第1課時誘導公式二、三、四課后訓練鞏固提升A組1.若sin(5π-α)=-13,則sinα的值為(A.-13 B.13 C.-22解析:∵sin(5π-α)=sin(4π+π-α)=sin(π-α)=sinα=-13,∴sinα=-1答案:A2.化簡sin2(π+α)-cos(π+α)·cos(-α)+1的值為()A.1 B.2sin2α C.0 D.2解析:原式=(-sinα)2-(-cosα)·cosα+1=sin2α+cos2α+1=2.答案:D3.已知α是三角形的一個內角,cos(π+α)=23,則tan(π-α)的值為(A.-52 B.255 C.52解析:∵cos(π+α)=23,∴cosα=-2又α是三角形的一個內角,∴sinα=53∴tan(π-α)=-tanα=-sinα答案:C4.已知sin51°=m,則cos2109°=()A.m B.-m C.1-m2 D解析:因為sin51°=m,所以cos2109°=cos(5×360°+309°)=cos309°=cos(360°-51°)=cos51°=1-答案:C5.化簡cos(-α)tan解析:cos=cosαtan(答案:16.已知n為整數,化簡sin(nπ+A.tannα B.-tannα C.tanα D.-tanα解析:當n=2k(k∈Z)時,sin=sin(2kπ當n=2k+1(k∈Z)時,sin=sin=-sinα-cos答案:C7.已知a=tan-7π6,b=cos23π4,c=sin-33π4,則a解析:∵a=-tan7π6=-tanπ6=-33,b=cos6π-π4=cosπ4=22,c=-sin33π4=-sin8答案:b>a>c8.下列三角函數:①sinnπ+4π3(n∈Z);②cos2nπ+π6(n∈Z);③sin2nπ+π3(n∈Z);④cos(2n+1)π-π6(n∈Z);⑤sin(2n+1)π-π3(n∈Z).其中與sinπ3數值相同的是.(填序號解析:①sinn②cos2nπ+π6=cos③sin2nπ+π④cos(2n+1=cosπ-π6=-cosπ6⑤sin(2n+1=sinπ-π3=因此與sinπ3數值相同的是②③⑤答案:②③⑤9.已知角α的終邊經過點P45(1)求sinα的值;(2)求cos(2解:(1)∵點P在單位圓上,∴sinα=-35(2)原式=cos=sinα由三角函數的定義,得cosα=45,故原式=510.已知f(α)=sin((1)化簡f(α);(2)若α是第三象限角,且sin(α-5π)=15,求f(α)的值(3)若α=-2220°,求f(α)的值.解:(1)f(α)=sinαcosαcos(2)∵sin(α-5π)=-sinα=15∴sinα=-15又α是第三象限角,∴cosα=-25∴f(α)=-cosα=25(3)∵-2220°=-6×360°-60°,∴f(α)=f(-2220°)=-cos(-2220°)=-cos(-6×360°-60°)=-cos60°=-12B組1.cos(2π+A.1 B.-1 C.tanα D.-tanα解析:原式=cosαtanαsin答案:C2.已知0<α<π2,sinα=45,則sin(αA.4 B.7 C.8 D.9解析:因為0<α<π2,sinα=4所以cosα=35,tanα=4所以原式=-sinα-答案:B3.若sin(180°+α)+sin(360°-α)=-a,則sin(-180°+α)+2sin(720°-α)的值為()A.-2a3 B.-C.2a3 D解析:因為sin(180°+α)+sin(360°-α)=-a,所以-sinα-sinα=-a,即sinα=a2所以原式=-sinα-2sinα=-3sinα=-32a答案:B4.cos(-585°)解析:原式=cos=cos225=cos=-cos45°sin45答案:2-25.已知f(x)=sinπx,x<0,f(解析:因為f-116==sin-2π+πf116=f56-1=f-=sin-π6-2=-12-2所以f-116+f116答案:-26.求值:sin-29π6+cos12π5tan2020π-cos解:原式=sin-4π-5π6+cos2π+2π5tan2024π=-sin5π6+cos2π5×0+cos=-sinπ6+cosπ3-1=-12+17.已知角α是其次象限角,且sinα=35(1)化簡sin(π+(2)若sinπ2-α=a,請推斷實數a的符號,計算cos13π2-α解:(1)因為α是其次象限角,所以