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

(edited 11 years ago)

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Reply 1

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..

Reply 2

raheem94

Original post by yclicc

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?

Reply 3

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.

Reply 4

byfieldp12

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)

Reply 5

raheem94

Original post by marcusmerehay

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

Reply 6

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.)

(edited 11 years ago)

Reply 7

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}

Reply 8

Hasufel

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!

Reply 9

Intriguing Alias

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.

Reply 10

raheem94

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.)

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}

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!

Original post by hassi94

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...

(edited 11 years ago)

Reply 11

nuodai

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.

(edited 11 years ago)

Reply 12

yclicc

Original post by raheem94

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

The newer casio calculators can give it directly in terms of pi. I don't know if they know this one though...

Reply 13

raheem94

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?

Reply 14

nuodai

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?

Reply 15

Intriguing Alias

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.

Reply 16

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 :frown:

(edited 11 years ago)

Reply 17

raheem94

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 :tongue: