已知cos(x)=-2√5/5,x∈(-π,0)求sin2x的值,求tan(2x+π/4)

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已知cos(x)=-2√5/5,x∈(-π,0)求sin2x的值,求tan(2x+π/4)

已知cos(x)=-2√5/5,x∈(-π,0)求sin2x的值,求tan(2x+π/4)
已知cos(x)=-2√5/5,x∈(-π,0)求sin2x的值,求tan(2x+π/4)

已知cos(x)=-2√5/5,x∈(-π,0)求sin2x的值,求tan(2x+π/4)
cos(x)=-2√5/5,
x∈(-π,0)
所以
sinx=-√5/5

sin2x=2sinxcosx=4/5
cos2x=2cos²x-1=3/5
tan2x=4/3

tan(2x+π/4)
=(tan2x+tanπ/4)/(1-tan2xtanπ/4)
=(tan2x+1)/(1-tan2x)
=(4/3+1)/(1-4/3)
=-7