Question

Simplify sine theta over square root of the quantity 1 minus sine squared theta.

Answer 1

recall 1-sin^2(x)=cos^2(x)

Answer 2

you having the problem that you need to leave to write again?

Answer 3

no

Answer 4

are you?

Answer 5

yes its annoying

Answer 6

hmm... not sure why you are having that problem

Answer 7

but do you see that you can use the identity 1-sin^2(x)=cos^2(x)

Answer 8

yes; so the sqare root of 1-sin^2(x) becomes square root of cos^2(x)

Answer 9

\[\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}=\frac{\sin(x)}{\sqrt{\cos^2(x)}}\]
now if cos(x) is positive then sqrt(cos^2(x))=cos(x)
or if cos(x) is negative then sqrt(cos^2(x))=-cos(x)

Answer 10

\[\text{ so you could say } =\frac{\sin(x)}{\pm \cos(x)} \]

Answer 11

now sin(x)/cos(x)= to what?

Answer 12

exacly because you don't know whats x

Answer 13

wait that's...

Answer 14

nope lost it. it only says to simplify and I think we did

Answer 15

sin(x)/cos(x)=tan(x)

Answer 16

I knew it was something like SOH CAH TOA

Answer 17

=tan(x) if cos(x) is positive
=-tan(x) is cos(x) is negative

Answer 18

its both though...

Answer 19

it isn't both
it is or

Answer 20

it just depends on what x is

Answer 21

cos(x) is positive in the 1st and 4th quadrant
so if x is in the 1st or 4th quadrant then you have tan(x)
cos(x) is negative in the 2nd and 3rd quadrant
so if x is in the 2nd or 3rd quadrant then you have -tan(x)

Answer 22

it still both negative and positive, but after figuring out the quadrants for sin, would it be negative? im taking a stab at it here

Answer 23

we don't do to consider where sin is pos or neg here

Answer 24

need to *

Answer 25

so then still confused

Answer 26

we had sqrt(cos^2(x))
not sqrt(sin^2(x))

Answer 27

\[\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}=\frac{\sin(x)}{\sqrt{\cos^2(x)}}=\frac{\sin(x)}{\cos(x)} \text{ if } \cos(x)>0 \\ \frac{\sin(x)}{\sqrt{1-\sin^2(x)}}=\frac{\sin(x)}{\sqrt{\cos^2(x)}}=\frac{\sin(x)}{-\cos(x)} \text{ if } \cos(x)<0\]
then we can say since sin(x)/cos(x)=tan(x) we have:


\[\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}=\frac{\sin(x)}{\sqrt{\cos^2(x)}}=\frac{\sin(x)}{\cos(x)} =\tan(x) \text{ if } \cos(x)>0 \\ \frac{\sin(x)}{\sqrt{1-\sin^2(x)}}=\frac{\sin(x)}{\sqrt{\cos^2(x)}}=\frac{\sin(x)}{-\cos(x)}=-\tan(x) \text{ if } \cos(x)<0\]
We can also do one more case...
if cos(x)=0 then our fraction we started with doesn't exist you know since we can't divide by zero

Answer 28

so you have three cases:
\[\frac{\sin(x)}{\sqrt{1-\sin^2(x)}} =\tan(x) \text{ when } \cos(x)>0 \\ \frac{\sin(x)}{\sqrt{1-\sin^2(x)}}=-\tan(x) \text{ when } \cos(x)<0 \\ \frac{\sin(x)}{\sqrt{1-\sin^2(x)}} \text{ doesn't exist when } \cos(x)=0\]

Answer 29

but we cant say which are true because we don't know cos(x). so then how do we figure it out?

Answer 30

as i said you have the three cases above

Answer 31

all three cases must be mentioned

Answer 32

you know what, I think we went to far. my answers are tan(theta), tan^2(theta),1, and -1.

Answer 33

based on your answers they assumed cos was positive

Answer 34

so then if its positive then your first case is true?

Answer 35

yes since cos is positive

Answer 36

and we know cos is pos because only pos tan was listed

Answer 37

gotta go

Answer 38

ok. thx so much

Answer 39

and you were right too.