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Thread: trigo proving 4

  1. #1
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    trigo proving 4

    $\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}$

    I started with the left hand side :
    $\displaystyle

    \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}
    $

    $\displaystyle =\frac{2cos2AcosA+2cos2A}{2cos2AcosA+cos2A}$

    Then i bring out the common factor .

    $\displaystyle =\frac{2cosA+2}{2cosA+1}$

    I tried the half angle formula ..

    $\displaystyle \frac{2(2cos^2\frac{A}{2}-1+1)}{2(a-sin^2\frac{A}{2}+1)}$

    but to no avail .
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    $\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}$

    I started with the left hand side :
    $\displaystyle

    \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}
    $

    $\displaystyle =\frac{2cos2AcosA+2cos2A}{2cos2AcosA+cos2A}$

    Then i bring out the common factor .

    $\displaystyle =\frac{2cosA+2}{2cosA+1}$

    I tried the half angle formula ..

    $\displaystyle \frac{2(2cos^2\frac{A}{2}-1+1)}{2(a-sin^2\frac{A}{2}+1)}$

    but to no avail .
    Thanks for showing us your working. But I think there must be something wrong with the question. For any angle $\displaystyle A < 30^o$, the cosines of $\displaystyle A, 2A$ and $\displaystyle 3A$ are all positive. So the result can't possibly equal $\displaystyle -\cot^2(\tfrac12A)$, can it?

    When you've got as far as $\displaystyle \frac{2\cos A+2}{2\cos A+1}$, did you think about using the half-angle formula using $\displaystyle t = \tan(\tfrac12A)$? Then $\displaystyle \cos A = \frac{1-t^2}{1+t^2}$

    So $\displaystyle \frac{2\cos A+2}{2\cos A+1}= \frac{\dfrac{2(1-t^2)}{1+t^2}+2}{\dfrac{2(1-t^2)}{1+t^2}+1} = ...$ ?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello thereddevilsThanks for showing us your working. But I think there must be something wrong with the question. For any angle $\displaystyle A < 30^o$, the cosines of $\displaystyle A, 2A$ and $\displaystyle 3A$ are all positive. So the result can't possibly equal $\displaystyle -\cot^2(\tfrac12A)$, can it?

    When you've got as far as $\displaystyle \frac{2\cos A+2}{2\cos A+1}$, did you think about using the half-angle formula using $\displaystyle t = \tan(\tfrac12A)$? Then $\displaystyle \cos A = \frac{1-t^2}{1+t^2}$

    So $\displaystyle \frac{2\cos A+2}{2\cos A+1}= \frac{\dfrac{2(1-t^2)}{1+t^2}+2}{\dfrac{2(1-t^2)}{1+t^2}+1} = ...$ ?

    Grandad

    Thanks Grandad , ya , this identity seems to be untrue , it cant be negative .
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by thereddevils View Post
    $\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}$
    Isn't it $\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A-2cos2A}=-cot^2\frac{A}{2}$ ?

    In this case we have

    $\displaystyle \frac{\cos A+1}{\cos A-1}=\frac{2\cos^2\frac{A}{2}}{-2\sin^2\frac{A}{2}}=-\cot^2\frac{A}{2}$
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