第三节两角和与差及二倍角三角函数公式题号1234567答案1.计算1-2sin222.5°的结果等于()A.B.C.D.解析:原式=cos45°=.故选B.答案:B2.设tan(α+β)=,tan=,则tan的值是()A.B.C.D.解析:tan=tan=.答案:B3.求值:=()A.-B.-C.D.答案:D4.若tanθ+=4,则sin2θ=()A.B.C.D.解析:由tanθ+=4得,+==4,即=4,∴sin2θ=.故选D.答案:D5.coscoscos=()A.B.C.D.解析:coscoscos=·2sincoscos·cos=·sincoscos=sincos=sin=sin=.故选D.答案:D6.若sin=,则cos等于()A.-B.-C.D.答案:C7.若sin=,则cos等于()A.-B.1C.-D.解析:∵sin=,∴cos=1-2sin2=1-2×=.又cos=cos=-cos[π-(-2θ)]=-cos,∴cos=-.故选C.答案:C8.函数y=的最小正周期为________.解析:y===tanx,所以最小正周期T=π.答案:π9.若角α的终边经过点P(1,-2),则tan2α的值为______.解析:∵tanα==-2,∴tan2α==.答案:10.已知tanα=2,则=________.解析:====.答案:11.若sin(π-α)=,α∈,则sin2α-cos2的值等于________.解析:∵sin(π-α)=,∴sinα=.又∵α∈,∴cosα=.∴sin2α-cos2=2sinαcosα-=2××-=.答案:12.已知向量a=(cos2x,1),b=(1,sin2x),x∈R,函数f(x)=a·b.(1)求函数f(x)的最小正周期;(2)若f=,求cos2α的值.解析:(1)f(x)=a·b=cos2x+sin2x==sin所以,f(x)的最小正周期T==π.(2)由已知得f=sin=cosα=,则cosα=,所以cos2α=2cos2α-1=2×-1=-.213.在△ABC中,已知cosA=,cos(A-B)=,且B<A.(1)求角B和sinC的值;(2)若△ABC的边AB=5,求边AC的长.解析:(1)由cosA=>0,cos(A-B)=>0,得0<A<且0<A-B<.可得sinA===,sin(A-B)===,∴cosB=cos[A-(A-B)]=cosAcos(A-B)+sinA·sin(A-B)=×+×=,∵0<B<π,且B