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Math Help - trigo proving 4

  1. #1
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    trigo proving 4

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

    I started with the left hand side :
     <br /> <br />
\frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}<br />

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

    Then i bring out the common factor .

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

    I tried the half angle formula ..

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

    I started with the left hand side :
     <br /> <br />
\frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}<br />

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

    Then i bring out the common factor .

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

    I tried the half angle formula ..

    \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 A < 30^o, the cosines of A, 2A and 3A are all positive. So the result can't possibly equal -\cot^2(\tfrac12A), can it?

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

    So \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 A < 30^o, the cosines of A, 2A and 3A are all positive. So the result can't possibly equal -\cot^2(\tfrac12A), can it?

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

    So \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
    \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}
    Isn't it \frac{cosA+cos3A+2cos2A}{cosA+cos3A-2cos2A}=-cot^2\frac{A}{2} ?

    In this case we have

    \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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