limit of cos[x]-1 / sin[x]

[Kristy]

(This was the other problem that I'm having trouble with. It is similar to the question on the one I just posted. I don't know if they are related, or not, so I just put each one separately. )
Directions: Find the limit.
Note: they are using the angle theta here but I don't know how to put the symbol for that so I just used x instead.

limit as x ->0 Cos[x]-1 / Sin[x]
What I tried:

Limit Cos[x]/Sin[x] – 1/Sin[x]

Here I tried to plug in 0 for x and got 1/0 – 1/0 which didn’t work

Using Trig identities:

Limit Cot[x] – Csc[x]

Here when I try to plug in 0 I get infinity – infinity.
Those are all the things I knew to try.

I tried graphing it and it looks like the answer should be zero.

 

[galactus]

\(\displaystyle \L\\\frac{cos(x)-1}{sin(x)}=-tan(\frac{x}{2})\)

Now, the limit is easy, ain't it?.

 

[Kristy]

\(\displaystyle \L\\\frac{cos(x)-1}{sin(x)}=-tan(\frac{x}{2})\)

Now, the limit is easy, ain't it?.

Wow, I totally don't understand how you got from the left hand side of the equal sign to the right. I'll look at it some more...
I know Sin[x]/Cos[x] = Tan[x] but that doesn't seem to be what is going on in what you did.

 

[galactus]

That's a half angle identity.

\(\displaystyle \L\\\frac{cos(x)-1}{sin(x)}=\frac{cos(x)}{sin(x)}-\frac{1}{sin(x)}=cot(x)-csc(x)=-tan(\frac{x}{2})\)

The half angle identity is \(\displaystyle \frac{1-cos(x)}{sin(x)}=tan(\frac{x}{2})\), therefore, \(\displaystyle \L\\\frac{cos(x)-1}{sin(x)}=-tan(\frac{x}{2})\)

 

[Kristy]

[...]

\(\displaystyle \L\\\=\=cot(x)-csc(x)=-tan(\frac{x}{2})\)

Is that something that should be obvious? (Or is the half angle thing that I just memorize? ) I did not follow this step, if I did, it would ALL make sense. I can see how it would just change signs in my problem. I don't understand how you can do cotangent - cosecant and get tangent of a half angle. I guess I could just memorize it, but I don't understand it and like to.

 

[galactus]

No, it's not obvious. Just know the half-angle identity. You don't have to derive it. It's not that obvious. Though it can be derived. I would suggest just memorizing it.

We can start with 2 famous identities:

\(\displaystyle \L\\sin^{2}(u)=\frac{1-cos(2u)}{2} \;\ and \;\ cos^{2}(u)=\frac{1+cos(2u)}{2}\)

\(\displaystyle \L\\tan^{2}(u)=\frac{sin^{2}(u)}{cos^{2}(u)}\)

Sub in the above and get:

\(\displaystyle \L\\tan^{2}(u)=\frac{1-cos(2u)}{1+cos(2u)}\)

Now let \(\displaystyle \frac{v}{2}=u\) in the above identity:

\(\displaystyle \L\\tan^{2}(\frac{v}{2})=\frac{1-cos(v)}{1+cos(v)}\)

\(\displaystyle \L\\tan(\frac{v}{2})=\pm\sqrt{\frac{1-cos(v)}{1+cos(v)}}\)

Multiply num and den in the radicand by \(\displaystyle 1-cos(v)\):

\(\displaystyle \L\\tan(\frac{v}{2})=\pm\sqrt{\frac{1-cos(v)}{1+cos(v)}\cdot\frac{1-cos(v)}{1-cos(v)}}\)

=\(\displaystyle \L\\\pm\sqrt{\frac{(1-cos(v))^{2}}{sin^{2}(v)}}\)

=\(\displaystyle \L\\\pm\frac{1-cos(v)}{sin(v)}=\pm{tan(\frac{v}{2})}\)

See?. The cotx-cscx is rather redundant here. I shouldn't have used that.

 

[Deleted member 4993]

You know

cos(x) = cos(2*x/2) = 1 - 2{sin(x/2)}^2
and

sin(x) = sin(2*x/2) = 2*sin(x/2) * cos(x/2)

Then

{cos(x) - 1}/sin(x)

= [ 1 - 2{sin(x/2)}^2 -1 ] / {2*sin(x/2) * cos(x/2)}

= - [2{sin(x/2)}^2] / {2*sin(x/2) * cos(x/2)}

= - sin(x/2) / cos(x/2)

= - tan(x/2)

You can also prove by same method:

{1 + cos(x)}/sinx = cot(x/2)

 

[Kristy]

No, it's not obvious. Just know the half-angle identity. You don't have to derive it. It's not that obvious. Though it can be derived. I would suggest just memorizing it.

Thanks. The problem seems a little less intimidating with just memorizing the half angle identity. The proof looks interesting but HARD!

 

[galactus]

That's why we just memorize them, unless specifically asked to derive.

 

[Kristy]

You know

cos(x) = cos(2*x/2) = 1 - 2{sin(x/2)}^2
and

sin(x) = sin(2*x/2) = 2*sin(x/2) * cos(x/2)

Then

{cos(x) - 1}/sin(x)

= [ 1 - 2{sin(x/2)}^2 -1 ] / {2*sin(x/2) * cos(x/2)}

= - [2{sin(x/2)}^2] / {2*sin(x/2) * cos(x/2)}

= - sin(x/2) / cos(x/2)

= - tan(x/2)

You can also prove by same method:

{1 + cos(x)}/sinx = cot(x/2)

Thanks. I'm going to print that out, and copy the problem by hand. It is a little hard to understand. I'm not sure if that is because it is a hard problem, or because it is hard when computer notation doesn't look like how I write it out. I'll see if that helps me understand it better!

 

[Deleted member 4993]

My opinion -

Remeber to derive them. I surely mess-up sign for half-angle formulae.

 

[Kristy]

That's why we just memorize them, unless specifically asked to derive.

That makes sense. Especially when I'm taking a test I'd rather not run out of time doing so many steps if its something I can memorize. Thank you for your explanations on these.

 

[Kristy]

I tried writing out all the steps [I also did copy and paste to try to use the tex noration but I couldn't get it to totally look okay. It just seems like such a short problem this way that it makes me think it is wrong. I guess I should just be glad it isn't longer?

\(\displaystyle \L\\\frac{cos(x)-1}{sin(x)}\)

=\(\displaystyle \L\\\frac{(-1)*(1-cos(x)}{sin(x)}\)

The half angle identity is \(\displaystyle \frac{1-cos(x)}{sin(x)}=tan(\frac{x}{2})\)

So the answer is \(\displaystyle \(-1) * tan(\frac{x}{2})\)

=\(\displaystyle \ -tan(\frac{x}{2})\)

 

[Kristy]

My opinion -

Remeber to derive them. I surely mess-up sign for half-angle formulae.

Okay, I guess I'm a little confused. The directions just say find the limit, and we didn't do any like this in class (we used some more basic trig identities but not the half angle one.) So I don't know whether or not to try to write up the steps of deriving the half angle formula.

 

[galactus]

I would venture to say you do not have to derive the formula. Just use the identity given and it is easy to see \(\displaystyle \L\\-\lim_{x\to\0} \;\ \tan(\frac{x}{2})=0\)

As for all that derivation we done, just thought it would be helpful to show you were it comes from.

You could show off if you wish and derive it. :wink:

 

[soroban]

Hello, Kristy!!

\(\displaystyle \L\lim_{\theta\to0} \frac{\cos\theta \,-\,1}{\sin\theta}\)

You may be expeccted to solve it like this . . .

We have: \(\displaystyle \L\:\frac{-(1\,-\,\cos\theta)}{\sin\theta}\)

Multiply by \(\displaystyle \frac{1\,+\,\cos\theta}{1\,+\,\cos\theta}:\L\;\frac{-(1\,-\,\cos\theta)}{\sin\theta}\,\cdot\,\frac{1\,+\,\cos\theta}{1\,+\,\cos\theta} \;=\;\frac{-(1\,-\,\cos^2\theta)}{\sin\theta(1\,+\,\cos\theta)}\)

. . \(\displaystyle \L=\;\frac{-\sin^2\theta}{\sin\theta(1\,+\,\cos\theta)} \;=\;\frac{-\sin\theta}{1\,+\,\cos\theta}\)


Therefore: \(\displaystyle \L\:\lim_{\theta\to0}\left(\frac{-\sin\theta}{1\,+\,\cos\theta}\right) \;=\;\frac{0}{1\,+\,1}\;=\;0\)

 

[Kristy]

I would venture to say you do not have to derive the formula. Just use the identity given and it is easy to see \(\displaystyle \L\\-\lim_{x\to\0} \;\ \tan(\frac{x}{2})=0\)

As for all that derivation we done, just thought it would be helpful to show you were it comes from.

You could show off if you wish and derive it. :wink:

Thanks. I think I'll just use the identity. You are right I was so busy trying to figure out how to simplify the expression that I forgot about taking the limit! I think I'll skip the showing off for today, these problems are hard enough to DO!

 

[Kristy]

Hello, Kristy!!

\(\displaystyle \L\lim_{\theta\to0} \frac{\cos\theta \,-\,1}{\sin\theta}\)

You may be expeccted to solve it like this . . .

We have: \(\displaystyle \L\:\frac{-(1\,-\,\cos\theta)}{\sin\theta}\)

Multiply by \(\displaystyle \frac{1\,+\,\cos\theta}{1\,+\,\cos\theta}:\L\;\frac{-(1\,-\,\cos\theta)}{\sin\theta}\,\cdot\,\frac{1\,+\,\cos\theta}{1\,+\,\cos\theta} \;=\;\frac{-(1\,-\,\cos^2\theta)}{\sin\theta(1\,+\,\cos\theta)}\)

. . \(\displaystyle \L=\;\frac{-\sin^2\theta}{\sin\theta(1\,+\,\cos\theta)} \;=\;\frac{-\sin\theta}{1\,+\,\cos\theta}\)


Therefore: \(\displaystyle \L\:\lim_{\theta\to0}\left(\frac{-\sin\theta}{1\,+\,\cos\theta}\right) \;=\;\frac{0}{1\,+\,1}\;=\;0\)

Okay, I think I understand all of the steps in solving it this way. Wow, talk about a weird problem. It seems like there are as many ways to solve it as there are people explaining it to me!