第3课时二倍角的正弦、余弦、正切公式课后篇巩固提升基础巩固1.(cosπ12-sinπ12)(cosπ12+sinπ12)=()A.-❑√32B.-12C.12D.❑√32解析原式=cos2π12-sin2π12=cosπ6=❑√32,故选D.答案D2.若tanα=3,则sin2αcos2α的值等于()A.2B.3C.4D.6解析sin2αcos2α=2sinαcosαcos2α=2tanα=2×3=6.答案D3.已知sin(π4-x)=35,则cos(π2-2x)的值为()A.1925B.1625C.1425D.725解析cos(π2-2x)=cos[2(π4-x)]=1-2sin2(π4-x)=1-2×(35)2=725.答案D4.若α为锐角,3sinα=tanα=❑√2tanβ,则tan2β等于()A.34B.43C.-34D.-43解析因为α为锐角,3sinα=tanα,所以cosα=13,则tanα=2❑√2,即tanβ=2,所以tan2β=2tanβ1-tan2β=-43.答案D5.若sinα+cosαsinα-cosα=12,则tan2α=()A.-34B.34C.-43D.43解析等式sinα+cosαsinα-cosα=12左边分子、分母同时除以cosα(显然cosα≠0),得tanα+1tanα-1=12,解得tanα=-3,∴tan2α=2tanα1-tan2α=34.答案B6.已知α∈(π2,π),sinα=❑√55,则tan2α=.解析由α∈(π2,π),sinα=❑√55,得cosα=-2❑√55,tanα=sinαcosα=-12,tan2α=2tanα1-tan2α=-43.答案-437.化简:2sin2α1+cos2α·cos2αcos2α=.解析原式=2sin2α2cos2α·cos2αcos2α=tan2α.答案tan2α8.若cos(75°+α)=13,则sin(60°+2α)=.解析依题意,cos(75°+α)=13,则cos(150°+2α)=2cos2(α+75°)-1=2×(13)2-1=-79,sin(60°+2α)=-cos(90°+60°+2α)=-cos(150°+2α)=79.答案799.求下列各式的值:(1)2cos2α-12tan(π4-α)sin2(π4+α);(2)2❑√3tan15°+tan215°;(3)sin10°sin30°sin50°sin70°.解(1)原式=cos2α2tan(π4-α)cos2(π2-π4-α)=cos2α2tan(π4-α)cos2(π4-α)=cos2α2sin(π4-α)cos(π4-α)=cos2αsin(2×π4-2α)=cos2αcos2α=1.(2)原式=❑√3tan30°(1-tan215°)+tan215°=❑√3×❑√33(1-tan215°)+tan215°=1.(3)(方法一)sin10°sin30°sin50°sin70°=12cos20°cos40°cos80°=2sin20°cos20°cos40°cos80°4sin20°=sin40°cos40°cos80°4sin20°=sin80°cos80°8sin20°=116·sin160°sin20°=116.(方法二)令x=sin10°sin50°sin70°,y=cos10°cos50°cos70°.则xy=sin10°cos10°sin50°cos50°sin70°cos70°=12sin20°·12sin100°·12sin140°=18sin20°sin80°sin40°=18cos10°cos50°cos70°=18y.∵y≠0,∴x=18.从而有sin10°sin30°sin50°sin70°=116.10.已知sinα+cosα=3❑√55,α∈(0,π4),sin(β-π4)=35,β∈(π4,π2).(1)求sin2α和tan2α的值;(2)求cos(α+2β)的值.解(1)由题意得(sinα+cosα)2=95,即1+sin2α=95,∴sin2α=45,又易知2α∈(0,π2),∴cos2α=❑√1-sin22α=35,∴tan2α=sin2αcos2α=43.(2)∵β∈(π4,π2),β-π4∈(0,π4),sin(β-π4)=35,∴cos(β-π4)=45,∴sin2(β-π4)=2sin(β-π4)cos(β-π4)=2425.又sin2(β-π4)=-cos2β,∴cos2β=-2425.又易知2β∈(π2,π),∴sin2β=725.又cos2α=1+cos2α2=45,∴cosα=2❑√55,∴sinα=❑√55,∴cos(α+2β)=cosαcos2β-sinαsin2β=2❑√55×(-2425)−❑√55×725=-11❑√525.能力提升1.4sin80°-cos10°sin10°=()A.❑√3B.-❑√3C.❑√2D.2❑√2-3解析4sin80°-cos10°sin10°=4cos10°sin10°-cos10°sin10°=2sin20°-cos10°sin10°=2sin(30°-10°)-cos10°sin10°=2(sin30°cos10°-cos30°sin10°)-cos10°sin10°=-❑√3.答案B2.若α∈(0,π2),且cos2α+cos(π2+2α)=310,则tanα=()A.12B.14C.13D.13或-7解析cos2α+cos(π2+2α)=cos2α-sin2α=cos2α-2sinαcosα=cos2α-2sinαcosαsin2α+cos2α=1-2tanαtan2α+1=310,整理得3tan2α+20tanα-7=0,解得tanα=13或tanα=-7.又α∈(0,π2),所以tanα=13,故选C.答案C3.设sin2α=-sinα,α∈(π2,π),则tan2α的值是.解析∵sin2α=2sinαcosα=-sinα,∴cosα=-12,又α∈(π2,π),∴sinα=❑√32,tanα=-❑√3,∴tan2α=2tanα1-tan2α=-2❑√31-(-❑√3)2=❑√3.答案❑√34.已知角α,β为锐角,且1-cos2α=sinαcosα,tan(β-α)=13,则β=.解析由1-cos2α=sinαcosα,得1-(1-2sin2α)=sinαcosα,即2sin2α=sinαcosα.∵α为锐角,∴sinα≠0,∴2sinα=cosα,即tanα=12.(方法一)由tan(β-α)=tanβ-tanα1+tanβtanα=tanβ-121+12tanβ=13,得tanβ=1.∵β为锐角,∴β=π4.(方法二)tanβ=tan(β-α+α)=tan(β-α)+tanα1-tan(β-α)tanα=13+121-13×12=1.∵β为锐角,∴β=π4.答案π4
