Solve the following inverses

[Math_Junkie]

sin[sup:3sw9elud]-1[/sup:3sw9elud](-1/sqrt(2))

cos[sup:3sw9elud]-1[/sup:3sw9elud](cos(2))

cos(cos[sup:3sw9elud]-1[/sup:3sw9elud](2))

These are my guesses:

sin[sup:3sw9elud]-1[/sup:3sw9elud](-1/sqrt(2)) = 5pi/4 and 7pi/4

cos[sup:3sw9elud]-1[/sup:3sw9elud](cos(2)) = 2 (since the inverse undoes what cos(x) does)

cos(cos[sup:3sw9elud]-1[/sup:3sw9elud](2)) = does not exist (since cos-1(2) is an imaginary number)

 

[Deleted member 4993]

sin[sup:3lx15og1]-1[/sup:3lx15og1](-1/sqrt(2))

cos[sup:3lx15og1]-1[/sup:3lx15og1](cos(2))

cos(cos[sup:3lx15og1]-1[/sup:3lx15og1](2))

These are my guesses:

sin[sup:3lx15og1]-1[/sup:3lx15og1](-1/sqrt(2)) = 5pi/4 and 7pi/4

cos[sup:3lx15og1]-1[/sup:3lx15og1](cos(2)) = 2 (since the inverse undoes what cos(x) does)

cos(cos[sup:3lx15og1]-1[/sup:3lx15og1](2)) = does not exist (since cos-1(2) is an imaginary number)

Looks good to me...

 

[fasteddie65]

cos^-1 2 is undefined; it is NOT an imaginary number. The range of y = cos x is [-1, 1].