What does #sin(arccos(4))-cot(arccos(1))# equal?

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Answer 1 · 2016-09-27T21:25:09

This expression cannot be calculated. See explanation.

The trigonimetric functions #sin# and #cos# have range #<-1;1>#, so there are NO values of #x# for which #cosx=4#, so #arccos(4)# is NOT defined.

Answer 2 · 2016-09-27T21:44:33

This is undefined on two distinct counts:

(1) If dealing with Real valued functions, then #4# is not in the range of #cos(x)#, so #arccos(4)# is undefined.

(2) #cot(arccos(1))# is always undefined.

As a Real valued function of Reals, the range of the function #cos(x)# is #[-1, 1]#, so #arccos(4)# is undefined.

But...

Note that #e^(i theta) = cos theta + i sin theta#

Hence:

#cos theta = (e^(i theta) + e^(-i theta))/2#

#sin theta = (e^(i theta) - e^(-i theta))/(2i)#

This yields definitions for #cos z# and #sin z# for Complex values of #z# using the formulae:

#cos z = (e^(iz) + e^(-iz))/2#

#sin z = (e^(iz) - e^(-iz))/(2i)#

With these definitions, we find that the Pythagorean identity still holds:

#cos^2 z + sin^2 z = 1" "# for all #z in CC#

Hence:

#sin(arccos(4)) = sqrt(1-4^2) = sqrt(-15) = sqrt(15)i#

How about #cot(arccos(1))# ?

Here we still get an undefined value since:

#arccos(1) = 0# and #cot(0) = cos(0)/sin(0) = 1/0# which is undefined.