:(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)(2)sin(25π/6)+cos(25π/3)+tan(-27π/4)

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:(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)(2)sin(25π/6)+cos(2

:(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)(2)sin(25π/6)+cos(25π/3)+tan(-27π/4)
:(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)
(2)sin(25π/6)+cos(25π/3)+tan(-27π/4)

:(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)(2)sin(25π/6)+cos(25π/3)+tan(-27π/4)
答案:1. 原式=sin(-a)(-cosa)(-sina)sina/(-cosa)sinasinacosa
=(-sin^3a)cosa/(-sin^2acos^2a)
=sina/cosa
=tana
2. 原式=sinπ/6+cosπ/3+tan(-3π/4)
=1/2+1/2+tanπ/4
=1+1=2

1、=-sina*-cosa*-sina*sina/-cosa*sina*sina*cosa=tana
2、=sin(pi/6)+cos(pi/3)+tan(5pi/4)=0.5+0.5+1=2

-1和2

(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)
=-sinα【-cosα】【-sinα】sinα/【-cosα】sinα【sinα】cosα
=sinα/osα
=tanα
(2)sin(25π/6)+cos(25π/3)+tan(-27π/4)

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(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)
=-sinα【-cosα】【-sinα】sinα/【-cosα】sinα【sinα】cosα
=sinα/osα
=tanα
(2)sin(25π/6)+cos(25π/3)+tan(-27π/4)
=sin(π/6)+cos(π/3)+tan(-3π/4)
=1/2+1/2-1
=0

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(1)sin(2π-α)=sin(-α)=-sinα
cos(π+α)=-cosα
cos(π/2+α)=sin(-α)=-sinα
cos(13/2π-α)=cos(π/2-α)=sinα
cos(π-α)=-cosα
sin(5π-α)=sin(π-α)=...

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(1)sin(2π-α)=sin(-α)=-sinα
cos(π+α)=-cosα
cos(π/2+α)=sin(-α)=-sinα
cos(13/2π-α)=cos(π/2-α)=sinα
cos(π-α)=-cosα
sin(5π-α)=sin(π-α)=sinα
sin(-π-α)=sinα
sin(9/2π+α)=sin(π/2+α)=cos(-α)=cosα
那么,原式=tanα
(2)sin(25/6π)=sin(π/6)=1/2
cos(25/3π)=cos(π/3)=1/2
tan(-27/4π)=tan(-π/4)=-1
那么,原式=0

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(1) tan α
(2) 2

(1)原式=(-sin α cos α sin α sin α)/(-cos α sin α sin α cos α)=tan α
(2)原式=sin (兀/6)+cos(兀/3)+tan(兀/4)=1/2+1/2+1=2

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