求值arctan( cot2),tan(1/2arccos4/5),cos[arcsin(-3/5)],arcsin(c
1个回答

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