2024-2025學(xué)年高中數(shù)學(xué)第五章三角函數(shù)5.5.1兩角和與差的正弦余弦和正切公式第3課時(shí)二倍角的正弦余弦正切公式課后篇鞏固提升含解析新人教A版必修1

PAGEPAGE1第3課時(shí)二倍角的正弦、余弦、正切公式課后篇鞏固提升基礎(chǔ)鞏固1.cosπ12-sinA.-32 B.-12 C.1解析原式=cos2π12-sin2π12=cos故選D.答案D2.若tanα=3,則sin2αcosA.2 B.3 C.4 D.6解析sin2αcos2α=2sinαcos答案D3.已知sinπ4-x=3A.1925 B.1625 C.14解析cosπ2-2=1-2sin2π4-x=1-2答案D4.若α為銳角,3sinα=tanα=2tanβ,則tan2β等于()A.34 B.43 C.-34解析因?yàn)棣翞殇J角,3sinα=tanα,所以cosα=13,則tanα=22,即tanβ=2,所以tan2β=2tanβ1答案D5.若sinα+cosαsinαA.-34 B.34 C.-4解析等式sinα+cosαsinα-cosα=12∴tan2α=2tanα答案B6.已知α∈π2,π,sinα=55解析由α∈π2,π,sinα=55,得cosα=-255,tanα=sinαcos答案-47.化簡:2sin2α1+cos2α解析原式=2sin2α2cos答案tan2α8.若cos(75°+α)=13,則sin(60°+2α)=解析依題意,cos(75°+α)=13,則cos(150°+2α)=2cos2(α+75°)-1=2×132-1=-79,sin(60°+2α)=-cos(90°+60°+2α)=-cos(150°+2α答案79.求下列各式的值:(1)2cos(2)23tan15°+tan215°;(3)sin10°sin30°sin50°sin70°.解(1)原式=cos2=cos2=cos2=cos2αcos2α(2)原式=3tan30°(1-tan215°)+tan215°=3×33(1-tan215°)+tan215°(3)(方法一)sin10°sin30°sin50°sin70°=12cos20°cos40°cos80°=2sin20(方法二)令x=sin10°sin50°sin70°,y=cos10°cos50°cos70°.則xy=sin10°cos10°sin50°cos50°sin70°cos70°=12sin20°·12sin100°·=18=18cos10°cos50°cos70°=1∵y≠0,∴x=18從而有sin10°sin30°sin50°sin70°=11610.已知sinα+cosα=355,α∈0,π4,sinβ(1)求sin2α和tan2α的值;(2)求cos(α+2β)的值.解(1)由題意得(sinα+cosα)2=95,即1+sin2α=95,∴sin2α=45,又易知2α∴cos2α=1-sin22α(2)∵β∈π4,π2,β-∴cosβ-∴sin2β-π4=2sinβ又sin2β-π4=-cos2β,∴cos2又易知2β∈π2,π,∴sin2又cos2α=1+cos2α2=45,∴cosα=25∴cos(α+2β)=cosαcos2β-sinαsin2β=255×實(shí)力提升1.4sin80°-cos10°sin10°=A.3 B.-3 C.2 D.22-3解析4sin80°-cos10=2sin20=2(sin30°答案B2.若α∈0,π2,且cos2α+cosπ2+2αA.12 B.14 C.13 D.1解析cos2α+cosπ2+2α=cos2α-sin2α=cos2α-2sinαcosα=cos2α-2sinαcosαsin2α+cos2α=1-2tan答案C3.設(shè)sin2α=-sinα,α∈π2,π,則tan2α解析∵sin2α=2sinαcosα=-sinα,∴cosα=-12又α∈π2,π,∴sinα=32∴tan2α=2tanα答案34.已知角α,β為銳角,且1-cos2α=sinαcosα,tan(β-α)=13,則β=解析由1-cos2α=sinαcosα,得1-(1-2sin2α)=sinαcosα,即2sin2α=sinαcosα.∵α為銳角,∴sinα≠0,∴2sinα=cos