sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?

题目

sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?
sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 = cos(-π/2) + j.sin (-π/2) = -j ,
tan (ωt) = -j 那么 tan (ωt) = -j
但是
sin (ωt) = (e j.ωt - e -j.ωt ) / 2j
cos(ωt) = (e j.ωt + e -j.ωt ) / 2
sin (ωt) / cos (ωt) = (e j.ωt - e -j.ωt )/ (e j.ωt + e -j.ωt ) .(1 / j) = (e j.ωt - e -j.ωt )/ (e j.ωt + e -j.ωt ) .(-j) ≠ -j
看这里 wiki 里面就用的这个等式 Electrical_impedance
This tells us that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is ,and that the AC voltage leads the AC current across a capacitor by -90 degrees (or the AC current leads the AC voltage across a capacitor by 90 degrees)

答案

你是不是不明白为什么tan (ωt) = -j 这个等式在数学上是不成立的,但我想你这里是要表示电路里的电流,电压的超前和滞后问题,所以在sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 这个...