设角a=-35π/6°,则2sin(π+α)cos(π-α)-cos(π+α)/1+sin^2α+sin(π-α)-co

设角a=-35π/6°,则2sin(π+α)cos(π-α)-cos(π+α)/1+sin^2α+sin(π-α)-cos^2(π+α)
米菲羲捷 1年前 已收到1个回答 举报

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a=35π/6
a=6π-π/6
a=-π/6
2sin(π+a)cos(π-a)-cos(π+a)/[1+sin^2a+sin(π-a)-cos^2(π+a)]
=2sin(π-π/6)cos(π+π/6)-cos(π-π/6)/[1+sin^2(-π/6)+sin(π+π/6)-cos^2(π-π/6)]
=2sinπ/6(-cosπ/6)-(-cosπ/6)/[1+sin^2(π/6)+(-sinπ/6)-(-cosπ/6)^2]
=-2sinπ/[6cosπ/6+cosπ/6/2sin^2(π/6)-sinπ/6]
=-cosπ/6(2sinπ/6-1)/[sinπ/6(2sinπ/6-1)]
=-cosπ/6/sinπ/6
=-(√3/2)/(1/2)
=-√3

1年前

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