已知cos(π/4-x)=4/5,x∈(-π/2,-π/4),求(sin2x-2sinx^2)/(1+tanx)的值

Question

已知cos(π/4-x)=4/5,x∈(-π/2,-π/4),求(sin2x-2sinx^2)/(1+tanx)的值

Answer 1

x∈(-π/2,-π/4)
2x∈(-π,-π/2)(第三象限角)
cos(π/4+x)=4/5
cos(π/2+2x)=2(cos(π/4+x))^2-1=2*(4/5)^2-1=7/25=-sin2x
即sin2x=-7/25
所以cos2x=-√[1-(-7/25)^2]=-24/25
所以[sin2x-2(sinx)^2]/(1+tanx)
=[2sinxcosx-2(sinx)^2]/(1+sinx/cosx)
=cosx*2sinx(cosx-sinx)/(cosx+sinx)
=sin2xcos2x/(1+sin2x)
=(-7/25)*(-24/25)/(1-7/25)
=28/75
可能方法有点烦
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