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Math Help - trigonometry problem

  1. #1
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    trigonometry problem

    using the substitution x= tan a, show that


    y = tan-1 2x /1+x2

    reduces to

    y = 2 tan-1 x

    this is what i have done:-


    tan y = 2x/1+x2

    tan y = 2 tan a / 1 + tan2 a

    tan y = 2 tan a / sec2 a

    tan y = 2sin a cos a = sin 2a


    i cant work this out further...can you help me plz?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slash View Post
    using the substitution x= tan a, show that


    y = tan-1 2x /1+x2

    reduces to

    y = 2 tan-1 x

    this is what i have done:-


    tan y = 2x/1+x2

    tan y = 2 tan a / 1 + tan2 a

    tan y = 2 tan a / sec2 a

    tan y = 2sin a cos a = sin 2a


    i cant work this out further...can you help me plz?
    do you mean: y = \tan^{-1} \left( \frac {2x}{1 + x^2} \right) ?

    please use the appropriate parentheses
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    do you mean: y = \tan^{-1} \left( \frac {2x}{1 + x^2} \right) ?

    please use the appropriate parentheses

    yes ur right..
    sorry im new to all this..i dont know how to use em..
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slash View Post
    yes ur right..
    sorry im new to all this..i dont know how to use em..
    first off, i think the denominator should be: 1 - x^2

    secondly, you would write it like this: y = arctan [(2x)/(1 - x^2)] .......arctan means the inverse tangent. or you could write tan^{-1} [(2x)/(1 - x^2)]

    Now here's how we do it:

    y = \tan^{-1} \left( \frac {2x}{1 - x^2} \right)

    \Rightarrow \tan y = \frac {2x}{1 - x^2}

    Let x = \tan a

    \Rightarrow \tan y = \frac {2 \tan a }{ 1 - \tan^2 a}

    But the right hand side is the double angle formula for tangent, so we get:

    \tan y = \tan 2a

    \Rightarrow y = \tan^{-1} ( \tan 2a ) = 2a

    Since, x = \tan a \implies a = \tan^{-1} x, we have:

    y = 2 \tan^{-1} x

    as desired
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  5. #5
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    Quote Originally Posted by slash View Post
    using the substitution x= tan a, show that


    y = tan-1 2x /1+x2

    reduces to

    y = 2 tan-1 x

    this is what i have done:-


    tan y = 2x/1+x2

    tan y = 2 tan a / 1 + tan2 a

    tan y = 2 tan a / sec2 a

    tan y = 2sin a cos a = sin 2a


    i cant work this out further...can you help me plz?
    You converted [1 +tan^2(a)] into sec^2(a). Although the conversion is true, it is not needed here.

    2tan(a) / 1+tan^2(a)
    is tan(2a).
    So,
    tan(y) = tan(2a)
    Get the arctans both sides,
    y = 2a -----------------------(i)

    But x = tan(a)
    So, getting the arctans of both sides,
    arctan(x) = a
    Substitute that into (i),
    y = 2[arctan(x)] --------------shown.
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  6. #6
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    Hello, slash!

    I agree with Jhevon . . . there is a typo.


    Using the substitution x\,= \,\tan a,

    show that: . y \:= \:\tan^{-1}\left(\frac{2x}{1-x^2}\right) reduces to: . y \:= \:2\tan^{-1}(x)
    . . . . . . . . . . . . . . . . . . {\color{red}\uparrow}

    We're expected to know this identity: . \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta}


    Making that substitution, we have:

    . . y \;=\;\tan^{-1}\left(\frac{2\tan a}{1 - \tan^2\!a}\right) \;=\;\tan^{-1}(\tan2a) \;=\;2a


    Since x = \tan a\!:\;\;a \,=\,\tan^{-1}(x)

    . . Therefore: . y \;=\;2\tan^{-1}(x)

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