化简：cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)RT、quickly~
问题描述：
化简：cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)
RT、
quickly~
答
cos(π-α)sin(-π-α) / cos(11/2π-β)tan(3π+β)
=cos(π-α)sin[-(π+α)] / cos[4π+(3/2π-β)]tan[2π+(π+β)]
=-cosαsinα / (-sinβ)tanβ
=cosαsinαcosβ / (sinβ)^2
或许问题就这么简单，但是对于数学问题还是要自己算了，知道过程，下次就不知道了！
答
=cos(π-a)sin(2π-π-a)/cos(6π-π/2-β)tanβ
=cos(π-a)sin(π-a)/cos(-π/2-β)tanβ
=0.5sin(2π-2a)/cos(π/2+β)tanβ
=0.5sin(-2a)/(-sinβ)tanβ
=0.5sin2a/sinβtanβ
答
cos(π-α)sin(-π-α) / cos(11/2π-β)tan(3π+β)
=cos(π-α)sin[-(π+α)] / cos[4π+(3/2π-β)]tan[2π+(π+β)]
=-cosαsinα / (-sinβ)tanβ
=cosαsinαcosβ / (sinβ)^2