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Trigonometric identity

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  1. Sep 10, 2015 #1

    Rectifier

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    The problem

    Show that the left side is equal to right side
    ## tan (\frac{x}{2}) = \frac{1-cos(x)}{sin(x)} ##​

    The attempt
    ##\tan(\frac{x}{2}) = \frac{ sin(\frac{x}{2}) }{ cos (\frac{x}{2}) } = \frac{ sin^2(\frac{x}{2}) }{ cos ^2 (\frac{x}{2}) } = \frac{\frac{1-cos(x)}{2}}{\frac{1+cos(x)}{2}} = \frac{1-cos(x)}{1+cos(x)} ## Fail :,(​
     
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  3. Sep 10, 2015 #2

    andrewkirk

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    Your second inequality is invalid.
     
  4. Sep 10, 2015 #3

    Rectifier

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    Made one more attempt but failed too:
    ##\frac{1-\cos(x)}{\sin(x)} \\ \frac{\sin^2(x)+\cos^2(x)-\cos(x)}{\sin(x)} \\ \frac{\sin^2(x)+(\cos(x)-1)\cos(x)}{\sin(x)}##
     
  5. Sep 10, 2015 #4

    andrewkirk

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    The left hand side of your identity is written in terms of trig functions of x/2. Don't you think it might help to start by trying to express the right-hand side in terms of trig functions of x/2?
     
  6. Sep 10, 2015 #5

    Mark44

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    He means at the 2nd equals sign. It looks like you squared the numerator and denominator, which almost always gives you an expression with a different value. So ##\frac{ sin(\frac{x}{2}) }{ cos (\frac{x}{2}) } \ne \frac{ sin^2(\frac{x}{2}) }{ cos ^2 (\frac{x}{2}) }##, in general.
     
  7. Sep 11, 2015 #6

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    Thank you for the clarification I got that part and abandoned the idea - hence attempt 2.


    Yeah, I am not sure what identities to use since these are the only ones I have on my mind right now
    ##sin^2\frac{x}{2}=\frac{1-cos(x)}{2}## and
    ##cos^2\frac{x}{2}=\frac{1+cos(x)}{2}##

    Any tips?
     
  8. Sep 11, 2015 #7

    andrewkirk

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    Do you know the formulas for ##\sin 2\theta## and ##\cos 2\theta##? (If not, search them - they're everywhere on the net, or rearrange the identities you already have in your last post)

    How can you use those formulas to write ##\sin x## in terms of trig functions of ##\frac{x}{2}##?

    And yes, when I said your second inequality I meant your second equality. I have this odd thing with sometimes accidentally saying the opposite of what I mean. It's scary when I'm giving directions in a car.
     
  9. Sep 11, 2015 #8

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    Yes, I know these.


    Which ones whould I rewrite?

    is it
    ##sin^2\frac{x}{2}=\frac{1-cos(x)}{2}## and
    ##cos^2\frac{x}{2}=\frac{1+cos(x)}{2}## ?
     
  10. Sep 11, 2015 #9

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    So here is one more attempt (perhaps a solution):
    ##tan\frac{x}{2} = \frac{1-cos(x)}{sin(x)}##​

    substitute ## x = 2v \Leftrightarrow v=x/2 ##

    ## tan\frac{x}{2} = tan\frac{2v}{2} = tan(v) = \frac{sin(v)}{cos(v)} = \frac{1-cos(2v)}{sin(2v)} \\ \frac{cos^2v+sin^2v-(cos^2v-sin^2v)}{2sin(v)cos(v)} = \frac{2sin^2v}{2sin(v)cos(v)} = \frac{sin(v)}{cos(v)} = tan(v) = tan(\frac{x}{2}) ##​

    in the last step i substitute ##v=x/2 ##
     
  11. Sep 11, 2015 #10

    andrewkirk

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    I think you've got it, but I can't follow the way you've laid it out. You start with ##\tan\frac{x}{2}##, transform that several times and then end up with what you started with!

    The last item on your first line is equal to the RHS of the desired identity, so you could just start there, or with the equality ##\frac{1-\cos x}{\sin x}=\frac{1-\cos 2v}{\sin 2v}## and then carry on through the second line until you end up with the LHS of the identity.

    By the way, did you know that if you put a backslash \ in front of a trig function in latex (ie write \sin rather than sin) it recognises the trig function , writes it more nicely and spaces it out so you won't have to put brackets around whatever you are taking the sin of, ie ##\sin 2v## instead of ##sin 2v## or ##sin(2v)##?
     
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