1.化简：⑴【Cos(x—π) tan(x—2π) tan(2π—x)】/【sin(π+x)】

题目

1.化简：⑴【Cos(x—π) tan(x—2π) tan(2π—x)】/【sin(π+x)】
⑵((sin^2 )(—x))—tan(360°—x) *tan(—x)—sin(180°—x) cos(360°—x) tan(180°+x)
2.求三角函数值：
⑴sin(—（43π）/6)
⑵cos(—(83π)/6)
⑶tan(—（41π）/3)
⑷cos((11π)/4)
⑸sin930°

答案

1(1)⑴[Cos(x—π) tan(x—2π) tan(2π—x)]/[sin(π+x)]
=cos(π-x)tanx(-tanx)/(-sinx)
=-cosxtanx(-tanx)/(-sinx)
=-sinxtanx/sinx=-tanx
(2)((sin^2 )(—x))—tan(360°—x) *tan(—x)—sin(180°—x) cos(360°—x) tan(180°+x)
=sin²x -tan²x-sinxcosxtnax
=sin²x -tan²x-sin²x=-tan²x
2.(1)sin43π/6=sin(6π+7π/6)=sin7/6π=sin(π+π/6)=-sinπ/6=-1/2
(2)cos(-83π/6)=cos(-14π+π/6)=cosπ/6=√3/2
(3)tan(-41π/3)=tan(-14π+π/3)=tanπ/3=√3
(4)cos11π/4=cos(2π+3π/4)=cos3π/4=cos(π-π/4)=-cosπ/4=-√2/2
(5)sin930°=sin(720°+210°)=sin210°=sin(180°+30°)=-sin30°=-1/2