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Math Help - Double angle question

  1. #1
    RAz
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    Double angle question

    The question is

    If tanA=4 and tabB=3/5, and A and B are acute angles, prove that A-B=pi/4 (pi in radians, so really it's 45degrees).

    My answer is:

    tan(x-y)=\frac {tan4-tan \frac {3}{5}}{1 + tan4 * tan \frac {3}{5}}<br />
=1<br />
\therefore tan(x-y)=1=\frac {\pi}{4}

    I have a feeling there is more to add. My lecturer said something about 135 being a possible answer, and I was wondering why the question included "acute"; as in first quadrant?
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  2. #2
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    Hello RAz
    Quote Originally Posted by RAz View Post
    The question is

    If tanA=4 and tabB=3/5, and A and B are acute angles, prove that A-B=pi/4 (pi in radians, so really it's 45degrees).

    My answer is:

    \color{red}tan(x-y)=\frac {tan4-tan \frac {3}{5}}{1 + tan4 * tan \frac {3}{5}}\color{black}<br />
=1<br />
\therefore tan(x-y)=1\color{red}=\frac {\pi}{4}

    I have a feeling there is more to add. My lecturer said something about 135 being a possible answer, and I was wondering why the question included "acute"; as in first quadrant?
    Your general method is OK, but your use of notation is very sloppy. Say what you mean! What you mean, of course, is:
    \tan(A-B)=\frac {\tan A-\tan B}{1 + \tan A \tan B} = \frac{4-\tfrac35}{1+4\cdot\tfrac35}= 1

    \Rightarrow A-B = \frac{\pi}{4}
    I can't see where the possible angle of 135^o might come from. \tan135^o = -1, not +1. And if A and B are acute with \tan A > \tan B, then A>B. So A-B is also acute. Hence, \tan(A-B) = 1\Rightarrow A-B = 45^o.

    Grandad
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  3. #3
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    If your teacher meant 225^\circ, I can see the possibility but definitely not 135^\circ
    Last edited by Punch; February 16th 2010 at 06:30 AM.
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