Example
1 from our text : Evaluation Inverse Trig Function
(a) Evaluate arcsin (- ½)
(1)
By definition, y = arcsin (- ½) implies that sin y = (-
½).
(2)
In the
interval [- π/2, π/2], the correct value of y is -
π/6,
(3)
arcsin(- ½) = - π/6
(b) Evaluate arccos (0)
(1)
By definition, y = arccos (0) implies that cos y = (0).
(2)
In the
interval [ 0 , π], the correct value of y is π/2,
(3)
arccos(0)= π/2
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