Question

giving a medal!!
find the exact value of the real number y.
y=csc^-1(-1)

Answer 1

same as solving
\[\sin(x)=-1\] for \(x\)

Answer 2

what does the -1 exponent on csc mean?

Answer 3

inverse function

Answer 4

but csc is not the inverse on sin,correct?

Answer 5

\[\csc^{-1}(x)=y\iff csc(y)=x, -\frac{\pi}{2}\leq y\leq \frac{\pi}{2}\]

Answer 6

on no inverse as in inverse function, not the reciprocal!!

Answer 7

\[\csc(x)=\frac{1}{\sin(x)}\] but
\[\sin^{-1}(x)\neq \csc(x)\] arcsine is the inverse function, not the reciprocal of the sine function

Answer 8

so,what will we do after that?

Answer 9

just like if
\[f(x)=2x+1\] then
\[f^{-1}(x)=\frac{x-1}{2}\] not \(\frac{1}{2x+1}\)

Answer 10

ooooooh,u r a genius!

Answer 11

lets go slow

Answer 12

arccosecant is the inverse of cosecant
you are being asked for the arccosecant of -1
that means you want a number whose cosecant is -1

Answer 13

yes:)

Answer 14

now you have no such button on your calculator so now we have to figure out what this really means

Answer 15

since the cosecant is the reciprocal of the sine function (not the inverse, the reciprocal) if the cosecant of a number is -1, then the sine of it must also be -1

Answer 16

yup

Answer 17

so,do we look at the unit circle for the answer?

Answer 18

yes

Answer 19

is he answer 3pi/2?

Answer 20

but on the unit circle there are an infinite number of numbers for which the sine is -1
one number is what you said, so it is tempting to say the answer is \(\frac{3\pi}{2}\) but that would be wrong

Answer 21

do we want the general solutions,then?

Answer 22

your answer has to lie in the interval from \(-\frac{\pi}{2}\) up to \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) is not in that interval

Answer 23

no, not a general solution
arccosecant is supposed to be a well defined function, so we only want one number not a list of numbers

Answer 24

do we find a coterminal angle?

Answer 25

yes
but don't think too hard

Answer 26

ok,so:
3pi/2-2pi=-pi/2?

Answer 27

yes, it should be \(-\frac{\pi}{2}\)
look in your book to find the range of arccosecant
it is a strange one

Answer 28

the range of cosecant is:all real numbers?

Answer 29

what is the range, @satellite73 ?

Answer 30

hold on
not the range of cosecant, the range of arccosecant

Answer 31

oh,ok,i'm looking for it:)

Answer 32

my book doesn't have it:(

Answer 33

the range of cosecant is not all real numbers, it is \((-\infty, -1]\cup [1,\infty)\)
the range of cosecant is
\[(-\pi,-\frac{\pi}{2}]\cup (0,\frac{\pi}{2})\]

Answer 34

sorry that second one should be the range of ARCCOSECANT

Answer 35

thankyou,soooo much!

Answer 36

yw