课后作业(十八)同角三角函数的基本关系与诱导公式一、选择题1.记cos(-80°)=k,那么tan100°=()A.B.-C.D.-2.等于()A.sin2-cos2B.cos2-sin2C.±(sin2-cos2)D.sin2+cos23.(2013·厦门模拟)已知α∈(-,0),sin(-α-)=则sin(-π-α)=()A.B.C.-D.-4.(2013·惠州模拟)已知tanθ=2,则sin2θ+sinθcosθ-2cos2θ=()A.-B.C.-D.5.若sinα是5x2-7x-6=0的根,则=()A.B.C.D.二、填空题6.已知sin(+α)=,则sin(-α)的值为________.7.已知tanα=2,则7sin2α+3cos2α=________.8.已知sin(x+)=,则sin(+x)+cos2(-x)=________.三、解答题9.已知函数f(x)=.(1)求函数y=f(x)的定义域;(2)设tanα=-,求f(α)的值.10.已知sin(π-α)-cos(π+α)=(<α<π).求下列各式的值:(1)sinα-cosα;(2)sin3(-α)+cos3(+α).11.已知向量a=(sinθ,cosθ),b=(2,1)满足a∥b,其中θ∈(0,).(1)求tanθ的值;(2)求的值.解析及答案一、选择题1.【解析】由cos(-80°)=k,得cos80°=k,∴sin80°=,∴tan100°=tan(180°-80°)=-tan80°=-.【答案】B2.【解析】原式=1==|sin2-cos2|,∵sin2>0,cos2<0,∴原式=sin2-cos2.【答案】A3.【解析】∵sin(-α-)=-sin(+α)=cosα=,且α∈(-,0),∴sinα=-=-=-,∴sin(-π-α)=-sin(π+α)=sinα=-.【答案】D4.【解析】sin2θ+sinθcosθ-2cos2θ====.【答案】D5.【解析】方程5x2-7x-6=0的两根为x1=-,x2=2,则sinα=-.原式==-=.【答案】B二、填空题6.【解析】sin(-α)=sin[π-(+α)]=sin(+α)=.【答案】7.【解析】7sin2α+3cos2α====.【答案】8.【解析】原式=-sin(+x)+cos2(+x)=-+(1-)=.【答案】三、解答题9.【解】(1)由cosx≠0,得x≠+kπ,k∈Z,所以函数的定义域是{x|x≠+kπ,k∈Z}.(2)∵tanα=-,∴f(α)====-1-tanα=.10.【解】由sin(π-α)-cos(π+α)=,得sinα+cosα=,两边平方,得1+2sinα·cosα=,故2sinα·cosα=-.又<α<π,∴sinα>0,cosα<0.(1)(sinα-cosα)2=1-2sinα·cosα=1-(-)=,∴sinα-cosα=.(2)sin3(-α)+cos3(+α)=cos3α-sin3α=(cosα-sinα)(cos2α+cosα·sinα+sin2α)=-×(1-)=-.11.【解】(1)∵a∥b,∴=,所以tanθ=2.(2)=2=====-4.3