※ 引述《coolbetter33 (香港3345678)》之銘言:
: 1 1 1
: ex: ----------- + ----------- + ... + ------------- = ?
: cos0'cos1' cos1'cos2' cos88'cos89'
由sin0/cos0 - sin1/cos1
=(sin0cos1-cos0sin1)/cos0cos1 = -sin1/cos0cos1
因此原式= [1/(-sin1)] * [sin0/cos0-sin1/cos1+sin1/cos1-sin2/cos2+...]
=[1/(-sin1)]*[sin0/cos0 - sin89/cos89]
=1/(-sin1) * (-cos1/sin1) = cos1/(sin1)^2
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