Question

计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6

cosπ/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+10cosπ
tan10°tan20°tan30°tan45°tan60°tan70°tan80°
过程要详细

Answer 1

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

Answer 2

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