计算s=sin1+++sin2+++sin3++……++sin2022
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你好亲，证明：（1）2sin1(sin1+sin2+sin3+...+sinn)
=cos0-cos2+cos1-cos3+cos2-cos4+...
+cos(n-2)-cosn+cos(n-1)-cos(n+1)
=cos0+cos1-cosn-cos(n+1)
（2）=>S=sin1+sin2+sin3+...+sinn=[cos0+cos1-cosn-cos(n+1)]/2sin1
=>S=[1+cos1-(cosn+cos(n+1))]/[2sin(1/2)*cos(1/2)]
=>S=[2*cos(1/2)*cos(1/2)-2cos(n+1/2)*cos(1/2)]/[2sin(1/2)*cos(1/2)]
=>S=[cos(1/2)-cos(n+1/2)]/sin(1/2)
（3）因为-cos(n+1/2)最大为1
所以[cos(1/2)-cos(n+1/2)]/2sin(
咨询记录 · 回答于2022-06-11
计算s=sin1+++sin2+++sin3++……++sin2022
你好亲，证明：（1）2sin1(sin1+sin2+sin3+...+sinn)=cos0-cos2+cos1-cos3+cos2-cos4+...+cos(n-2)-cosn+cos(n-1)-cos(n+1)=cos0+cos1-cosn-cos(n+1)（2）=>S=sin1+sin2+sin3+...+sinn=[cos0+cos1-cosn-cos(n+1)]/2sin1=>S=[1+cos1-(cosn+cos(n+1))]/[2sin(1/2)*cos(1/2)]=>S=[2*cos(1/2)*cos(1/2)-2cos(n+1/2)*cos(1/2)]/[2sin(1/2)*cos(1/2)]=>S=[cos(1/2)-cos(n+1/2)]/sin(1/2)（3）因为-cos(n+1/2)最大为1所以[cos(1/2)-cos(n+1/2)]/2sin(
所以[cos(1/2)-cos(n+1/2)]/2sin(1/2)
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