Question

Prove that (cotX-cscX)(cosX+1)= -sinX

Answer 1

I worked out the problem and I got it to sinx^2X/sinx= -sinX
I just don't know how or where the negative should be

Answer 2

u shld get a -sin^2 x

Answer 3

how do you get -sin^2x though

Answer 4

Oh ok I see where it's coming from,
You should be getting something like this, assuming you took the same approach,\[\large \frac{\cos^2x-1}{\sin x}\]Did you get something like that half way through?

Answer 5

yes!

Answer 6

(cosx-1)(cosx+1) = cos^2x - 1 = -sin^2 x

Answer 7

factor a negative out of each top term :)
You have your identity backwards heh!

Answer 8

\[\large 1-\cos^2x=\sin^2x\]

Answer 9

\[\large \cos^2x-1=-(1-\cos^2x) \quad = \quad -(\sin^2x)\]

Answer 10

do you have a program called onenote?

Answer 11

No :o

Answer 12

*Googling...* Ooo that looks neat :D

Answer 13

You still confused on this one? <:o

Answer 14

haha well i am going to attach a pic of what i have. i just dont know how i get the negative. I understand where you are coming from but I just dont know what the heck i did

Answer 15

yes

Answer 16

k

Answer 17

\[\huge \frac{\cos^2x-1}{\sin x}\quad \color{red}{\neq} \quad \frac{\sin^2x}{\sin x}\]

Answer 18

You're ok up to that point.

Answer 19

will you just do the whole problem for me and send me a pic hahah

Answer 20

Recall your identity,\[\huge \cos^2x+\sin^2x=1\]Moving cosine to the other side gives us,\[\large \huge \sin^2x \quad =\quad \color{purple}{1-\cos^2x}\]

Answer 21

See the mistake or no? :)

Answer 22

See how your 1 and cosine are in opposite spots in your step? :D

Answer 23

That's not necessary, you're only missing one tiny tiny step :D Just reeaddd!! lol

Answer 24

hold on let me compare what i did to what youre saying!!!

Answer 25

So my Cos^2X-1/sinx= -sinx is WRONG

Answer 26

Yes.

Answer 27

how should it look like

Answer 28

Because,\[\large \cos^2x-1 \color{red}{\neq} \sin^2x\]

Answer 29

The top is all we need to worry about.

Answer 30

Factoring out a negative one gives us,\[\large \cos^2x-1 \quad \rightarrow \quad -(-\cos^2x+1)\]Which we can write like this,\[\large -(1-\cos^2x)\]From here we can apply the identity, \[\large \color{salmon}{1-\cos^2x\quad=\quad \sin^2x}\]Giving us,\[\large -(\color{salmon}{1-\cos^2x}) \quad \rightarrow \quad -(\color{salmon}{\sin^2x})\]

Answer 31

okayy. let me do this when i get back home around 3. be back online. I MEAN IT

Answer 32

\[\large \cos^2x-1 \neq \sin^2x\]
\[\large 1-\cos^2x=\sin^2x\]

Having trouble understanding the difference between those two statements? :C

Answer 33

lol

Answer 34

yes I do beucase the one cant be negative

Answer 35

bye!

Answer 36

bye :3 go home lol

Answer 37

okay i am back. but i am trying to work out the problem now

Answer 38

lol k :)

Answer 39

hehe i get it now:) THANK YOU

Answer 40

you were so helpful

Answer 41

yay team \c:/

Answer 42

i have another problem i dont know how to do

Answer 43

Close this thread, Open a new one.
You can type @zepdrix in the description somewhere and it will make it easier for me to find it :D

Answer 44

okay boss