cosπ 9*cos2π 9*cos(-23π 9)

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

原式=-sin2a+tanacos2a-sinacosacos2a=-2sinacosa+(sina/cosa)*(2cos²a-1)-(2sinacosacos2a)/2=-2sinaco

yge请稍候.再答：

α-β∈(0,π)cos(α-β)=-4/5,所以sin(α-β)=3/5α+β∈(3π/2,5π/2)sin(α+β)=-5/13,所以cos(α+β)=12/13和差化积公式：sin(α+β)+s

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