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证明:cosπ/11-cos2π+cos3π/11-cos4π/11+cos5π/11=1/2
题目详情
证明:cosπ/11-cos2π+cos3π/11-cos4π/11+cos5π/11=1/2
▼优质解答
答案和解析
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
=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
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