化简:cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)RT、quickly~
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化简:cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)RT、quickly~
化简:cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)
RT、
quickly~
化简:cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)RT、quickly~
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
=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
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!
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