Exact value of arctan[1+sqrt(2)]

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  1. raheem94's Avatar
    1 30-05-2012  18:05
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    Exact value of arctan[1+sqrt(2)]
    Hi,

    I was doing an integration question, which required use of substitution. I took a substitution which made the working very long.

    The answer is \displaystyle \frac{\pi}{32}

    I get the answer as \displaystyle \frac14 \left( arctan(1+ \sqrt2 ) - \frac{\pi}{4} \right)

    Now i have no idea how to convert arctan(1+ \sqrt2) into an exact value in terms of \pi

    Though wolfram is able to do this, it gives, arctan(1+ \sqrt2) = \dfrac{3 \pi}8

    I have tried the easier substitution and got the right answer, but i just want to know how can i make this more difficult approach work.

    Thanks in advance
    Last edited by raheem94; 30-05-2012 at 18:19.
  2. yclicc's Avatar
    2 30-05-2012  18:32
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    Re: Exact value of arctan[1+sqrt(2)]
    If you substitute wolframalpha's result into your equation you get the right answer.

    I don't think there is a direct way to prove this value of arctan. You can always try entering it into a calculator and dividing by pi..
  3. raheem94's Avatar
    3 30-05-2012  18:42
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by yclicc)
    If you substitute wolframalpha's result into your equation you get the right answer.

    I don't think there is a direct way to prove this value of arctan. You can always try entering it into a calculator and dividing by pi..
    Thanks for your reply.

    I know wolframalpha's result gives the right answer, but how does wolfram calculates it?

    For artanh we have a formula, is there a similar formula for arctan, any idea?
  4. marcusmerehay's Avatar
    4 30-05-2012  18:47
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    Re: Exact value of arctan[1+sqrt(2)]
    Unless you happen to know the tan function very well, I don't think there's an 'easy' way of doing it.

    If you're allowed a calculator then it's a simple matter of checking against different multiples of pi really.
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  5. byfieldp12's Avatar
    5 30-05-2012  18:49
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    Re: Exact value of arctan[1+sqrt(2)]
    No, as tanh(x) is defined as ((e^2x)-1)/((e^2x)+1) which is arranged to make x the subject, giving a formula for artanh(x). This is not the case with tan(x)/arctan(x), although they may be approximated to a polynomial by Maclaurin/taylor series, which is how a computer can calculate values of tan(x)/arctan(x)
  6. raheem94's Avatar
    6 30-05-2012  18:52
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by marcusmerehay)
    Unless you happen to know the tan function very well, I don't think there's an 'easy' way of doing it.

    If you're allowed a calculator then it's a simple matter of checking against different multiples of pi really.
    Ok, thanks.

    I was thinking that there might be a formula for it.

    I just did \displaystyle arctan( 1 + \sqrt2) = 1.178 \ldots = \pi \times \frac{ 1.178 \ldots }{ \pi } = \pi \times 0.375 = \frac{3\pi}8
  7. nuodai's Avatar
    7 30-05-2012  19:02
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    Re: Exact value of arctan[1+sqrt(2)]
    The + hints at using some kind of half-angle formula (it's not like to be a double-angle formula because they normally give fractions, for tan at least).

    You can write \tan \theta = \tan \dfrac{\varphi}{2} where \varphi = 2\theta. Using the formula \tan \dfrac{\varphi}{2} = \operatorname{cosec} \varphi - \cot \varphi, we have

    \operatorname{cosec} \varphi - \cot \varphi = 1 + \sqrt{2}

    We don't know any tan~cot angles that give \sqrt{2}, so it looks like we should use the sin~cosec for that. We're done if we can find \varphi such that \sin \varphi = \dfrac{1}{\sqrt{2}} and \tan \varphi = -1. (And we can.)
    Last edited by nuodai; 30-05-2012 at 19:04.
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  8. jack.hadamard's Avatar
    8 30-05-2012  19:07
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    Re: Exact value of arctan[1+sqrt(2)]
    You can derive all \displaystyle \tan\left(\frac{\pi}{2^n}\right) by making use of \displaystyle\tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} and \displaystyle \tan^2\theta = \frac{1 - \cos2\theta}{1 + \cos2\theta}.
  9. Hasufel's Avatar
    9 30-05-2012  19:13
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    ok, thanks.

    I was thinking that there might be a formula for it.

    I just did \displaystyle arctan( 1 + \sqrt2) = 1.178 \ldots = \pi \times \frac{ 1.178 \ldots }{ \pi } = \pi \times 0.375 = \frac{3\pi}8
    ooh! Nice one!
  10. Intriguing Alias's Avatar
    10 30-05-2012  19:15
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    • Location: Yorkshire
    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    Hi,

    I was doing an integration question, which required use of substitution. I took a substitution which made the working very long.

    The answer is \displaystyle \frac{\pi}{32}

    I get the answer as \displaystyle \frac14 \left( arctan(1+ \sqrt2 ) - \frac{\pi}{4} \right)

    Now i have no idea how to convert arctan(1+ \sqrt2) into an exact value in terms of \pi

    Though wolfram is able to do this, it gives, arctan(1+ \sqrt2) = \dfrac{3 \pi}8

    I have tried the easier substitution and got the right answer, but i just want to know how can i make this more difficult approach work.

    Thanks in advance
    Whilst replies from Nuodai and Jacques etc. are great - surely your calculator should be able to give you an exact answer (assuming this is a-level)? Which is obviously a lot easier in an exam situation.
  11. raheem94's Avatar
    11 30-05-2012  19:36
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by nuodai)
    The + hints at using some kind of half-angle formula (it's not like to be a double-angle formula because they normally give fractions, for tan at least).

    You can write \tan \theta = \tan \dfrac{\varphi}{2} where \varphi = 2\theta. Using the formula \tan \dfrac{\varphi}{2} = \operatorname{cosec} \varphi - \cot \varphi, we have

    \operatorname{cosec} \varphi - \cot \varphi = 1 + \sqrt{2}

    We don't know any tan~cot angles that give \sqrt{2}, so it looks like we should use the sin~cosec for that. We're done if we can find \varphi such that \sin \varphi = \dfrac{1}{\sqrt{2}} and \tan \varphi = -1. (And we can.)
    :ta:
    You are a genius!

    (Original post by jack.hadamard)
    You can derive all \displaystyle \tan\left(\frac{\pi}{2^n}\right) by making use of \displaystyle\tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} and \displaystyle \tan^2\theta = \frac{1 - \cos2\theta}{1 + \cos2\theta}.
    :ta:
    I am not quite sure how i will apply your method to my question and will it take long time?

    (Original post by Hasufel)
    ooh! Nice one!
    :ta:

    (Original post by hassi94)
    Whilst replies from Nuodai and Jacques etc. are great - surely your calculator should be able to give you an exact answer (assuming this is a-level)? Which is obviously a lot easier in an exam situation.
    I did one way above which gave the answer in terms of \pi , but can the calculator directly find the answer in terms of \pi
    Though i hope to avoid such stupid mistakes in exam, the question could have been done very much quickly had i taken a different substitution but the substitution i took required too much working and at the end there was the arctan case...
    Last edited by raheem94; 30-05-2012 at 19:37.
  12. nuodai's Avatar
    12 30-05-2012  19:52
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    I am not quite sure how i will apply your method to my question and will it take long time?
    In fact all you need is the first identity. If you rearrange

    \tan 2\theta = \dfrac{2\tan \theta}{1-\tan^2 \theta}

    then, using the quadratic formula, you get

    \tan \theta = \dfrac{-1 \pm \sqrt{1+\tan^2 2\theta}}{\tan 2\theta}

    If we write t_n = \tan \dfrac{\pi}{2^n} (for n \ge 2), then we must have t_n > 0 and so we take the + sign, and the above equation gives

    t_n = \dfrac{-1 + \sqrt{1+t_{n-1}^2}}{t_{n-1}}

    which is quite nice.
    Last edited by nuodai; 30-05-2012 at 19:54. Reason: LaTeX tags...
  13. yclicc's Avatar
    13 30-05-2012  20:25
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    :ta:
    You are a genius!



    :ta:
    I am not quite sure how i will apply your method to my question and will it take long time?



    :ta:



    I did one way above which gave the answer in terms of \pi , but can the calculator directly find the answer in terms of \pi
    Though i hope to avoid such stupid mistakes in exam, the question could have been done very much quickly had i taken a different substitution but the substitution i took required too much working and at the end there was the arctan case...
    The newer casio calculators can give it directly in terms of pi. I don't know if they know this one though...
  14. raheem94's Avatar
    14 30-05-2012  20:27
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by yclicc)
    The newer casio calculators can give it directly in terms of pi. I don't know if they know this one though...
    I have fx-991ES, can this one give answers in terms of pi?
  15. nuodai's Avatar
    15 30-05-2012  20:38
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    I have fx-991ES, can this one give answers in terms of pi?
    Yes - it does by default, doesn't it?
  16. Intriguing Alias's Avatar
    16 30-05-2012  20:44
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    I have fx-991ES, can this one give answers in terms of pi?
    Yep this should give answers in terms of pi by default. Make sure you're not in Stat mode.
  17. JonathanM's Avatar
    17 30-05-2012  20:45
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    Re: Exact value of arctan[1+sqrt(2)]
    If it doesn't, just use radians and divide your answer by Pi and stick pi on the end of it. Or use degrees and convert yourself.

    What exam board are you? As I do edexcel and I don't know about the tanh stuff .
    Last edited by JonathanM; 30-05-2012 at 20:47.
  18. raheem94's Avatar
    18 30-05-2012  20:48
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by JonathanM)
    If it doesn't, just use radians and divide your answer by Pi and stick pi on the end of it. Or use degrees and convert yourself.
    (Original post by hassi94)
    Yep this should give answers in terms of pi by default. Make sure you're not in Stat mode.
    (Original post by nuodai)
    Yes - it does by default, doesn't it?
    Oh yes, it does it.

    I never use the calc on Mthlo mode, i always prefer LineIO, i just set it into the other and it gave the answer in terms of pi.

    Regarding the stat mode, i never use it, i did the stats exam but i never tried to take some time to learn how to use the stats mode
  19. raheem94's Avatar
    19 30-05-2012  20:49
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by JonathanM)
    If it doesn't, just use radians and divide your answer by Pi and stick pi on the end of it. Or use degrees and convert yourself.

    What exam board are you? As I do edexcel and I don't know about the tanh stuff .
    I also do edexcel, but this tanh stuff is in FP3.
  20. Intriguing Alias's Avatar
    20 30-05-2012  20:59
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    Re: Exact value of arctan[1+sqrt(2)]
    (Original post by raheem94)
    Oh yes, it does it.

    I never use the calc on Mthlo mode, i always prefer LineIO, i just set it into the other and it gave the answer in terms of pi.

    Regarding the stat mode, i never use it, i did the stats exam but i never tried to take some time to learn how to use the stats mode
    Not going to lie, I'm a budding mathematician and a tech fanatic but I have no idea what the modes on a calculator are whatsoever
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