Question

simplify 1-sin^2x/sinxcsc x

Answer 1

IDENTITIES!!! did this last week in calculus.

Answer 2

does it help to know that \(1-\sin^2(x)=\cos^2(x)\) ?

Answer 3

also need to know that cosecant is the reciprocal of sine
so \(\sin(x)\csc(x)=1\)

Answer 4

I typed the equation wrong, Its 1-sin^2x/ sinx -cscx

Answer 5

lol ok lets start again

Answer 6

the numerator is still \(\cos^2(x)\)

Answer 7

the denominator is
\[\sin(x)-\frac{1}{\sin(x)}=\frac{\sin^2(x)-1}{\sin(x)}\]

Answer 8

would you multiply the numerator by the reciprocal?

Answer 9

you good from there, or you want to walk through the algebra

Answer 10

yes

Answer 11

and when you do that, you will have a term that looks like
\[\frac{1-\sin^2(x)}{\sin^2(x)-1}\] which is just \(-1\)

Answer 12

clear or no?

Answer 13

I know the answer is supposed to be -sinx, but I got cos^2 x

Answer 14

i misled you about changing \(1-\sin^2(x)\) to \(\cos^2(x)\)
i mean it is true but not necessary

Answer 15

\[\frac{(1-\sin^2(x))\times \sin(x)}{\sin^2(x)-1}=-\sin(x)\] because \[\frac{1-\sin^2(x)}{\sin^2(x)-1}=-1\]

Answer 16

ok with that?

Answer 17

I started over, and I'm stuck on sin^2x-1 * 1-sin^2x

Answer 18

ahh because you have to multiply the the reciprocal!

Answer 19

\[\frac{1-\sin^2(x)}{\frac{\sin^2(x)-1}{\sin(x)}}=1-\sin^2(x)\times \frac{\sin(x)}{\sin^2(x)-1}\]

Answer 20

you gotta flip the bottom one
that is why it is
\[\frac{(1-\sin^2(x))\sin(x)}{\sin^2(x)-1}\]

Answer 21

This is where I get really confused, I have \[1-\sin ^{2}x * \frac{ sinx }{ \sin ^{2}x-1 }\]

Answer 22

ok good

Answer 23

wouldn't I need to multiply, so i have a common denominator? \[\frac{ 1-\sin ^{2}x }{ 1-\sin ^{2}x } * \frac{ 1-\sin ^{2} x}{ 1 }\]

Answer 24

multiply means multiply in the numerator, not in the denominator
i.e. it is
\[\frac{1-\sin^2(x)}{1}\times \frac{\sin(x)}{\sin^2(x)-1}\] you don't find a common denominator when you multiply fractions, that is for addition

Answer 25

suppose you had numbers, like
\[\frac{3}{\frac{2}{5}}\]
that is the same as
\[3\times \frac{5}{2}=\frac{15}{2}\]

Answer 26

Ohh! Thats where I was going wrong. So it would be
\[\frac{ 1-\sin ^{3}x }{ \sin ^{2}x-1 }\]

Answer 27

you missed the parentheses when you multiplied
it is
\[\frac{(1-\sin^2(x))\sin(x)}{\sin^2(x)-1}\]

Answer 28

don't multiply by \(\sin(x)\)
but rather cancel the top and the bottom as \(-1\)

Answer 29

is this step okay?\[\frac{1-\sin^2(x)}{1}\times \frac{\sin(x)}{\sin^2(x)-1}\]

Answer 30

Yes, that's what I have. I don't understand how it is -1 though. The answer is supposed to be -sinx

Answer 31

ok lets go slow
we got that part, now lets make it look like this
\[\frac{1-\sin^2(x)}{\sin^2(x)-1}\times \frac{\sin(x)}{1}\] is that okay? i just rearranged it a bit

Answer 32

yes

Answer 33

ahh damn typo sorry,i meant the first part is
\[\frac{1-\sin^2(x)}{\sin^2(x)-1}\]

Answer 34

and that is always \(-1\) just like
\[\frac{1-5}{5-1}=-1\] no matter what, that is \(-1\)

Answer 35

so
\[\frac{1-\sin^2(x)}{\sin^2(x)-1}\times \sin(x)=-1\times \sin(x)=-\sin(x)\]

Answer 36

Okay, I got it! Thank you so much!

Answer 37

whew
yw
now i have a question
what is "46"?

Answer 38

It was my entry number for competition

Answer 39

hope you won

Answer 40

I did, thanks.