数学计算：
（1）arcsin{sin[(-4π)/5]}
(2)arcsin(cos4)
(3)arccos(sin4)
（1）arcsin{sin[(-4π)/5]} =-π/5
详解:-π/2≤arcsinx≤π/2,
sin[(-4π)/5]}=-sin[(4π)/5]
=-sin[(π-4π)/5]=-sin(π/5)
∴arcsin{sin[(-4π)/5]}= arcsin{-sin(π/5]}
-arcsinsin(π/5=-π/5
(2)arcsin(cos4)=4-3π/2
详解:-π/2≤arcsinx≤π/2,
π<4<3π/2,0<3π/2-4<π/2,cos4=-sin(3π/2-4)
∴arcsin(cos4)
=arcsin[-sin(3π/2-4)]
=-arcsinsin(3...全部
（1）arcsin{sin[(-4π)/5]} =-π/5
详解:-π/2≤arcsinx≤π/2,
sin[(-4π)/5]}=-sin[(4π)/5]
=-sin[(π-4π)/5]=-sin(π/5)
∴arcsin{sin[(-4π)/5]}= arcsin{-sin(π/5]}
-arcsinsin(π/5=-π/5
(2)arcsin(cos4)=4-3π/2
详解:-π/2≤arcsinx≤π/2,
π<4<3π/2,0<3π/2-4<π/2,cos4=-sin(3π/2-4)
∴arcsin(cos4)
=arcsin[-sin(3π/2-4)]
=-arcsinsin(3π/2-4)
=-(3π/2-4)
=4-3π/2
(3)arccos(sin4)=4-π/2
详解:0≤arccosx≤π,
π<4<3π/2,0<3π/2-4<π/2,sin4=-cos(3π/2-4)
∴arccos(sin4)
=arccos[-cos(3π/2-4)]
=π-arccoscos(3π/2-4)
=π-(3π/2-4)
=4-π/2
。
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