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Math Help - trig identity

  1. #1
    Super Member bigwave's Avatar
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    trig identity

    prove that:

     <br />
\tan^{-1}{\left(\frac{1}{5}\right)}<br />
+\tan^{-1}{\left(\frac{2}{3}\right)}<br />
=\frac{\pi}{4}<br />

    first, the coordinates of  \frac{\pi}{4} \Rightarrow \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)<br />

    and the coordinates of
    <br />
\tan^{-1}\left(\frac{1}{5}\right)<br />
\Rightarrow \left(\frac{5}{\sqrt{26}},\frac{1}{\sqrt{26}}\righ  t)<br />
    and
    <br />
\tan^{-1}\left(\frac{2}{3}\right)<br />
\Rightarrow \left(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\righ  t)<br />

    so thot from these that using \cos{\left(\alpha + \beta\right)} I could get the coordinates of \frac{\pi}{4}

    but did not unless I made an error somewhere....
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by bigwave View Post
    prove that:

     <br />
\tan^{-1}{\left(\frac{1}{5}\right)}<br />
+\tan^{-1}{\left(\frac{2}{3}\right)}<br />
=\frac{\pi}{4}<br />

    first, the coordinates of  \frac{\pi}{4} \Rightarrow \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)<br />

    and the coordinates of
    <br />
\tan^{-1}\left(\frac{1}{5}\right)<br />
\Rightarrow \left(\frac{5}{\sqrt{26}},\frac{1}{\sqrt{26}}\righ  t)<br />
    and
    <br />
\tan^{-1}\left(\frac{2}{3}\right)<br />
\Rightarrow \left(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\righ  t)<br />

    so thot from these that using \cos{\left(\alpha + \beta\right)} I could get the coordinates of \frac{\pi}{4}

    but did not unless I made an error somewhere....
    Hmm, the way I'm coming up with is pretty long.

    Looking at the unit circle, you can define point P at \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) and Q at \left(\frac{5\sqrt{26}}{26},\frac{\sqrt{26}}{26}\r  ight). O is the origin. Then find the line perpendicular to line OQ going through P, call it line m. Find the intersection of m and OQ, call it point Z. Then verify that the length of line segement PZ divided by the length of line segment OZ is 2/3.

    Someone care to share a better way?
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  3. #3
    MHF Contributor undefined's Avatar
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    Okay, I should have focused more on the angle sum identity than on the unit circle aspect. We have this identity for tangent (we could also use the ones for sine and cosine, but this seems faster):

    \tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}

    Here, \tan\alpha = \frac{1}{5} and \tan\beta = \frac{2}{3}.

    So \tan(\alpha + \beta) = \frac{\frac{1}{5} + \frac{2}{3}}{1 - \frac{1}{5} \cdot \frac{2}{3}}=1

    \alpha+\beta=\tan^{-1}(1)=\frac{\pi}{4}

    Much simpler than my first post.
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  4. #4
    Super Member bigwave's Avatar
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    thanks, I see how you got it...

    I did go back to the \cos\left(\alpha + \beta\right)

    we have:

    <br />
\Rightarrow <br />
\frac{5}{\sqrt{26}}\frac{2}{\sqrt{13}}<br />
+\frac{1}{\sqrt{26}}\frac{3}{\sqrt{13}}<br />
\Rightarrow<br />
\frac{10}{13\sqrt{2}}+\frac{3}{13\sqrt{2}}<br />
\Rightarrow<br />
\frac{13}{13\sqrt{2}}<br />
\Rightarrow<br />
\frac{1}{\sqrt{2}} \Rightarrow \frac{\sqrt{2}}{2} = \frac{\pi}{4}<br />

    looks like just needed to be more carefull
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