计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπtan10°t

来源：学生作业帮助网编辑：作业帮时间：2024/05/05 18:05:47

计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπtan10°t
答：
1)
cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6
=1/2-1+(3/4)*(√3/3)²-1/2+(√3/2)²
=-1+(3/4)*(1/3)+3/4
=-1+1/4+3/4
=0
2)
[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）
=(tan30°cos210°cos60°) / [cot(-60°)sin(-30°)]
=( -tan30°cos30°cos60°) / (cot60°sin30°)
=-sin30°cos60°tan60°/sin30°
=-sin60°
=-√3/2
3）
cos25/6π+cos25/3π+tan(-25/4π)
=cos(π/6)+cos(π/3)-tan(π/4)
=√3/2+1/2-1
=(√3-1)/2
4）
5sinπ/2+2cos0-3sin3/2π+10cosπ
=5*1+2-3*(-1)+10*(-1)
=5+2+3-10
=0
5）
tan10°tan20°tan30°tan45°tan60°tan70°tan80°
=tan10°tan20°tan30°tan45°ctan30°ctan20°ctan10°
=(tan10°ctan10°)*(tan20°ctan20°)*(tan30°ctan30°)tan45°
=1×1×1×1
=1

全部展开

cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6=1/2-1+3/4x1/3-1/2+3/4=0
[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）
=[-(√3/3)(-√3/2)(1/2)]/[(-√3/3)(1/2)]=-√3/2
cos25/6π+cos25/3π+tan(-25/4π)
=cos(4π+π/6)+cos(8π+π/3)+tan(-6π-π/4)=√3/2+1/2-1=(√3-1)/2
5sinπ/2+2cos0-3sin3/2π+10cosπ=5+2+3-10=0
tan10°tan20°tan30°tan45°tan60°tan70°tan80°
=tan10°cot10° tan20°cot20° tan30°cot30°
=1

收起

计算（1）cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπt 计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπtan10°t cosπ/3－tanπ/4+cosπ－sin3π/2 化简计算计算sin10π-√2cos(-19π/4)-tan(-13π/3)=? 计算sin10π-√2cos(-19π/4)-tan(-13π/3)= tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1求cosa*|tana| 计算2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2同上写出过程 Sinπ/3Tanπ/3+tanπ/6COSπ/6－Tanπ/4COSπ/2 2cos π/2-tan π/4+3/4tan的平方 π/6-sin π/6+cos的平方 π/6 +sin 3π/2 求计算过程及...2cos π/2-tan π/4+3/4tan的平方 π/6-sin π/6+cos的平方 π/6 +sin 3π/2 ①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)②计算(1)tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5) (2)sin(-60°)+cos(225°)+tan135°(3)cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)（4）tan10°+tan170°+sin1866°-sin(-606°) sin29π/6+cos(-29π/3)+tan(-25/4) 计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)], cosπ|3-tanπ|4+3|4tan²π|6+cos²π|6-sinπ|4 cosπ/3+tanπ/4+3tan²π/6+sinπ/2+cosπ+sin3π/2等于?. 化简:(sin(α-3π)*cos(-α))/(tan(α-5π)*cos^2(α-5π)*tan(4π-α)) 已知(1+tanα)(1+tanβ)=4cos（π/3）0 高一必修4三角函数sin420*cos750+sin(-330)*cos(-660)tan675+tan765-tan(-330)+tan(-690)sin(25π/6)+cos(25π/3)+tan(-25π/4)2.已知sin(π+α)=-1/2 计算(1)sin(5π-α) (2)sin(π/2+α) (3)cos(α-3π/2) (4)tan(π/2-α) 已知tan（α+π/4）=3,计算2sinαcosα+6cos方α-3/5cos方α-6sinαcosα-5sin方α