Question

cosx^2x-1 divided by cosx+1

Answer 1

oops i mean cosx^(2x)?

Answer 2

no, cosine squared x minus one divided by cosine x plus one

Answer 3

lol

Answer 4

(cosx)^2-1?

Answer 5

the exponent 2 is after the cos lol

Answer 6

\[\frac{\cos^2x-1}{cosx+1}=\frac{(cosx+1)(cosx-1)}{cosx+1}\]

Answer 7

So you have:
\[\frac{\cos^2(x)-1}{\cos(x)+1}\]

Answer 8

so = cosx+1

Answer 9

ah the method of myininaya!

Answer 10

\[(cosx)^2=\cos^2x\]
just so you know these are the same

Answer 11

The top is a difference of squares which factors giving:
\[\frac{(\cos(x)-1)(\cos(x)+1)}{cos(x)+1}\rightarrow \cos(x)+1\]

Answer 12

the answer is cosx-1

Answer 13

Yeah, I mistyped, should read cos(x)-1

Answer 14

motherofgod :( :( :( is it cosx-1 or cosx+1 ?

Answer 15

lol sure male
you just copied me ;)

Answer 16

cos(x)-1

Answer 17

LOL kk, sorry guys

Answer 18

I think I am going to put "difference of squares" as a text shortcut on my computer so I can just paste it with a keypress...

Answer 19

hello estudier. did you finish expected waiting time question?

Answer 20

Haha estudier, too true. Around here at least xDD

Answer 21

LOL - i know the diff of squares i'm just confused about the exponentiation convention with trig functions

Answer 22

I think so, I am just trying to convince myself...

Answer 23

@canuckish it has nothing at all to do with trig

Answer 24

\[(trig(x))^2=trig^2(x)\]

Answer 25

\[\frac{z^2-1}{z+1}=z-1\]

Answer 26

@estudier i think problem is hard. maybe have to break up into even and odd tosses

Answer 27

Oh yeah, same with the logarithms.
\[(\cos(x))^n=\cos^n(x); (\ln(x))^n=\ln^n(x)\]
Just don't confuse it with the arctrig functions.
\[\frac{1}{\sin(x)}=\sin^{-1}(x)=\csc(x) \ne \arcsin(x)=\sin^{-1}(x)\]
So you see the difference.

Answer 28

i will go to post and tell you my thoughts

Answer 29

honestly i never seen \[\ln^n(x)\]

Answer 30

that's all my book does is square the function :( it's very confusing

Answer 31

but i would think it would be equal to \[(\ln(x))^n\] if such a notation does exist

Answer 32

Really? I use it all the time:
\[\ln^2(x)\]

Answer 33

\[\frac{1}{sinx}\neq \sin^{-1}x\]

Answer 34

Well, kinda, just don't misinterpret it:
\[\frac{1}{\sin(x)}=(\sin(x))^{-1}\]

Answer 35

how are you guys getting the divisor symbol to show in the eq. builder? I just see (ab) and it doesn't have the line....

Answer 36

\[\left(\begin{matrix}1 \\ sinx\end{matrix}\right) = cscx ? yeah?\]

Answer 37

see - no line :(

Answer 38

\frac{a}{b}

Answer 39

\[\frac{1}{sinx} = cscx\] there we go

Answer 40

And yes it does equal that. :P