Prove a Trigonometry identity

Discussion in 'General Off-Topic Chat' started by shakirmoledina, Dec 1, 2012.

Dec 1, 2012
  1. shakirmoledina
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    Member shakirmoledina Legend

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    Hey guys,
    For those who are fans of pure maths, help me prove the following identity

    cosx/ tanx(1-sinx) = 1 + 1/sinx

    I tested it through a calculator and they match so its only the theoretical matching that I cannot do.

    things you can use are:
    sin 2 x + cos 2 x = 1
    and tanx = sinx/cosx

    Happy headaches!
     
  2. Veho

    Global Moderator Veho The man who cried "Ni".

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    Oh all right. One clarification needed, is the (1-sinx) part of the denominator, or is the whole fraction (cosx/tanx) multiplied by it?


    Here you go:

    cosx/[(sinx(1-sinx))/cosx] = sinx/sinx + 1/sinx

    cos^2(x)/[sinx(1-sinx)] = (1+sinx)/sinx ///multiply all by sinx(1-sinx)

    cos^2x = (1+sinx)(1-sinx) = 1 - sin^2x.

    Which is true.
     
  3. shakirmoledina
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    Member shakirmoledina Legend

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    uhuh so can u prove it from one side only so that the multiplication of the denominator doesnt come into place?
    Its a different approach where u consider both sides at once.

    Its for a boy who has his exams on monday.
     
  4. Veho

    Global Moderator Veho The man who cried "Ni".

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    :rolleyes:

    cosx/[(sinx(1-sinx))/cosx] = cos^2x/[sinx(1-sinx)] =

    = (1-sin^2x)/[sinx(1-sinx)] =///using the equivalence that a^2-b^2 = (a+b)(a-b) /// =

    = [(1+sinx)(1-sinx)]/[sinx(1-sinx)] =

    =(1+sinx)/sinx =

    = 1/sinx + sinx/sinx =

    = 1 + 1/sinx
     
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  5. shakirmoledina
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    equivalence... awesome veho
    vehi awesome
     
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  13. leic7

    Member leic7 GBAtemp Regular

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    Another strategy for proving something like that is to separate all the 'complex' looking fractions to ONE side of the equation, so that the other side contains only a 'simple' term. In general, it's easier to reduce a complex formula to something simpler, than it's to try and match one complex formula with another complex formula. so for example, move the last term 1/sinx to the left side, and solve for
    cosx/[tanx(1-sinx)] - 1/sinx

    If you get 1, you're done.
     
  14. shakirmoledina
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    YOU might wanna decomplexify your statement there. no full stop in that sentence.
     
  15. leic7

    Member leic7 GBAtemp Regular

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    Sorry, I just meant that for an equation such as:

    cosx/tanx(1-sinx) = 1 + 1/sinx

    ...there are fractions on both sides of "=". Here, both "cosx/tanx(1-sinx)" and "1/sinx" are fractions.

    If you can isolate these fractions to the left side of "=", the right side of "=" is left with the integer "1":

    cosx/tanx(1-sinx) = 1 + 1/sinx (1)
    cosx/tanx(1-sinx) - 1/sinx = 1 (2)

    To prove (1) is equivalent to proving (2), to prove (2) you can simply reduce "cosx/tanx(1-sinx) - 1/sinx" to "1". I think a student would find it relatively easier if they know the "answer" is "1", and all they have to do is "solve" it.
     
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  16. shakirmoledina
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    uhuh i like that since thats exactly the things that are required for solving such questions
     

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