求值arctan（ cot2）,tan(1/2arccos4/5),cos[arcsin(-3/5)],arcsin(cos1/3)
问题描述：
求值arctan（ cot2）,tan(1/2arccos4/5),cos[arcsin(-3/5)],arcsin(cos1/3)
答
关键是饭三角函数的值域
arcsin[-π/2,π/2]
arccos[0,π]
arctan[-π/2,π/2]
arccot[0,π]
arctan(tan(π/2-2))
-π/2
tan(1/2arccos4/5)=1/3
x=arxsin(-3/5)
x属于[-π/2,π/2]
cos(arcsin(-3/5))
=cosx
=4/5
arcsin(cos1/3)
arcsin(sin(π/2-1/3))
=π/2-1/3