Question

find
\(cos^{-1}(2)\)

Answer 1

make graph of arccos(x)

Answer 2

http://www.wolframalpha.com/input/?i=arccos%282%29

Answer 3

its in comple , cant graph it

Answer 4

i need to how convert it :-\
dnt need only answer :(

Answer 5

that is like cos x =2
how would cos x exist when it lies only in range [-1,1] ??????

Answer 6

complex *

Answer 7

cosy = 2

Answer 8

Oh ok

Answer 9

e^iy + e^(-iy) = 4

Answer 10

ok then ?

Answer 11

no clue, im here to learn how to solve this from you :)

Answer 12

Is the answer 1.32i

Answer 13

ok i might continue mmm
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the funny thing is a car was going to hit me , but God it stoped at the last sec !
when the driver stoped he looked at me and said are u oki miss :O
i said no :( idk how to solve arccos 2 :'(
haha :P
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ok ill try

Answer 14

ok i got it
\(cos ^{-1} z= -i[z+i(1-z^2)^\frac{1}{2}\)]
mmm

Answer 15

Do you know
\[\huge \cos(z)=\frac{ e ^{iz} + e ^{-iz} }{ 2 }\]

Answer 16

ik it mmm but im trying to derive arccos from it

Answer 17

Multiply by \[\huge e ^{iz}\]

Answer 18

but since i got the formula :D
thx all :D

Answer 19

And solve the quadratic

Answer 20

e^iy + e^(-iy) = 4

x+1/x = 4

x = \(2 \pm \sqrt3\)

e^iy = x

y = - i log x

Answer 21

and take the log u will get the answer
as 1.32i

Answer 22

y = - i ln x

Answer 23

sry i made a typo
\(cos ^{-1} z= -i\log [z+i(1-z^2)^\frac{1}{2}]\)
\(cos ^{-1} z= -i\log [2+i(1-2^2)^\frac{1}{2}]\)
\(cos ^{-1} z= -i\log [ 2-\sqrt 3]\)

Answer 24

the hartnn answer :D
ty !