证明:cosπ/11-cos2π+cos3π/11-cos4π/11+cos5π/11=1/2
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证明:cosπ/11-cos2π+cos3π/11-cos4π/11+cos5π/11=1/2
cos(pi/11)-cos(2pi/11)+cos(3pi/11)-cos(4pi/11)+cos(5pi/11)
=cos(pi/11) + cos(3pi/11) + cos(5pi/11) + cos(7pi/11) + cos(9pi/11)
= [sin(5pi/11)][cos(pi/11) + cos(3pi/11) + cos(5pi/11) + cos(7pi/11) + cos(9pi/11)] / [sin(5pi/11)]
= (1/2)[sin(6pi/11) + sin(4pi/11) + sin(8pi/11) + sin(2pi/11) + sin(10pi/11) + 0 + sin(12pi/11) + sin(-2pi/11) + sin(14pi/11) + sin(-4pi/11)] / [sin(5pi/11)] (积化和差:sinAcosB = [sin(A+B) + sin(A-B)]/2)
= (1/2)[sin(6pi/11) + sin(4pi/11) + sin(8pi/11) + sin(2pi/11) + sin(10pi/11) + sin(-10pi/11) + sin(-2pi/11) + sin(-8pi/11) + sin(-4pi/11)] / [sin(5pi/11)]
= (1/2)[sin(6pi/11)] / [sin(5pi/11)]
= (1/2)[sin(5pi/11)] / [sin(5pi/11)]
= 1/2
求COSπ/5*COS2π/5
cosπ/5×cos2π/5=
COSπ/5乘以COS2π/5
cosπ/5×cos2π/5
cosπ/5cos2π/5
cosπ/5cos2π/5=?
证明:cosπ/11-cos2π+cos3π/11-cos4π/11+cos5π/11=1/2
cosπ/5 - cos2π/5 怎样变成 cosπ/5 + cos2π/5rt
分别计算:cosπ/3,cosπ/5cos2π/5,cosπ/7cos2π/7cos3π/7根据上述结果猜想一个一般的结论,并给予证明
化简cos2α+cos2(π/3+α)+cos2(π/3-α)化简cos²α+cos²(π/3+α)+cos²(π/3-α)
证明:cos2α+cos2β=2cos(α+β)cos(α-β)
sinπ.cosπ.sin2π.cos2π分别是多少
cosπ/9*cos2π/9*cos4π/9=?
求值:cosπ/7*cos2π/7*cos3π/7
cosπ/7cos2π/7cos3π/7=?
cosπ/7cos2/7πcos4/7π=
cosπ/9cos2π/9cos23π/9
(cosπ/5)(cos2π/5)的值等于