Question

HELP! how do you prove sin^-1 = cscx ?

Answer 1

\[\sin ^{-1} x \ne \csc x\]
however if you meant \[(\sin x)^{-1} = \csc x\]
then that's right

Answer 2

\[(\sin x)^{-1} \implies \frac{1}{\sin x} \implies \csc x\]
does that help?

Answer 3

\[\frac{1}{\sin}=cscx\]

Answer 4

@best.shakir what just happened? where did 1-cos x come from?

Answer 5

WAIT if csc^-1 = sin^-1 (1/x) AM i correct?

Answer 6

sorry

Answer 7

...that doesnt sound right

Answer 8

@lsugano

Answer 9

having the power of -1 the number will come as reciprocal &reminding the property that \[1/\sin=\csc\]

Answer 10

\[(\csc x)^{-1} \implies \sin x \ne \sin (\frac 1x)\]

Answer 11

@lgbasallote
are they equal:

\[\large \sin^2(x) = (\sin(x))^2\]

Answer 12

no im talking about identities for the inverse functions! so its not an exponent!

Answer 13

@waterineyes yes

Answer 14

i know @lsugano

Answer 15

do you mean arc cosecant?

Answer 16

Are they equal ??

\[\large \sin^{-1}x = (\sin(x))^{-1}\]

Answer 17

no

Answer 18

How ??

Answer 19

\[\sin(x)=\frac{
\text{opposite}}{\text{hypotenuse}}\]

\[\csc(x)=\frac{\text{hypotenuse}}{
\text{opposite}}\]

Answer 20

\[\sin^{-1} x\] that is arcsin
\[(\sin x)^{-1} \implies \frac{1}{\sin x} \implies \csc x\]

Answer 21

u can prove it as since sinx=p/h\[\color{red}{{1\over \frac{p}{h}}={h \over p}}\]

Answer 22

h/p=csc

Answer 23

anyway..i have to go now...good luck on all of you because you'll need it considering how vague the question is ;)

Answer 24

You have just replaced 2 by -1 and you are saying no ??

Answer 25

nonono i havent reached arc cosecant yet. the book said that if x≤-1 or x≥1 then csc^-1x= sin^-1 (1/x) i want to know the opposte!!

Answer 26

im so lost....

Answer 27

first of all u have to know that sin=p/h

Answer 28

sin^-1=1/sin

Answer 29

is it clear to u @Isugano ??????????/

Answer 30

yes okay

Answer 31

1/p/h=