小猪快跑520
幼苗
共回答了1788个问题 举报
∵∴1/2*ln[(1-cosx)/(1+cosx)]
=1/2*ln[2sin^2(x/2)]/2cos^2(x/2)]
=1/2*ln tan^2(x/2)
=ln tan(x/2).............(1)
ln(1-cosx)/sinx
=ln(1-(1-2sin²x/2)/[2sinx/2cosx/2)
=lnsinx/2/cosx/2
=lntanx/2
∴(1/2)ln|(1-cosx)/(1+cosx)|=ln|(1-cosx)/sinx|
1年前
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