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Math Help - arctan ?

  1. #1
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    arctan ?

    Find all the values of x which satisfy the equation

    (a) \tan^{-1}x+\tan^{-1}2x=\frac{1}{2}\pi

    (b) Show that the sum of the three angles whose tangent have the following values
    \frac{3}{8} , \frac{1}{3} , \frac{2}{19} is \frac{1}{4}\pi
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Find all the values of x which satisfy the equation

    (a) \tan^{-1}x+\tan^{-1}2x=\frac{1}{2}\pi

    (b) Show that the sum of the three angles whose tangent have the following values
    \frac{3}{8} , \frac{1}{3} , \frac{2}{19} is \frac{1}{4}\pi
    (a) Let \tan^{-1} (x) = \alpha \Rightarrow \tan \alpha = x and \tan^{-1} (2x) = \beta \Rightarrow \tan \beta = 2x.

    Then you need the value(s) of x such that

    \alpha + \beta = \frac{\pi}{2}

    \Rightarrow \sin ( \alpha + \beta ) = 1 \Rightarrow \sin \alpha \cos \beta + \sin \beta \cos \alpha = 1

    \Rightarrow \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{4x^2 + 1}} + \frac{2x}{\sqrt{4x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = 1

    \Rightarrow \frac{3x}{\sqrt{x^2 + 1} \sqrt{4x^2 + 1}} = 1

    \Rightarrow 3x = \sqrt{x^2 + 1} \sqrt{4x^2 + 1} \Rightarrow 9x^2 = 4x^4 + 5x^2 + 1

    \Rightarrow (2x^2 - 1)^2 = 0.

    Solve this equation carefully - one of the solutions is extraneous (Google it).


    I might have time for (b) later (although I'd prefer to see if you now have some ideas on how to do it ....)
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  3. #3
    MHF Contributor red_dog's Avatar
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    Use the identity: \arctan a+\arctan b=\arctan\frac{a+b}{1-ab}

    a) \arctan\frac{3x}{1-2x^2}=\frac{\pi}{2}\Rightarrow 1-2x^2=0\Rightarrow x=\pm\frac{\sqrt{2}}{2}

    The correct solution is x=\frac{\sqrt{2}}{2}

    b) arctan\frac{3}{8}+\arctan\frac{1}{3}+\arctan\frac{  2}{19}=\arctan\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{1}{8}}+\arctan\frac{2}{19}=

    =\arctan\frac{17}{21}+\arctan\frac{2}{19}=\arctan 1=\frac{\pi}{4}
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    (a) Let \tan^{-1} (x) = \alpha \Rightarrow \tan \alpha = x and \tan^{-1} (2x) = \beta \Rightarrow \tan \beta = 2x.

    Then you need the value(s) of x such that

    \alpha + \beta = \frac{\pi}{2}

    \Rightarrow \sin ( \alpha + \beta ) = 1 \Rightarrow \sin \alpha \cos \beta + \sin \beta \cos \alpha = 1

    \Rightarrow \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{4x^2 + 1}} + \frac{2x}{\sqrt{4x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = 1

    \Rightarrow \frac{3x}{\sqrt{x^2 + 1} \sqrt{4x^2 + 1}} = 1

    \Rightarrow 3x = \sqrt{x^2 + 1} \sqrt{4x^2 + 1} \Rightarrow 9x^2 = 4x^4 + 5x^2 + 1

    \Rightarrow (2x^2 - 1)^2 = 0.

    Solve this equation carefully - one of the solutions is extraneous (Google it).


    I might have time for (b) later (although I'd prefer to see if you now have some ideas on how to do it ....)

    thanks , i will give it a shot !
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  5. #5
    Senior Member pacman's Avatar
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    arctan x + arctan 2x = pi/2

    below is the graph of arctan x + arctan 2x = pi/2;

    (sqrt of 2)/2 is indeed correct
    Attached Thumbnails Attached Thumbnails arctan ?-.gif  
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  6. #6
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    Quote Originally Posted by thereddevils View Post
    Find all the values of x which satisfy the equation


    (b) Show that the sum of the three angles whose tangent have the following values
    \frac{3}{8} , \frac{1}{3} , \frac{2}{19} is \frac{1}{4}\pi
    Lets call the 3 angles a , b, and c .

    \tan [(a+b)+c] = \frac{\frac{\tan a +\tan b}{1-\tan a\tan b}+\tan c}{1-\frac{\tan a+\tan b}{1-\tan a\tan b}(\tan c)}

     <br />
=\frac{\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}+\frac{2}{19}}{1-\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}(\frac{2}{19})}=1 <br />

    and arctan 1 gives \frac{\pi}{4}

    Am i correct ?
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  7. #7
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    Quote Originally Posted by thereddevils View Post
    Lets call the 3 angles a , b, and c .

    \tan [(a+b)+c] = \frac{\frac{\tan a +\tan b}{1-\tan a\tan b}+\tan c}{1-\frac{\tan a+\tan b}{1-\tan a\tan b}(\tan c)}

     <br />
=\frac{\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}+\frac{2}{19}}{1-\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}(\frac{2}{19})}=1 <br />

    and arctan 1 gives \frac{\pi}{4}

    Am i correct ?
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