赞 fengxue1996 幼苗
共回答了18个问题采纳率：88.9% 举报
2tan(π/4-α)*sin(π/4+α)
= 2sin(π/4-α)*sin(π/4+α)/[cos(π/4 - α)]
而2sin(π/4-α)*sin(π/4+α)
= 2cos[(π/2) - (π/4-α)]·cos[(π/2) - (π/4+α)]
= 2cos[(π/4)+α]·cos[(π/4)-α]
∴原式 = (2cosα - 1)/2cos[(π/4)+α]
= 2sin(π/4-α)*sin(π/4+α)/[cos(π/4 - α)]
而2sin(π/4-α)*sin(π/4+α)
= 2cos[(π/2) - (π/4-α)]·cos[(π/2) - (π/4+α)]
= 2cos[(π/4)+α]·cos[(π/4)-α]
∴原式 = (2cosα - 1)/2cos[(π/4)+α]
1年前
1
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