pwam520
幼苗
共回答了19个问题采纳率：78.9% 举报
(2 Cos[x]^2 - 1)/( 2 Tan[pi/4 - x] Sin[pi/4 + x]^2)
= (2 Cos[x]^2 - 1)/( 2 Tan[pi/4 - x] Cos[pi/4 - x]^2)
= (2 Cos[x]^2 - 1)/( 2 Sin[pi/4 - x] Cos[pi/4 - x])
= (2 Cos[x]^2 - 1)/ Sin[pi/2 - 2 x]
= (2 Cos[x]^2 - 1)/Cos[2 x]
= 1
1年前
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