设角=35/6兀,则2sin(兀+a)cos(兀-a)-cos(兀+a)/1+sin^2a+sin(兀-a)-cos^2

flora_nn2022-10-04 11:39:542条回答

设角=35/6兀,则2sin(兀+a)cos(兀-a)-cos(兀+a)/1+sin^2a+sin(兀-a)-cos^2(兀+a)值=

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何在在21 共回答了18个问题 | 采纳率72.2%
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
lovediving 共回答了430个问题 | 采纳率
a=35π/6 ->a=-π/6
(2sin(π+a)cos(π-a)-cos(π+a))/(1+sin^2a+sin(π-a)-cos^2(π+a))
=(2sinacosa+cosa)/(1+sin^2a+sina-cos^2a)
=((2sina+1)cosa)/(2sin^2a+sina)
=((2sina+1)cosa)/((2sina+1)sina)
=cota=cot(-π/6)=-√3
1年前

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