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Trigonometric Identity?

  1. Jan 29, 2006 #1
    Trigonometric Identity??

    I just can't figure this out. I don't think i've covered enough material to do this. Can anyone help? I've put the entire question but im sure all i need is a little explanation and maybe the first answer and i could do the rest. :confused:

    [tex] Let z = tan(\theta/2).// Show that

    cos\theta = \frac{1-z^2}{1+z^2}

    sin\theta = \frac{2z}{1+z^2}

    \frac{d\theta}{dz} = \frac{2}{1+z^2}

    [/tex]

    Thanks. Sorry i tried to use latex but cant figure out spaces.
     
    Last edited: Jan 29, 2006
  2. jcsd
  3. Jan 29, 2006 #2

    TD

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    Start off with plugging it in and use that [itex]\sec ^2 a = 1 + \tan ^2 a[/itex].

    For example, to start the first one:

    [tex]\frac{{1 - z^2 }}{{1 + z^2 }} = \frac{{1 - \tan ^2 \left( {x/2} \right)}}{{1 + \tan ^2 \left( {x/2} \right)}} = \frac{{2 - \sec ^2 \left( {x/2} \right)}}{{\sec ^2 \left( {x/2} \right)}} = \frac{2}{{\sec ^2 \left( {x/2} \right)}} - 1[/tex]

    You're almost there, switch from sec to cos and use the double angle formula. Can you try the other ones now?
     
  4. Jan 29, 2006 #3

    VietDao29

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    You should look up some double-angle formulae in your textbook (note that: [tex]x = 2 \times \frac{x}{2}[/tex]), and some identities like:
    [tex]1 + \tan ^ 2 x = \frac{1}{\cos ^ 2 x} = \sec ^ 2 x[/tex]
    I'll do the sin(x) for you:
    [tex]\frac{2 \tan \left( \frac{x}{2} \right)}{1 + \left( \tan ^ 2 \frac{x}{2} \right)} = \frac{2 \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)}}{\frac{1}{\cos ^ 2 \left( \frac{x}{2} \right)}} = 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) = \sin (x)[/tex].
    You can do the rest, right? :smile:
    --------------
    Ooops, I didn't know you have posted, TD. Sorry to say nearly what you have said...
     
    Last edited: Jan 29, 2006
  5. Jan 29, 2006 #4

    TD

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    No problem, just a little less work for jamesbob but I'll doubt he'll mind :smile:
     
  6. Jan 29, 2006 #5
    Thanks so much for your help so far. I really appreciate the time you've spent to help me.

    For the first one my complete working was (including every line, needed or not) :

    [tex] \frac{{1 - z^2 }}{{1 + z^2 }} = \frac{{1 - \tan ^2 \left( {\theta/2} \right)}}{{1 + \tan ^2 \left( {\theta/2} \right)}} [/tex]

    here i said that

    [tex] \tan^2\theta = \sec^2\theta - 1 [/tex]

    so

    [tex] 1 - \tan^2(\frac{\theta}{2}) = 1 - (-1 + \sec^2(\frac{\theta}{2}))[/tex]
    and
    [tex] 1 + \tan^2(\frac{\theta}{2}) = 1 + \sec^2(\frac{\theta}{2}) [/tex]

    so continuing we have:

    [tex] \frac{2 - \sec^2(\frac{\theta}{2})}{\sec^2(\frac{\theta}{2})} = /frac{2}{\sec^2(\frac{\theta}{2})} = /frac{2}{\sec^2(\frac{\theta}{2})} - 1 [/tex]

    here i said that [tex] \sec(x) = \frac{1}{\cos(x)} [/tex]

    so we have, switching to cos
    [tex] \frac{2}{\frac{1}{\cos^2(\frac{\theta}{2})}} = 2\cos^2(\frac{\theta}{2}) - 1 [/tex]

    and using [tex] 2\cos^2x-1 = \cos2x [/tex]
    we have
    [tex] \cos(2 x (\frac{\theta}{2}) = cos(\theta) [/tex]

    I hope this LaTeX coding works. I'll do another post for the next two answers.
     
    Last edited: Jan 29, 2006
  7. Jan 29, 2006 #6

    TD

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    Apart from a few LaTeX-troubles, this seems alright :smile:
     
  8. Jan 29, 2006 #7
    Thanks, yeah for some reason i just couldnt get those fractions to work when they seem identical to others. I just started using LaTeX today so i aint that experienced. For [tex]\sin(\theta) = \frac{2z}{1 + z^2}[/tex] i said that this is equal to

    [tex] \frac{2\tan(\frac{\theta}{2})}{1 = \tan^2(\frac{\theta}{2})} [/tex] and i'm now realising its just the same as whats already been posted so i'm not going to do it again.

    I'm having difficulty with the last one tho. Is it anything to do with

    [tex] \int\frac{2}{1 + z^2} = 2\tan^-1\tan(\frac{\theta}{2}) [/tex] ?
     
  9. Jan 29, 2006 #8

    TD

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    You want to find [itex]d\theta /dz[/itex] but we know that [itex]z = \tan \left( {\theta /2} \right)[/itex]. Try to rewrite this last expression so you have [itex]\theta[/itex] as a function of x instead of the other way arround it is now. Then you can find its derivative with respect to z which is just what you want.
     
    Last edited: Jan 29, 2006
  10. Jan 29, 2006 #9
    Yeah thats what i had tried to do originally but got totally lost. best i could do was

    [tex]\theta=\tan^-1 \times \frac{2\tan(\frac{\theta}{2})}{1-\tan^2(\frac{\theta}{2})}[/tex]

    and i reckon this is totally the wrong direction.
     
  11. Jan 29, 2006 #10

    VietDao29

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    This looks fine (as TD has already confirmed you). But your 4th LaTeX part is wrong, it reads:
    [tex]1 + \tan ^ 2 \left( \frac{\theta}{2} \right) = 1 + \sec ^ 2 \left( \frac{\theta}{2} \right)[/tex], when it should actually reads:
    [tex]1 + \tan ^ 2 \left( \frac{\theta}{2} \right) = \sec ^ 2 \left( \frac{\theta}{2} \right)[/tex].
    Other than that, everything looks good. :smile:
    -------------------
    Since you've posted your answer, I'd like to show you a little faster way:
    Multiply both numerator, and denominator by cos2(theta / 2)
    [tex]\frac{1 - \tan ^ 2 \left( \frac{\theta}{2} \right)}{1 + \tan ^ 2 \left( \frac{\theta}{2} \right)} = \frac{\cos ^ 2 \left( \frac{\theta}{2} \right)}{\cos ^ 2 \left( \frac{\theta}{2} \right)} \times \frac{1 - \tan ^ 2 \left( \frac{\theta}{2} \right)}{\frac{1}{\cos ^ 2 \left( \frac{\theta}{2} \right)}} = \frac{\cos ^ 2 \left( \frac{\theta}{2} \right) - \sin ^ 2 \left( \frac{\theta}{2} \right)}{1} = \cos (\theta)[/tex].
    -------------------
    [tex]z = \tan \left( \frac{\theta}{2} \right) \Rightarrow \frac{\theta}{2} = \arctan z \Rightarrow \theta = 2 \arctan z[/tex]
    Can you differentiate arctan(x) with respect to x?
    ie, what's:
    [tex]\frac{d(\arctan(x))}{dx} = ?[/tex]
    Have you covered it yet?
    You can go from here, right?
     
  12. Jan 30, 2006 #11
    umm not done arctan yet, any otherr way to do it?
     
  13. Jan 30, 2006 #12
    This leads onto a question saying: a standard substitution to convert integrals of trigonometric functions into integrals of rational functions is the substitution
    [tex]z = \tan(\frac{\theta}{2})[/tex] ​
    Use this substitution to show that

    [tex] \int\sec\thetad\theta = ln(\sec\theta + \tan\theta) + C [/tex]

    I dont know if this helps in the direction we need to take it. I can't see how to do this question either.
     
  14. Jan 30, 2006 #13

    VietDao29

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    Okay, 1 other way is to note that:
    arctan(tan(x)) = x (this is true for all [itex]x \in ]- \pi / 2 ; \ \pi / 2[[/itex], right?)
    Let u = tan(x). We have:
    arctan(u) = x
    Differentiate both sides with respect to x, we have:
    arctan'(u) u'(x) = 1 (We use the chain rule for the LHS)
    So:
    [tex]\arctan ' (u) = \frac{1}{u'(x)} = \frac{1}{\tan '(x)} = \frac{1}{\frac{1}{\cos ^ 2 x}} = \frac{1}{1 + \tan ^ 2 x} = \frac{1}{1 + u ^ 2}[/tex]. So that means:
    [tex]\arctan ' (x) = \frac{1}{1 + x ^ 2}[/tex]. Okay, I think I've told you the answer. Do you get this? :cool:
    ---------------
    For the next problem. Let [tex]t = \tan \left( \frac{\theta}{2} \right)[/tex]
    Before you integrate that expression. First, answer these question:
    1. What's [itex]\cos \theta[/itex] in terms of t?
    2. What's dx in terms of t, and dt?
    3. What does the integral finally become?
     
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