i.d. proofs

**stealthmaths** · February 19th 2010, 01:43 PM

Hi:
I am asked to prove the following identities:
a) 2cot2θ≡cotθ-tanθ
b) tan3θ≡3tanθ-tan³θ/1-3tan²θ

I think I should tackle the most complicated side as it is easier to aim for the easier looking side. (hope that makes sense )
I should warn you that I'm not good at this so please excuse my fumbling
a) 2cot2θ≡cotθ-tanθ
I will attempt the RHS, cotθ-tanθ
cotθ-sinθ/cosθ
so I'm stuck! Told you

**e^(i*pi)** · February 19th 2010, 02:05 PM

Originally Posted by stealthmaths

Hi:
I am asked to prove the following identities:
a) 2cot2θ≡cotθ-tanθ
b) tan3θ≡3tanθ-tan³θ/1-3tan²θ

I think I should tackle the most complicated side as it is easier to aim for the easier looking side. (hope that makes sense )
I should warn you that I'm not good at this so please excuse my fumbling
a) 2cot2θ≡cotθ-tanθ
I will attempt the RHS, cotθ-tanθ
cotθ-sinθ/cosθ
so I'm stuck! Told you

I'd work from the left to avoid the double angle. It seems you are unaware that $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$

$\tan (2 \theta) = \frac{2\tan \theta}{1- \tan ^2 \theta}$

$2\cot (2 \theta) = \frac{2}{\tan(2 \theta)} = \frac{2}{\frac{2\tan \theta}{1- \tan ^2 \theta}} = \frac{2(1-\tan^2 \theta)}{2 \tan \theta} = \frac{1-\tan ^2 \theta}{\tan \theta}$

Splitting the fraction:

$\frac{1}{\tan \theta} - \frac{\tan ^2 \theta}{\tan \theta} = \cot \theta - \tan \theta = \text{ RHS }$

**stealthmaths** · February 19th 2010, 02:41 PM

Hi:
thank you:
I was aware of cotθ=1/tanθ=cosθ/sinθ
But I find this a mine field. And never know where to start nor which replacements to use. It's really confusing to me.

I would like to return to this a little later and ask you some questions about your thinking, if I may?

But first:
b) tan3θ≡(3tanθ-tan³θ)/1-3tan²θ
so the LHS has only one thing to play with. So I should tackle this side?

tan3θ=sin3θ/cos3θ

On the right track?

**Punch** · February 19th 2010, 11:26 PM

Originally Posted by stealthmaths

Hi:
thank you:
I was aware of cotθ=1/tanθ=cosθ/sinθ
But I find this a mine field. And never know where to start nor which replacements to use. It's really confusing to me.

I would like to return to this a little later and ask you some questions about your thinking, if I may?

But first:
b) tan3θ≡(3tanθ-tan³θ)/1-3tan²θ
so the LHS has only one thing to play with. So I should tackle this side?

tan3θ=sin3θ/cos3θ

On the right track?

In my opinion, for b), tackle the left hand side by writing $tan3\theta=(tan2\theta+\theta)$ and use the addition formulae.

**e^(i*pi)** · February 20th 2010, 04:16 AM

Originally Posted by Punch

In my opinion, for b), tackle the left hand side by writing $tan3\theta=(tan2\theta+\theta)$ and use the addition formulae.

I agree.

Also thank goodness for LaTeX. It's not that hard to write on paper though

$\tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B}$

$\therefore \: \: \tan (2 \theta) = \frac{2 \tan \theta}{1-\tan ^2 \theta}$

Write $\tan (3 \theta) as \tan (2 \theta + \theta)$

Note that $A = 2 \theta$ and $B = \theta$

Hence $\tan (2 \theta + \theta) = \frac{\tan (2 \theta) + \tan \theta}{1- \tan (2 \theta) \tan \theta}$

Subbing in our double angle identity for \tan wherever $\tan (2 \theta)$ occurs:

$\frac{\frac{2 \tan \theta}{1-\tan ^2 \theta}+\tan \theta}{1 - \frac{2 \tan \theta}{1-\tan ^2 \theta} \tan \theta}<br />$

Looking at the numerator:

$\frac{2 \tan \theta}{1-\tan ^2 \theta}+\tan \theta$

Get the same denominator by multipling $\tan \theta$ by $\frac{1-\tan^2 \theta}{1-\tan^2 \theta}$

$\frac{2 \tan \theta + \tan \theta (1- \tan ^2 \theta)}{1-\tan^2 \theta}$

Distribute the $\tan \theta$ and simplify:

$<br /> \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta}$

Note the similarity of this to the right hand side's numerator. Therefore, leave it like this and work on the denominator

Looking at the denominator (of the overall fraction)

$1 - \frac{2 \tan \theta}{1-\tan^2 \theta} \tan \theta$

Rewrite 1 as $\frac{1-\tan^2 \theta}{1-\tan^2 \theta}$

The second term can be simplified to $\frac{2 \tan ^2 \theta}{1-\tan^2 \theta}$

Since the two terms now have the same denominator we can combine them:

$\frac{(1- \tan^2 \theta)- 2 \tan^2 \theta}{1- \tan^2 \theta}$

The numerator of this fraction can be simplified to $1-3 \tan^2 \theta$

$\frac{1-3 \tan^2 \theta}{1- \tan^2 \theta}$

As this is also similar to the RHS leave it in this form.

Bringing both parts together

Numerator: $\frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta}$

Denominator: $\frac{1-3 \tan^2 \theta}{1- \tan^2 \theta}$

Which makes our fraction equal to $\frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \div \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta}$

Dividing by a fraction is the same as 'flipping' it (taking the reciprocal) and multiplying.

$<br /> \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \div \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta} = \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \times \frac{1- \tan^2 \theta}{1-3 \tan^2 \theta}$

$1- \tan^2 \theta$ cancels to leave us with:

$\frac{3 \tan \theta - \tan^3 \theta}{1- 3 \tan^2 \theta} = \text{ RHS}$

**stealthmaths** · February 20th 2010, 05:17 AM

Well. That was a rather bigger than expected process.
Thanks for that

Your right - I should learn Latex, and will make an effort to do so after this weekend, which is this coming Monday. Promise Promise.
For anybody reading this who is interested, I just did a search on Latex video tutorials and came up with a couple of sites instantly. It seems there is a lot of help available online, as always.

TechScreencast - LaTeX Intro :: tutorial, videos, articles, screencast to learn technology"

latex video tutorial rapidshare, megaupload, torrent download

Though I haven't looked closely at them, yet, I'm pretty certain they will be helpful and are reputable sites.
Is the last link ok?

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