5.5.2簡(jiǎn)單的三角恒等變換

5.5.2簡(jiǎn)單的三角恒等變換A級(jí)必備知識(shí)基礎(chǔ)練1.cos2π8-14A.2-14 B.2+14 C2.已知α為第一象限角,且tanα=43,則sinα2的值為(A.55 B.55 C.±553.(多選題)已知函數(shù)f(x)=sinxcosx+sin2x,則下列說法正確的是()A.f(x)的最大值為2B.f(x)的最小正周期為πC.f(x)的圖象關(guān)于直線x=π8D.f(x)在0,4.函數(shù)f(x)=sin2x+sinxcosx+1的最小正周期為.

5.若tanα=17,則1+cos2αsin2α6.證明:2sinxB級(jí)關(guān)鍵能力提升練7.若函數(shù)f(x)=(1+3tanx)cosx,則fπ12=()A.6-22C.1 D.28.若3π<x<4π,則1+cosx2+1-A.2cosπ4-x2 B.2cosπC.2sinπ4-x2 D.2sinπ9.設(shè)函數(shù)f(x)=2cos2x+3sin2x+a(a為實(shí)常數(shù))在區(qū)間0,π2上的最小值為4,那么a的值等于A.4 B.6C.4 D.310.已知等腰三角形的頂角的余弦值等于725,則它的底角的余弦值為(A.34 B.C.12 D.11.(多選題)有以下四個(gè)關(guān)于三角函數(shù)的命題,其中正確的是()A.?x∈R,sin2x2+cos2B.?x,y∈R,sin(xy)=sinxsinyC.?x∈[0,π],1-cos2xD.sinx=cosy,則x+y=π12.已知sinx+π3=33,則cosx+cosxπ3=.

13.若cosθ=725,θ∈(π,2π),則sinθ2+cosθ2=,sinθ2cosθ14.已知sinα=1213,sin(α+β)=45,α,β均為銳角,求cosβC級(jí)學(xué)科素養(yǎng)創(chuàng)新練15.已知sinA+sinB+sinC=0,cosA+cosB+cosC=0,求證:cos2A+cos2B+cos2C=32

5.5.2簡(jiǎn)單的三角恒等變換1.Bcos2π8-2.C因?yàn)棣翞榈谝幌笙藿?且tanα=43,所以cosα=35,而α2是第一或第三象限角.當(dāng)α2是第一象限角時(shí),sinα2=1-cosα2=55;當(dāng)α2是第三象限角時(shí),sin3.BCD∵f(x)=12sin2x+1-cos2x2=12(sin2xcos∴f(x)max=22+12=2+1當(dāng)x=π8時(shí),sin2x-π4=1,∴直線x=π8為當(dāng)x∈0,π4時(shí),2∴f(x)在0,π綜上B,C,D正確,A不正確.4.πf(x)=sin2x+sinxcosx+1=1-cos2x2+12sin2x+1=12(sin2xcos2x)+32=22sin2x5.7因?yàn)閠anα=sin2α1+cos2α=176.證明左邊=2sinxcosx所以原等式成立.7.D∵f(x)=1+3·sinxcosxcosx=cosx+3sinx=2sinx+π6,∴8.C因?yàn)?π<x<4π,所以3π2<x2<2π,sinx2于是1+cosx2+1-cosx2=cosx2+sinx2=cosx2sinx2=2229.Cf(x)=2cos2x+3sin2x+a=1+cos2x+3sin2x+a=2sin2x+π當(dāng)x∈0,π2時(shí),2∴f(x)min=2·-12+a+1∴a=4.10.B設(shè)等腰三角形的頂角為α,底角為β,則cosα=725.又β=π2-α2,即cosβ=cosπ211.BC因?yàn)閟in2x2+cos2x2=1≠12,所以A為假命題;當(dāng)x=y=0時(shí),sin(xy)=sinxsiny,所以B為真命題;因?yàn)?-cos2x2=1-(1-2sin2x)2=|sinx|=sinx,x∈[0,π],所以C為真命題;當(dāng)x=π2,y=12.1因?yàn)閟inx+π3=33,所以cosx+cosxπ3=cosx+12cosx+32sin=32cosx+32sinx=332cosx+1=3sinx+π3=1.13.1575因?yàn)棣取?π,2π所以sinθ2cosθ2=1+cosθ2所以sinθ2+cosθ2=1514.解∵0<α<π2∴cosα=1-∵0<α<π2,0<β<π∴0<α+β<π,若0<α+β<π2∵sin(α+β)<sinα,∴α+β<α,∴β<0,與已知矛盾,∴π2<α+β<π∴cos(α+β)=35∴cosβ=cos[(α+β)α]=cos(α+β)cosα+sin(α+β)sinα=35∵0<β<π2,∴0<β∴cosβ215.證明由已知,得sinA+sinB=sinC,①cosA+cosB=cosC.②和差化積,得2sinA+B2cosA-B2=2cosA+B2cosA-B2=∵當(dāng)cosA-B2=0時(shí),sinC=cosC=∴cosA-B2③÷④,得tanA+B2=tan∴cos(A+B)=1-tan