i.d. proofs

• Feb 19th 2010, 01:43 PM
stealthmaths
i.d. proofs
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
• Feb 19th 2010, 02:05 PM
e^(i*pi)
Quote:

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 $\displaystyle \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$

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

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

Splitting the fraction:

$\displaystyle \frac{1}{\tan \theta} - \frac{\tan ^2 \theta}{\tan \theta} = \cot \theta - \tan \theta = \text{ RHS }$
• Feb 19th 2010, 02:41 PM
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.(Headbang)
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?
• Feb 19th 2010, 11:26 PM
Punch
Quote:

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.(Headbang)
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 $\displaystyle tan3\theta=(tan2\theta+\theta)$ and use the addition formulae.
• Feb 20th 2010, 04:16 AM
e^(i*pi)
Quote:

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

I agree.

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

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

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

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

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

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

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

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

Looking at the numerator:

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

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

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

Distribute the $\displaystyle \tan \theta$ and simplify:

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

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)

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

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

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

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

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

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

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

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

Bringing both parts together

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

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

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

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

$\displaystyle \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \div \frac{1-3 \tan^2 \theta}{1- \tan^2 \theta} = \frac{3 \tan \theta - \tan^3 \theta}{1-\tan^2 \theta} \times \frac{1- \tan^2 \theta}{1-3 \tan^2 \theta}$

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

$\displaystyle \frac{3 \tan \theta - \tan^3 \theta}{1- 3 \tan^2 \theta} = \text{ RHS}$
• Feb 20th 2010, 05:17 AM
stealthmaths
Well. That was a rather bigger than expected process.
Thanks for that (Happy)

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.(Clapping)

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