1. ## PROVING IDENTITY..URGENT HELP

I NEED HELP PROVING THIS IDENTITY.......prove identity tan3Ѳ=3tanѲ-tan^3Ѳ/1-3tan^2Ѳ

2. Originally Posted by nithu
I NEED HELP PROVING THIS IDENTITY.......prove identity tan3Ѳ=3tanѲ-tan^3Ѳ/1-3tan^2Ѳ
to do $\tan 3 \theta \equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$ .................(1)

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

thus, we can start with the left hand side of equation (1), and note that $\tan 3 \theta = \tan (2 \theta + \theta ) = \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta}$ ..........................(2)

we can further break down $\tan 2 \theta$ using the same formula. $\tan 2 \theta = \tan ( \theta + \theta) = \frac {2 \tan \theta}{1 - \tan^2 \theta}$

plug this into equation (2) and simplify to make it look like the right hand side of equation (1)

3. Originally Posted by Jhevon
to do $\tan 3 \theta \equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$ .................(1)

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

thus, we can start with the left hand side of equation (1), and note that $\tan 3 \theta = \tan (2 \theta + \theta ) = \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta}$ ..........................(2)

we can further break down $\tan 2 \theta$ using the same formula. $\tan 2 \theta = \tan ( \theta + \theta) = \frac {2 \tan \theta}{1 - \tan^2 \theta}$

plug this into equation (2) and simplify to make it look like the right hand side of equation (1)

i got this part..i just dont know how to furthur break it down...can u plz show me all the steps how to do this equation

4. Hello, nithu!

Okay . . . from the top . . .

We need two identities: . $\tan2\theta \:=\: \frac{2\tan\theta}{1-\tan^2\!\theta} \qquad \tan(A + B) \:=\:\dfrac{\tan A + \tan B}{1 - \tan A\tan B}$

Prove: . $\tan3\theta \:=\:\frac{3\tan\theta -\tan^3\!\theta}{1-3\tan^2\!\theta}$

We have: . $\tan3\theta \;=\;\tan(\theta + 2\theta)$

. . . . . . . . . . . . $= \;\frac{\tan\theta + 2\tan\theta}{1 - \tan\theta\tan2\theta}$

. . . . . . . . . . . . $= \;\frac{\tan\theta + \dfrac{2\tan\theta}{1-\tan^2\!\theta}}{1 - \tan\theta\!\cdot\!\dfrac{2\tan\theta}{1 - \tan^2\!\theta}}$

Multiply by $\frac{1-\tan^2\theta}{1-\tan^2\!\theta}$

. . $\tan3\theta \;=\;\frac{(1-\tan^2\!\theta)\left(\tan\theta + \dfrac{2\tan\theta}{1-\tan^2\!\theta}\right)}
{(1-\tan^2\!\theta)\left(1 - \tan\theta\!\cdot\!\dfrac{2\tan\theta}{1-\tan^2\!\theta}\right)}$

. . . . . . $= \;\frac{(1-\tan^2\!\theta)\tan\theta + 2\tan\theta}{(1 - \tan^2\!\theta) - 2\tan^2\!\theta}$

. . . . . . $= \;\frac{\tan\theta - \tan^3\!\theta + 2\tan\theta}{1 - \tan^2\!\theta - 2\tan^2\!\theta}$

. . . . . . $= \;\frac{3\tan\theta - \tan^3\!\theta}{1 - 3\tan^2\!\theta}$ . . . There!

5. Okay, I wasn't right in the head when I chose to prove it from RHS. Here's my solution (I skipped some steps but you can follow up, lol):

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

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

$= \frac{\sin{\theta} (4 - \sec^2{\theta})}{\cos{\theta} (4 - 3\sec^2{\theta})}$

$= \frac{4\sin{\theta}\cos^2{\theta} - \sin{\theta}}{\cos^2{\theta}} \div \frac{4\cos^2{\theta} - 3}{\cos{\theta}}$

...

$= \frac{\sin{\theta}(1+2\cos{2\theta})}{\cos{\theta} (2\cos{2\theta} - 1)}$

$= \frac{2(\sin{\theta}\cos{2\theta}) + \sin{\theta}}{2(\cos{\theta}\cos{2\theta}) - \cos{\theta}}$

================

$\sin{\theta}\cos{2\theta} = \frac{\sin{(\theta + 2\theta)} + \sin{(\theta - 2\theta)}}{2} = \frac{\sin{(3\theta)} + \sin{(-\theta})}{2} = \frac{\sin{(3\theta)} - \sin{\theta}}{2}$

$\cos{\theta}\cos{2\theta} = \frac{\cos{(\theta - 2\theta)} + \cos{(\theta + 2\theta)}}{2} = \frac{\cos{(-\theta)} + \cos{(3\theta)}}{2} = \frac{\cos{(\theta)} + \cos{(3\theta)}}{2}$

================

$\frac{2(\frac{\sin{(3\theta)} - \sin{\theta}}{2}) + \sin{\theta}}{2(\frac{\cos{(\theta)} + \cos{(3\theta)}}{2}) - \cos{\theta}}$

$= \frac{\sin{(3\theta)} - \sin{\theta} + \sin{\theta}}{\cos{(\theta)} + \cos{(3\theta)} - \cos{\theta}}$

$= \frac{\sin{3\theta}}{\cos{3\theta}} = \tan{3\theta}$

I skipped some steps where I switched $cos^2{\theta}$into $\frac{1+cos{2\theta}}{2}$

I hope I didn't make a mistake or typo.

6. Originally Posted by Chop Suey
Okay, I wasn't right in the head when I chose to prove it from RHS. Here's my solution (I skipped some steps but you can follow up, lol):

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

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

$= \frac{\sin{\theta} (4 - \sec^2{\theta})}{\cos{\theta} (4 - 3\sec^2{\theta})}$

$= \frac{4\sin{\theta}\cos^2{\theta} - \sin{\theta}}{\cos^2{\theta}} \div \frac{4\cos^2{\theta} - 3}{\cos{\theta}}$

...

$= \frac{\sin{\theta}(1+2\cos{2\theta})}{\cos{\theta} (2\cos{2\theta} - 1)}$

$= \frac{2(\sin{\theta}\cos{2\theta}) + \sin{\theta}}{2(\cos{\theta}\cos{2\theta}) - \cos{\theta}}$

================

$\sin{\theta}\cos{2\theta} = \frac{\sin{(\theta + 2\theta)} + \sin{(\theta - 2\theta)}}{2} = \frac{\sin{(3\theta)} + \sin{(-\theta})}{2} = \frac{\sin{(3\theta)} - \sin{\theta}}{2}$

$\cos{\theta}\cos{2\theta} = \frac{\cos{(\theta - 2\theta)} + \cos{(\theta + 2\theta)}}{2} = \frac{\cos{(-\theta)} + \cos{(3\theta)}}{2} = \frac{\cos{(\theta)} + \cos{(3\theta)}}{2}$

================

$\frac{2(\frac{\sin{(3\theta)} - \sin{\theta}}{2}) + \sin{\theta}}{2(\frac{\cos{(\theta)} + \cos{(3\theta)}}{2}) - \cos{\theta}}$

$= \frac{\sin{(3\theta)} - \sin{\theta} + \sin{\theta}}{\cos{(\theta)} + \cos{(3\theta)} - \cos{\theta}}$

$= \frac{\sin{3\theta}}{\cos{3\theta}} = \tan{3\theta}$

I skipped some steps where I switched $cos^2{\theta}$into $\frac{1+cos{2\theta}}{2}$

I hope I didn't make a mistake or typo.
Man...you always have to do things the hard way...

Usually, I decide to tackle the simple looking side just for fun...to see if I can keep my sanity as I try to prove the ridiculous looking identity

--Chris

7. Originally Posted by Chris L T521
Man...you always have to do things the hard way...

Usually, I decide to tackle the simple looking side just for fun...to see I can keep my sanity as I try to prove the ridiculous looking identity

--Chris
It can't hurt to show different methods, eh? I guess I like to deviate from the norm.

8. If you want another method out of the norm....

$\tan(3 \theta)=\frac{\sin(3 \theta)}{\cos(3 \theta)}$

$\sin(3 \theta)=2 \sin(\theta) \cos^2(\theta)+\sin(\theta)(1-2\sin^2(\theta))=2 \sin(\theta)-2\sin^3(\theta)+\sin(\theta)-2\sin^3(\theta)$ $=3 \sin(\theta)-4\sin^3(\theta)$ (I skipped a few steps, you should know this formula )

$\cos(3 \theta)=\cos(\theta)(2 \cos^2(\theta)-1)-2 \cos(\theta) \sin^2(\theta)=2 \cos^3(\theta)-\cos(\theta)-2 \cos(\theta)+2 \cos^3(\theta)$ $=4 \cos^3(\theta)-3 \cos(\theta)$

$\tan(3 \theta)=\frac{3 \sin(\theta)-4 \sin^3(\theta)}{4 \cos^3(\theta)-3\cos(\theta)}$

Factor by $\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta)$ :

$\tan(3 \theta)=\tan(\theta) \cdot \frac{3-4 \sin^2(\theta)}{4 \cos^2(\theta)-3}$

Use identity $1+\tan^2(\theta)=\frac{1}{\cos^2(\theta)} \implies \cos^2(\theta)=\frac{1}{1+\tan^2(\theta)}$ (we'll use it for the denominator) and $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{1}{1+\tan^2(\theta)}=\frac{\tan^2(\theta)}{1 +\tan^2(\theta)}$ (we'll use it for the numerator).

This yields :

$\tan(3 \theta)=\tan(\theta)\cdot \frac{3- \tfrac{4\tan^2(\theta)}{1+\tan^2(\theta)}}{\tfrac{ 4}{1+\tan^2(\theta)}-3}$

Multiply by $1=\frac{1+\tan^2(\theta)}{1+\tan^2(\theta)}$ :

$\tan(3 \theta)=\tan(\theta) \cdot \frac{3+3\tan^2(\theta)-4\tan^2(\theta)}{4-3-3\tan^2(\theta)}$

$\boxed{\tan(3 \theta)=\frac{3 \tan(\theta)-\tan^3(\theta)}{1-3 \tan^2(\theta)}}$

If you want to do it a complicated way, do it COMPLETELY !

(A few more steps could've been skipped but... it's better leaving them )

Edit : similar to Chops', but he used inverse trig functions (boooo) and I did it the reverse order LOL