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2019-2020年高中数学第三章三角恒等变换3.2两角和与差的三角函数课后导练北师大版必修基础达标1.sin18等于( )A.cos20cos2+sin20sin2B.cos20cos2-sin20sin2C.sin20cos2+cos20sin2D.sin20cos2-cos20sin2解析:选项A为cos(20-2)=cos18;B为cos(20+2)=cos22;C为sin(20+2)=sin22;D为sin(20-2)=sin18.答案:D2.化简sincos-cossin的值是( )A. B. C.-sin D.sin解析:先用诱导公式将角转化,再逆用公式即得.原式=-sincos+cossin=sin(-)=sin=.答案:B3.满足coscos=+sinsin的一组、的值是( )A.= = B.= =C.= = D.= =解析:将原式变形:coscos-sinsin=cos(+)=,+=2k(kZ),只有A选项适合.答案:A4.计算的值等于( )A. B.- C. D.解析:将35拆成30+5,25拆成30-5展开化简.原式=-.答案:B5.的值是( )A.1 B.2 C.4 D.解析:原式=4.答案:C6.计算cos(-35)cos(25+)+sin(-35)sin(25+)的值为_.解析:根据原式可逆用两角差的余弦公式来求解.原式=cos(-35-25-)=cos(-60)=.答案:7.若tan=,则tan(+)=_.解析:tan(+)=3.答案:38.tan=,tan=,0,求+的值.解析:tan(+)=1.又0,+2,+=.9.求下面函数的值:(1)tan20+tan40+tan20tan40;(2)(3);(4)(tan10-)解析:(1)原式=tan(20+40)(1-tan20tan40)+tan20tan40=(1-tan20tan40)+tan20tan40=.(2)原式=tan15=tan(45-30)=.(3)原式=.(4)原式=(tan10-tan60)=-2.10.已知sin(+)=,sin(-)=,求的值.解析:由已知得+,得2sincos= -,得2cossin= 得=4,即=4.综合运用11.在ABC中,若cosA=,cosB=,则cosC的值是( )A. B.C. 或 D.解析:在ABC中,0A,0B0,cosB=0,得0A,0B0,0,02,求A、.解析:I2=sin(100t-)=cos(-100t+)=cos(-100t)=-cos(-+100t)=-cos(100t+),I3=I1+I2=sin(100t+)-cos(100t+)=2sin(100t+)cos(100t+)=2sin(100t+-)=2sin(100t+).A=2,=100,=.拓展探究16.如下图所示,工人师傅要把宽是4 cm和8 cm的钢板焊接成60角,下料时x应满足什么条件?思路分析:可寻找关于x的三角函数的某种关系,寻求x满足的条件.解:由题图可知CBD=60,则ABD=60-x,在ABC中,sinx=,在ABD中,sin(60-x)=,由得=2,即sin(60-x)=2sinx,cosxsinx=2sinx.sinx=cosx.tanx=.当x满足tanx=时,符合要求.
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