOTE=cosQUOTE=sinθ.A中sin(πθ)=sinθ;B中sin(π+θ)=sinθ;C中cosQUOTE=sinθ;D中cosQUOTE=sinθ.【補(bǔ)償訓(xùn)練】(多選題)角θ與φ都是任意角,若滿足θ+φ=90°,則稱θ與φ“廣義互余”.已知sin(π+α)=QUOTE,下列角β中,可能與角α“廣義互余”的是 ()A.sinβ=QUOTE B.cos(π+β)=QUOTEC.tanβ=QUOTE D.tanβ=QUOTE.【解析】選AC.因?yàn)閟in(π+α)=sinα,所以sinα=QUOTE,若α+β=90°,則β=90°α,故sinβ=sin(90°α)=cosα=±QUOTE,故A滿足;C中tanβ=QUOTE,即sinβ=QUOTEcosβ,又sin2β+cos2β=1,故sinβ=±QUOTE,即C滿足,而B、D不滿足.二、填空題(每小題5分,共10分)5.若sinQUOTE=QUOTE,則cosQUOTE=________.

【解題指南】根據(jù)題意先分析αQUOTE與α+QUOTE的關(guān)系,再由誘導(dǎo)公式即可化簡求值得解.【解析】因?yàn)閟inQUOTE=QUOTE,所以cosQUOTE=cosQUOTE=cosQUOTE=sinQUOTE=QUOTE.答案:QUOTE【補(bǔ)償訓(xùn)練】已知cos(75°+α)=QUOTE,則sin(α15°)+cos(105°α)的值是________.

【解析】sin(α15°)+cos(105°α)=sin[(75°+α)90°]+cos[180°(75°+α)]=sin[90°(75°+α)]cos(75°+α)=cos(75°+α)cos(75°+α)=2cos(75°+α),因?yàn)閏os(75°+α)=QUOTE,所以原式=QUOTE.答案:QUOTE6.已知cosQUOTE=QUOTE,則cosQUOTE=________,sinQUOTE=________.

【解析】cosQUOTE=cosQUOTE=cosQUOTE=QUOTE.sinQUOTE=sinQUOTE=sinQUOTE=sinQUOTE=cosQUOTE=QUOTE.答案:QUOTEQUOTE三、解答題7.(10分)已知sinα是方程5x27x6=0的根,α是第三象限角,求QUOTE·tan2(πα)的值.【解析】方程5x27x6=0