sin kπ/4(k从1到n)的和为什么等于{cos[π/8]-cos[(n+1/2)π/4]}/2sin(π/8)?
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sin kπ/4(k从1到n)的和为什么等于{cos[π/8]-cos[(n+1/2)π/4]}/2sin(π/8)?
sin(nπ/4)=[2sin(nπ/4)sin(π/8)]/[2sin(π/8)],
然后对[2sin(nπ/4)sin(π/8)]用积化和差,
2sin(nπ/4)sin(π/8)=cos(nπ/4-π/8)-cos(nπ/4+π/8)=cos[(n-1)π/4+π/8]-cos(nπ/4+π/8)
所以sin(nπ/4)=[cos[(n-1)π/4+π/8]-cos(nπ/4+π/8)]/[2sin(π/8)],
然后∑sin kπ/4=[cosπ/8-cos(π/4+π/8)+cos(π/4+π/8)-cos(2*π/4+π/8)+.
+cos[(n-1)π/4+π/8]-cos(nπ/4+π/8)]/[2sin(π/8)]
={cos[π/8]-cos(nπ/4+π/8)]}/2sin(π/8)
={cos[π/8]-cos[(n+1/2)π/4]}/2sin(π/8)
然后对[2sin(nπ/4)sin(π/8)]用积化和差,
2sin(nπ/4)sin(π/8)=cos(nπ/4-π/8)-cos(nπ/4+π/8)=cos[(n-1)π/4+π/8]-cos(nπ/4+π/8)
所以sin(nπ/4)=[cos[(n-1)π/4+π/8]-cos(nπ/4+π/8)]/[2sin(π/8)],
然后∑sin kπ/4=[cosπ/8-cos(π/4+π/8)+cos(π/4+π/8)-cos(2*π/4+π/8)+.
+cos[(n-1)π/4+π/8]-cos(nπ/4+π/8)]/[2sin(π/8)]
={cos[π/8]-cos(nπ/4+π/8)]}/2sin(π/8)
={cos[π/8]-cos[(n+1/2)π/4]}/2sin(π/8)
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