sin29π/6+cos(-29π/3)+tan(-25π/4) =sin(24π/6+5π/6)+cos(-30π/3+π/3)+tan(-24π/4-π/4) =sin5π/6+cosπ/3-tanπ/4
sin29π/6+cos(-29π/3)+tan(-25/4)
1个回答
相关问题
-
sin[25π/6]+cos[25π/3]+tan(-[25π/4])=______.
-
sin[25π/6]+cos[25π/3]+tan(-[25π/4])=______.
-
求值:2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/
-
计算:cox(π/3)-tan(π/4)+(3/4)tan(π/6)-sin(π/6)+cos(π/6)...
-
cos(-29兀/3)+tan(-25兀/4)+sin29兀/6的化简过程?
-
sin4π/3×cos25π/6×tan5π/4的值是
-
sin-3π/4= cos11π/6= tan13π/4= sin7π/2= cos-3π/2= sin10π/3=
-
sin(-21π/4) cos(-20π/3) tan(-19π/6)=
-
求值cos9π/4+tan(-11π/6)-sin9π/4+tan(--11π/3)
-
化简a²sin3/π+b²cosπ/6-absinπ/4cosπ/4tanπ/3+abcosπ/6= 过 程