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Math Help - General solutions.

  1. #1
    Super Member Showcase_22's Avatar
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    General solutions.

    The question is to find the general solution for the first line of the following picture:



    My method is included in that to but it differs from the book answer. One of the biggest problems is that I have two answers and they only have one!

    I figured that the second solution must be included in the first, but for my solutions this doesn't happen.

    Can anyone point out what i'm doing wrong?
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  2. #2
    Senior Member pankaj's Avatar
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    Your answer is as correct as can be.Try taking LCM and you will get a factor as 4n+1 and in the other you will get 4n+3.By giving different integral values to n you will find that they comprise of the same values as you will obtain when you will put them in 2n+1

    Another method can be
    \tan 5\theta=\cot 4\theta = \tan (\frac {\pi}{2}-4\theta)

    5\theta=n\pi +\frac {\pi}{2}-4\theta

    9\theta=n\pi +\frac{\pi}{2}

    \theta=\frac{n\pi}{9} +\frac{\pi}{18}

    \theta=(2n+1)\frac{\pi}{18}
    Last edited by pankaj; September 17th 2008 at 07:51 AM.
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  3. #3
    Super Member Showcase_22's Avatar
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    I sort-of-get what you've done, except the first line. Where does that come from?

    I'll also try and get their answer from mine.
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  4. #4
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    mr fantastic's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I sort-of-get what you've done, except the first line. Where does that come from?

    I'll also try and get their answer from mine.
    \cot A = \tan \left( \frac{\pi}{2} - A\right).

    The proof is simple:

    RHS = \tan \left( \frac{\pi}{2} - A\right) = \frac{\sin \left( \frac{\pi}{2} - A\right)}{\cos \left( \frac{\pi}{2} - A\right)} = \frac{\cos A}{\sin A} = \cot A = LHS.
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