第2课时两角和与差的正弦、余弦、正切公式课后篇巩固提升基础巩固1.sin35°cos5°-cos35°sin5°=()A.12B.1C.2D.2sin40°解析sin35°cos5°-cos35°sin5°=sin(35°-5°)=sin30°=12.答案A2.若sin(π6-α)=cos(π6+α),则tanα=()A.-1B.0C.12D.1解析由已知得12cosα-❑√32sinα=❑√32cosα-12sinα,因此1-❑√32sinα=❑√3-12cosα,于是tanα=-1.答案A3.若tan(α+β)=25,tan(α-β)=14,则tan2α=()A.16B.2213C.322D.1318解析tan2α=tan[(α+β)+(α-β)]=tan(α+β)+tan(α-β)1-tan(α+β)tan(α-β)=25+141-25×14=1318.答案D4.sin(θ+75°)+cos(θ+45°)-❑√3cos(θ+15°)的值等于()A.±1B.1C.-1D.0解析原式=sin[(θ+45°)+30°]+cos(θ+45°)-❑√3cos[(θ+45°)-30°]=❑√32sin(θ+45°)+12cos(θ+45°)+cos(θ+45°)-❑√3[❑√32cos(θ+45°)+12sin(θ+45°)]=❑√32sin(θ+45°)+32cos(θ+45°)-32cos(θ+45°)-❑√32sin(θ+45°)=0.答案D5.设α∈(0,π2),β∈(0,π2),且tanα=1+sinβcosβ,则()A.3α-β=π2B.3α+β=π2C.2α-β=π2D.2α+β=π2解析由tanα=1+sinβcosβ,得sinαcosα=1+sinβcosβ,得sinαcosβ-cosαsinβ=cosα,sin(α-β)=sin(π2-α).又α∈(0,π2),β∈(0,π2),故α-β=π2-α,即2α-β=π2.答案C6.化简:sin(α-150°)+cos(α-120°)cosα=.解析原式=sinαcos150°-cosαsin150°+cosαcos120°+sinαsin120°cosα=-❑√32sinα-12cosα-12cosα+❑√32sinαcosα=-1.答案-17.(一题多空题)已知锐角α,β满足(tanα-1)(tanβ-1)=2,则tan(α+β)=,α+β=.解析因为(tanα-1)(tanβ-1)=2,所以tanα+tanβ=tanαtanβ-1.因此tan(α+β)=tanα+tanβ1-tanαtanβ=-1,因为α+β∈(0,π),所以α+β=3π4.答案-13π48.已知α∈(0,π2),tanα=2,则cos(α-π4)=.解析由tanα=2,得sinα=2cosα.又sin2α+cos2α=1,α∈(0,π2),∴cosα=❑√55,sinα=2❑√55.∴cos(α-π4)=cosαcosπ4+sinαsinπ4=❑√55×❑√22+2❑√55×❑√22=3❑√1010.答案3❑√10109.tan23°+tan37°+❑√3tan23°tan37°的值是.解析∵tan60°=❑√3=tan23°+tan37°1-tan23°tan37°,∴tan23°+tan37°=❑√3−❑√3tan23°tan37°,∴tan23°+tan37°+❑√3tan23°tan37°=❑√3.答案❑√310.化简求值:(1)sin(α+β)cos(α-β)+cos(α+β)sin(α-β);(2)cos(70°+α)sin(170°-α)-sin(70°+α)cos(10°+α);(3)cos21°·cos24°+sin159°·sin204°.解(1)原式=sin(α+β+α-β)=sin2α.(2)原式=cos(70°+α)sin(10°+α)-sin(70°+α)cos(10°+α)=sin[(10°+α)-(70°+α)]=sin(-60°)=-❑√32.(3)原式=cos21°cos24°+sin(180°-21°)sin(180°+24°)=cos21°cos24°-sin21°sin24°=cos(21°+24°)=cos45°=❑√22.11.已知α,β均为锐角,且tanβ=cosα-sinαcosα+sinα,求tan(α+β)的值.解tanβ=cosα-sinαcosα+sinα=1-tanα1+tanα=tan(π4-α),因为α,β均为锐角,所以-π4<π4-α<π4,0<β<π2,又y=tanx在(-π2,π2)上是单调函数,所以β=π4-α,即α+β=π4,tan(α+β)=1.能力提升1.已知α∈(π2,3π2),tan(α-π4)=-3,则sinα=()A.❑√55B.-❑√55C.2❑√55D.±❑√55解析tanα=tan[(α-π4)+π4]=tan(α-π4)+tanπ41-tan(α-π4)tanπ4=-12,因为α∈(π2,3π2),所以α∈(π2,π),故sinα=1❑√5=❑√55.答案A2.设α,β都为锐角,且cosα=❑√55,sin(α+β)=35,则sinβ等于()A.2❑√525B.11❑√525C.❑√55D.-❑√55或11❑√525解析∵α为锐角,cosα=❑√55,∴sinα=2❑√55.∵α,β都为锐角,∴0<α+β<π.∵sin(α+β)=35,∴cos(α+β)=±45.当cos(α+β)=-45时,sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα=35×❑√55+45×2❑√55=11❑√525;当cos(α+β)=45时,sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα=35×❑√55−45×2❑√55=-❑√55,与已知β为锐角矛盾.∴sinβ=11❑√525.答案B3.已知cos(α+β)=45,cos(α-β)=-45,则cosαcosβ=.解析由已知得cosαcosβ-sinαsinβ=45,cosαcosβ+sinαsinβ=-45,两式相加得2cosαcosβ=0,故cosαcosβ=0.答案04.已知△ABC中,❑√3tanAtanB-tanA-tanB=❑√3,则C的大小为.解析依题意,tanA+tanB1-tanAtanB=-❑√3,即tan(A+B)=-❑√3,又0
