高中數(shù)學(xué)第七章三角函數(shù)7.2任意角的三角函數(shù)7.2.4誘導(dǎo)公式素養(yǎng)練含解析新人教B版必修第三冊

PAGEPAGE17.2.4誘導(dǎo)公式課后篇鞏固提升基礎(chǔ)鞏固1.sin-π3+2sin4π3+3sin2A.1 B.12C.0 D.-1答案C2.已知角α的頂點(diǎn)與原點(diǎn)O重合,始邊與x軸非負(fù)半軸重合,它的終邊過點(diǎn)P-35,-45,則sin(π-α)=()A.-45 B.4C.-35 D.解析由三角函數(shù)的定義可得sinα=-45,則sin(π-α)=sinα=-4答案A3.已知a=tan-7π6,b=cos23π4,c=sin-33π4,則aA.b>a>c B.a>b>c C.b>c>a D.a>c>b解析因?yàn)閍=-33,b=22,c=-所以b>a>c.答案A4.(多選)化簡1+2sin(π-2)A.sin2-cos2 B.|cos2-sin2|C.±(cos2-sin2) D.無法確定解析原式=|sin(π-2)+cos(π-2)|=|sin2-cos2|=sin2-cos2.答案AB5.已知tanα-π6=12,且α∈0,π2,則cos2π3-α=()A.-55 B.5C.-255 D解析由tanα-π6=12>0,且α∈0,π2可得0<α-π6<則sinα-π6=55,故cos2π3-α=cosπ2-α-π6=sinα-π6=55.答案B6.設(shè)tan(5π+α)=m,則sin(α-3解析由題意知tanα=m,原式=-sin答案m7.已知sinπ3-α=12,則解析∵π3∴cosπ6+α=答案18.化簡:(1)1+cosπ2+αsinπ2-(2)sin(解(1)原式=1+(-sinα)cosαtanα=1-sin2α=cos2α.(2)原式=(-=-=si=-sinαcosα=-tan9.已知sin(5π+α)=lg1310,求cos(2π+α)解∵sin(5π+α)=sin(π+α)=-sinα,lg1310=lg10-∴sinα=13,∴cos(2π+α)=cosα=±1-sin2α=實(shí)力提升1.已知函數(shù)f(x)=cosx2,則下列等式成立的是(A.f(2π-x)=f(x)B.f(2π+x)=f(x)C.f(-x)=f(x)D.f(-x)=-f(x)解析f(-x)=cos-x2=cosx2=f(答案C2.已知f(cosx)=sinx,設(shè)x是第一象限角,則f(sinx)為()A.secx B.cosx C.sinx D.1-sinx解析f(sinx)=fcosπ2-x=sinπ2答案B3.已知α為銳角,2tan(π-α)-3cosπ2+β+5=0,tan(π+α)+6sin(π+β)-1=0,則sinα的值是A.355 B.377 C.解析由已知可知-2tanα+3sinβ+5=0,tanα-6sinβ-1=0,所以tanα=3.又tanα=sinαcosα,所以9所以sin2α=910因?yàn)棣翞殇J角,所以sinα=310答案C4.已知sin(2π3sin2θA.1 B.2 C.3 D.6解析因?yàn)樵?sinθ·tanθ·tan所以3si=3tan2θ答案A5.已知tan(π-θ)=3,則sin(π2+θA.-1 B.-1C.1 D.1解析由tan(π-θ)=3,得-tanθ=3,即tanθ=-3,則sin(π答案D6.tan1°·tan2°·tan3°·…·tan89°=.

解析因?yàn)閠an1°·tan89°=1,所以tan1°·tan2°·…·tan89°=1×1×…×1答案17.(雙空)已知函數(shù)f(x)=cosxπ6,x≤0,log12(x+2解析由題意可得f(2)=log12(2+2)=log124=-2,則f[f(2)]=cos-π3=答案-218.已知α是第三象限的角,且f(α)=sin((1)化簡f(α);(2)若cosα-3π2=15,解(1)f(α)=sin(π-α)(2)∵cosα-3π2=-sinα,∴sinα=-15.又α是第三象限的角,∴cosα∴f(α)=259.已知tanα,1tanα是關(guān)于x的方程3x2-3kx+3k2-13=0的兩實(shí)根,且3π<α<7π2,求cos(3π+α)+sin(π+解tanα,1tanα是關(guān)于x的方程3x2-3kx+3k2-13=0的兩實(shí)根,1=tanα·1tanα=所以k2=163因?yàn)?π<α<7π2,所以tanα>0,sinα<0,cosα<又tanα+1tanα=--3k3=k,所以k>于是tanα+1tanα=sinαcosα+cosαsin所以(sinα+cosα)2=1+2sin