What does #sin(arccos(5))-3sec(arc sin(8))# equal?
2 Answers
Nothing meaningful.
Explanation:
The arguments for "arccos" and "arcsin" must be within the range
The argument of "arccos" must be a value which could be generated by the
Similarly for "arcsin".
Using definitions of Complex
#sin(arccos(5)) - 3sec(arcsin(8))=(2sqrt(6)+sqrt(7)/7)i#
Explanation:
While Alan's answer is correct for
#cos z = (e^(iz)+e^(-iz))/2#
#sin z = (e^(iz)-e^(-iz))/(2i)#
Note in passing that
With these definitions, it is possible to define
If
Hence
If
Hence
So:
#sin(arccos(5)) - 3sec(arcsin(8))=2sqrt(6)i - 3/(3 sqrt(7)i)=(2sqrt(6)+sqrt(7)/7)i#