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Homework Help: Prove the Trigonometric

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    prove : 1-cos x / sin x = sin x / 1+cosx

    1-cosx / 1+cos x = tan^2 (x/2)

    2. Relevant equations

    3. The attempt at a solution
    I have No idea how to solve it
  2. jcsd
  3. Nov 23, 2008 #2
    If you don't have any idea how to prove it, start by working on both sides (for the first expression).
    The first step should look like: (1-cosx)(1+cosx)=sin2x
  4. Nov 23, 2008 #3
    Nice i got it now sin^2 x + cos^2 x = 1

    wat about the 2nd ?
  5. Nov 23, 2008 #4


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    Homework Helper

    I can help you with the first one, since I have yet to learn about half angles. Mainly the [tex]tan^2(\frac{x}{2})[/tex]

    ok so we need to prove [tex]\frac{1-cosx}{sinx}=\frac{sinx}{1+cosx}[/tex]
    For questions like these, it is a good habit to only manipulate one side of the equation and necessary if you want all the marks.
    Lets take the Left Hand Side then:

    ok so we need to somehow convert the denominator from sine to cosine.
    You would've learnt the trigonometric identity [tex]sin^2x+cos^2x=1[/tex]
    Then lets multiply both the numerator and denominator by sinx:

    The denominator can be converted by the simple manipulation of the trig identity

    From here it is quite simple so I will let you take over :smile:
  6. Nov 23, 2008 #5
    we are proving not solving..
  7. Nov 23, 2008 #6
    @icystrike please stop using useless comments.
    My method is absolutely correct.
    Since sin2x+cos2x=1
    And we can transform it as 1-cos2x-cos2x=1
    So 1=1 which is correct.
    Maybe Mentallic method is better since uses one of the sides (LHS) to prove the other (RHS).
    For the second trigonometric identity:
  8. Nov 23, 2008 #7
    Дьявол Thank you so much :)
  9. Nov 23, 2008 #8
    something out of trigonometric :

    i have g(x): 23w4sxz.jpg

    i need (gog)x = but how can i do it ? I'll have 2 radicals
  10. Nov 23, 2008 #9
    Is that supposed to be [tex]g(x) = \sqrt[3]{x + 1}[/tex]?

    If so, I don't really see the problem with having two radicals.
  11. Nov 23, 2008 #10


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    Homework Helper

    lol at the picture! :rofl:

    Can I just ask what (gog)x is?
  12. Nov 24, 2008 #11
    I think, that he thinks :smile: of composition.
    [tex](g \circ g)x[/tex]
    If [tex]g(x) = \sqrt[3]{x + 1}[/tex], then g(g(x)) would be [tex]g(\sqrt[3]{x + 1})=\sqrt[3]{\sqrt[3]{x+1}+ 1}[/tex]
  13. Nov 24, 2008 #12
    It sounded to me like he was just worried about having a radical sign inside a radical sign in his composite function, but I don't think there really is a problem with that in this case. It should be fine the way Дьявол wrote it.
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