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

##### 2 Answers

This expression cannot be calculated. See explanation.

#### Explanation:

The trigonimetric functions

This is undefined on two distinct counts:

(1) If dealing with Real valued functions, then

(2)

#### Explanation:

As a Real valued function of Reals, the range of the function

But...

Note that

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 = (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

Here we still get an undefined value since:

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