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Math Help - PROVING IDENTITY..URGENT HELP

  1. #1
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    PROVING IDENTITY..URGENT HELP

    I NEED HELP PROVING THIS IDENTITY.......prove identity tan3Ѳ=3tanѲ-tan^3Ѳ/1-3tan^2Ѳ
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nithu View Post
    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)
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    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
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  4. #4
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    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)}<br />
{(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!

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  5. #5
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    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.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chop Suey View Post
    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
    Last edited by Chris L T521; August 22nd 2008 at 10:40 PM.
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  7. #7
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    Quote Originally Posted by Chris L T521 View Post
    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.
    Last edited by Chop Suey; August 22nd 2008 at 11:52 PM.
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  8. #8
    Moo
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    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
    Last edited by Moo; August 23rd 2008 at 12:40 AM.
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