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

来源：学生作业帮编辑：搜搜做题作业网作业帮分类：数学作业时间：2024/07/03 14:58:32

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°

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

1.化简：⑴【Cos(x—π) tan(x—2π) tan(2π—x)】/【sin(π+x)】化简：1.【（sin^2）(-X-π) *cos（π+X）cosX】/【tan(2π+X) *（cos^3 (-X-π) 三角函数已知f（x）=（2cos^4x—2cos²x+1/2）/（2tan（π/4—x）sin²（x 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π 已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)] 化简cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x) 化简2cos*2-1 / 2tan(π/4-x)sin*2（π/4+x）求证（1—tan^2X)/(1+tan^2X)=cos^2X—sin^2X f(a)=[sin(π-x)cos(2π-x)tan(-x+3π/2)]/cot(-x -π)sin(-π--x) 化简 f(x)=1/cos²x-tan²x+根号2*sin(2x-π/4) 求证 tan（2π-X）sin（-2π-X）cos（6π-X）/ sin（X+3π/2)*cos（X+3π/2)=-ta 化简[sin(180+x)-tan(-x)+tan(-360-x)]/[tan(x+180)+cos(-x)+cos(-