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

##### 1 Answer

#### Explanation:

If

However, inspired by Euler's formula

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

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

With these definitions, note that

Suppose

Let

#(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

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)#

On the other hand,

Consider a right angled triangle with sides

Let

Then

and

So:

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