Let's Redefine the assumption
Calculus
Level
pending
When we study about inverse trigonometric Functions , we assume that co-domain of arcsin and arccos is [-pi/2 , pi/2] and [0,pi] respectively.
Odd Function
Both Even And odd
Niether Even Nor Odd
Even Function
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1 solution
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Blue portion in the graph represents the output of the arcsin function discussed in the question, Red one represents the original one.(Ignore the small gap in between)
We will try to relate the arcsin function given in question and original arcsin function.
To avoid confusion, I will denote the arcsin function given in question with arcsin(p) and original one with arcsin(x).
Using graphs, it is easily observable that if we changed the sign of blue graph and shift our co-ordinate axes upwards by (pi) units, we will get red graph.
Applying the above transformations, we can write
arcsin(p)= pi - arcsin(x)
Using same concept, We can write,
arccos(p)= 2pi - arccos(x)
Now, since arcsin(x) and arccos(x) are in there original domains,we can use the result derived for them.
arcsin(p)= pi - arcsin(x) ,
this implies that arcsin(-p)= pi - arcsin(-x)
or arcsin(p)=pi+arcsin(x)
Similarily,
arccos(p) = 2pi- arccos(x)
this implies that arccos(-p) = 2pi - (pi-arccos(x))
or arccos(-p) = p+arccos(x)
Using above results, it is easy to determine that f(-x)=-f(x)