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Math Help - more trig help please!

  1. #1
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    Talking more trig help please!

    1) 2sin^2 x+sin2x=1 find x for 0<x<pi

    2) prove that (1+tan \theta)(1+tan( \pi/4- \theta))=2

    3) prove \frac{{sinx}}{{1-cosx}}=cot\frac{{x}}{{2}}

    I've been working on this all afternoon and my solution still doesn't match with the answers
    Last edited by snowball; September 17th 2008 at 12:00 PM.
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  2. #2
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    For problem 2, I used the trigonometric identity \tan(a - b) = \frac{\tan{a} - \tan{b}}{1 + \tan{a}\tan{b}}.

    This results in the expression (1 + \tan{\theta})(1 + \frac{1 - \tan{\theta}}{1 + \tan{\theta}}).

    When you multiply it out you get this:

    1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}},

    which simplifies to \frac{2 + 2 \tan{\theta}}{1 + \tan{\theta}} = 2.
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  3. #3
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    1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}},

    after this step I got \frac{1 - \tan^2{\theta}}{{1+tan\theta}}+1+tan\theta

    am I right or wrong? and could u explain how to simplify down to 2+2tan \theta/1+tan \theta please?

    sorry I'm not very familiar with the codes...


    I just worked that out... thanks again... now anyone could tell me where to start on the other 2 questions pls?
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  4. #4
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    Hello,
    1) 2sin^2 x+sin2x=1 find x for 0<x<pi


    This is the same as \sin(2x)=1-2\sin^2(x)

    \sin(2x)=\cos(2x)
    Now, you can think of several ways to solve it. Please show your work

    Quote Originally Posted by snowball View Post
    3) sinx/(1-cosx)=cotx/2
    Is it \cot \left(\frac x2\right) or \frac{\cot(x)}{2} ?

    Quote Originally Posted by snowball View Post
    1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}},

    after this step I got \frac{1 - \tan^2{\theta}}/{1+tan\theta}+1+tan\theta

    am I right or wrong? and could u explain how to simplify down to 2+2tan \theta/1+tan \theta please?

    sorry I'm not very familiar with the codes...
    (you don't need to write / if you use \frac{...}{...})

    Get the common denominator and simplify.

    But it is easier to see it this way :

    (1+\tan \theta)\left(1+\tfrac{1-\tan \theta}{1+\tan \theta}\right)=(1+\tan \theta)+({\color{red}1+\tan \theta}) \cdot \frac{1-\tan \theta}{\color{red}1+\tan \theta}=(1+\tan \theta)+(1-\tan \theta)

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  5. #5
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    I finally know how the codes work... thanks to Moo too
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