cosπ 9*cos2π 9
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 11:48:53
同底的对数相加,结果等于真数积的对数真数之积为cosπ/9cos2π/9cos4π/9=(8sinπ/9cosπ/9cos2π/9cos4π/9)/(8sinπ/9)=4sin2π/9cos2π/9c
cos2α/[sin(α-π/4)]=-根号2/2cos2α/[sin(π/4-α)]=根号2/2sin(π/2-2α)/[sin(π/4-α)]=根号2/2利用二倍角公式,即2cos(π/4-α)=
cosπ/7-cos2π/7+cos3π/7=cosπ/7+cos3π/7+cos5π/7=(2sinπ/7*cosπ/7+2sinπ/7*cos3π/7+2sinπ/7*cos5π/7)/(2sin
原式=sinπ7(cos2π7+cos4π7+cos6π7)sinπ7=sinπ7cos2π7+sinπ7cos4π7+sinπ7cos6π7sinπ7=12(sin3π7−sinπ7)+12(sin
sin@cos@cos2@=sin2@cos2@/2=sin4@/4
cosπ/5*cos2π/5=(2sinπ/5*cosπ/5*cos2π/5)/(2sinπ/5)=(sin2π/5*cos2π/5)/(2sinπ/5)=(2sin2π/5*cos2π/5)/2*(
先把它看成分母为1的分数,(cosπ/5)(cos2π/5)/1,然后分子分母同时乘sin(π/5),这样分子上可以用一下sin的二倍角公式式子变为:sin(2π/5)*cos(2π/5)/2sin(
若cos2a/sin((a-(π/4))=-(√2)/2,则cosa+sina的值为由cos2α/sin(α-π/4)=-(√2)/2,且cos2α=cos²α-sin²αsin(
这一类题可用对称法:设sinπ/5×sin2π/5=mcosπ/5×cos2π/5=n则4mn=(2sinπ/5×cosπ/5)×(2sin2π/5×cos2π/5)=sin2π/5×sin4π/5=
sinπ=0,cosπ=-1,tanπ=0,sin2π=0,cos2π=1,tan2π=0
利用黄金分割容易得COSπ/5=(根号5+1)/4COS2π/5=(根号5-1)/4乘起来是1/4
cos^2(π/2-a)+cos2^2(π/6+a)=sin^2a+1/2(1+cos(π/3+2a)=1/2(1-cos(π/3+2a)+1/2(1+cos(π/3+2a)=1再问:cos^2(3/
由公式:sin(a+b)=sina*cosb+sinb*cosa可得:sin(π/5+π/5)=2sin(π/5)*cos(π/5).4sin(π/5)*cos(π/5)=2sin(2π/5);2si
sinπ=0cosπ=-1sin2π=0cos2π=1
(cos2α-sin2α)/(2sinαcosα+cos^2α)=(cos^2a-sin^2a-2sina*cosa)/(2sina*cosa+cos^2a)(分子、分母同时除于cos^2a,这样函数
sin(π/2+a)=cosa=-√5/5∴a∈(π/2,π)∴sina=2√5/5[cos²(π/4+a/2)-cos²(π/4-a/2)]/[sin(π-a)+cos(3π+a
原式=-sin2a+tanacos2a-sinacosacos2a=-2sinacosa+(sina/cosa)*(2cos²a-1)-(2sinacosacos2a)/2=-2sinaco
yge请稍候.再答:
原式=2sinπ/5cosπ/5cosπ/5/(2sinπ/5)=sin2π/5cosπ/5/(2sinπ/5)=2sin2π/5cosπ/5/(4sinπ/5)=sin4π/5/(4sinπ/5)=
倍角公式sin2a=2sinacosa所以2sinπ/5*cosπ/5=sin2π/5再问:是另一部再答:那就是上下同乘以2sinπ/5再问:自己看懂了,嘿嘿