Question

How would this be simplified, if possible?

Answer 1

\[\frac{ 1-\cos^3x}{ 1-\cos^2x }\]

Answer 2

I think that I'm trying to get it to be something similar to \[\frac{ 1-cosx }{ x }\] but I'm not sure that it can get there.

Answer 3

For numerator use
(a-b)^3=(a-b)(a^2+b^2+ab)
And for denominator use
(a^2-b^2)=(a+b)(a-b)
Then simplify

Answer 4

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

Is this right so far?

Answer 5

Sorry its a^3-b^3

Answer 6

Yes correct till now

Answer 7

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

Answer 8

U can simplify it further if want...otherwise its fine till here
[1+cosx(1+cosx)]/(1+cosx)
So it will be
1/(1+cosx) +cosx

Answer 9

\[\frac{ (1+cosx)(1+cosx)+cosx }{ 1+cosx } = \frac{ (1+cosx)(1+cosx) }{ 1+cosx } + \frac{ cosx }{ 1+cosx }\]

Answer 10

No (1+cos^2x) is not equal to
(1+cosx)(1+cosx)
It is equal to (1+cosx)^2

Answer 11

\[\frac{ (1+cosx)(1+cosx) }{ 1+cosx } + \frac{ cosx }{ 1+cosx } = \frac{ 1+cosx }{ 1 } + \frac{ cosx }{ 1+cosx }\]

Answer 12

Is (1+cosx)(1+cosx) not the same as (1+cosx)^2 just expanded?

Answer 13

Yes they r same but it is not equal to
(1+cos^2 x)

Answer 14

Yeah, I just realized that much.

Answer 15

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

I don't understand how you got your final answer. It's right but it doesn't compute with me for some reason

Answer 16

Oh wait. Okay. So I get how you got here: \frac{ 1+ cosx(1+cosx) }{ 1+cosx } and then did you just break up the equation to \[\frac{ cosx(1+cosx) }{ 1+cosx }+\frac{ 1 }{ 1+cosx }\]

Answer 17

Yes u got it right!! :)

Answer 18

And then it gets simplified from \[\frac{ cosx(1+cosx) }{ 1+cosx }+\frac{ 1 }{ 1+cosx } = cosx +\frac{ 1 }{ 1+cosx }\]

Answer 19

Yupp

Answer 20

Thank you so much!

Answer 21

Yr wlcm ☺

Answer 22

Another question if I may. We're doing limits and it asks for what is the limit as x->0 \[\frac{ sinx-sinxcosx }{ x cosx }\] and I managed to get this down to the simplified \[\frac{ tanx }{ x } - \frac{ sinx }{ x }\]

Answer 23

However, this has to be simplified more because if you substitute 0 in for x, it will have an undefined fraction.

Answer 24

\[\frac{ tanx }{ x } - 1\] is what that means too because of the fact that there is a special thing saying that \[\frac{ sinx }{ x }\]\[=1\]

Answer 25

U did right till here.
Now Both tanx and sinx are 0 for x=0.
Their value gradually inc. as u inc. the value of x
So when x->0 means that x has values very near to zero.
And for values near zero tanx and sinx
are also near zero that is sinx is approximately same as x and tanx is also approximately same as x.
Therefore tanx/x will be equal to 1
And sinx/x will also be equal to 1