sin(-5π-α）×cos(3π+α)
=sin(6π-5π-α)×cos(2π+π+α)
=sin(π-α)×cos(π+α)
=sinα×(-cosα)
=-sinαcosα
=-1/2*sin2α
=-1/2*(2tanα)/[1+(tanα)^2]
=-1/2*(2*a)/[1+a^2]
=-a/(1+a^2)这位大哥-sinαcosα=-1/2*sin2α =-1/2*(2tanα)/[1+(tanα)^2]这看不懂啊能不能再细一些用的是什么方法呢3Q