已知cos(π/4+x)=4/5,x∈(-π/2,-π/4).求(sin2x-2sin²x)/(1+tanx)
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已知cos(π/4+x)=4/5,x∈(-π/2,-π/4).求(sin2x-2sin²x)/(1+tanx)
cos(π/4+X)=4/5,X∈(-π/2,-π/4),则π/4+X∈(-π/4,0),2X∈(-π,-π/2)
sin(π/4+X)=-3/5,
sin(π/2+2X)=2 sin(π/4+X)cos(π/4+X)=-24/25
cos2X =sin(π/2+2X)=sin(π/2-2X)=-24/25,
sin2X=-7/25
(sin2X-2sin²X)/(1+tanX)
=(2sinXcosX-2sin²X)/(1+sinX/cosX)
=2sinXcosX(cosX-sinX)/(cosX+sinX)
=sin2X(1-sin2X)/cos2X
=(-7/25)×(1+7/25)/(-24/25)
=28/75
sin(π/4+X)=-3/5,
sin(π/2+2X)=2 sin(π/4+X)cos(π/4+X)=-24/25
cos2X =sin(π/2+2X)=sin(π/2-2X)=-24/25,
sin2X=-7/25
(sin2X-2sin²X)/(1+tanX)
=(2sinXcosX-2sin²X)/(1+sinX/cosX)
=2sinXcosX(cosX-sinX)/(cosX+sinX)
=sin2X(1-sin2X)/cos2X
=(-7/25)×(1+7/25)/(-24/25)
=28/75
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