What does #sin(arccos(3))+csc(arc cot(4))# equal?

1 Answer
Mar 16, 2016

#sin(arccos(3))+csc(arc cot(4)) = 2sqrt(2) i + sqrt(17)#

Explanation:

If #sin#, #cos#, etc are being treated as Real valued functions of Real values, then #arccos(3)# is not defined, since #3# lies outside the range of #cos(x)# which is #[-1, 1]#.

However, inspired by Euler's formula #e^(ix) = cos x + i sin x#, it is possible to define #sin#, #cos# etc for Complex values as follows:

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

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

With these definitions, note that #sin^2 z + cos^2 z = 1# for any #z in CC#.

Suppose #cos z = a# for some #a in CC#. What is #z#?

Let #t = e^(iz)#. Then #e^(-iz) = 1/t# and we have:

#(t+1/t)/2 = cos z = a#

Hence:

#t+1/t = 2a#

Hence:

#t^2-2at+1 = 0#

Solving using the quadratic formula, we get:

#e^(iz) = t = (2a+-sqrt(4a^2-4))/2 = a+-sqrt(a^2-1)#

So:

#iz = ln(a+-sqrt(a^2-1))#

Hence:

#z = -i ln(a+-sqrt(a^2-1)) = +-i ln(a+sqrt(a^2-1))#

It is reasonable that the definition of #arccos(x)# for #x > 1# should choose the positive sign for this expression.

Hence:

#arccos(3) = i ln(3+sqrt(8)) = i ln(3+2sqrt(2))#

Then:

#sin(arccos(3))#

#= (e^(-ln(3+2sqrt(2)))-e^(ln(3+2sqrt(2))))/(2i)#

#=1/(2i)(1/(3+2sqrt(2)) - (3+2sqrt(2)))#

#=i/2((3+2sqrt(2))-(3-2sqrt(2)))#

#=color(blue)(2sqrt(2)i)#

#color(white)()#
On the other hand, #csc(arc cot(4))# is all about Real values and somewhat easier to construct.

Consider a right angled triangle with sides #1#, #4# and #sqrt(17)#

Let #theta# be the smallest angle, i.e. the one between the hypotenuse and the longer leg.

Then #cot(theta) = "adjacent"/"opposite" = 4/1 = 4#

and #csc(theta) = "hypotenuse"/"opposite" = sqrt(17)/1 = sqrt(17)#

So:

#csc(arc cot(4)) = color(blue)(sqrt(17))#