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# use Maclaurin to prove pi/4=arctan(1/2)+arctan(1/3) watch

1. Hi,

thanks for any help...

I am asked to prove pi/4=arctan(1/2)+arctan(1/3). The section I am studying is all about maclaurin series so I 'm sure I am supposed to use them.

I have already found the series for
arctanx = x - (x^3)/3 + (x^5)/5 - (x^7)/7...

and subbing 1 that arctan1 = pi/4 = 1 - 1/3 + 1/5 - 1/7...
I find this to be approx. equal to 0,72381

but then if I sub 1/2 and 1/3 for x in the maclaurin series for arctanx then summing these gives me 0,785213???

2. Do you have to use maclaurin series, as it can be quickly done without it?
3. actually it doesn't say that i definitely say I need I just thought the context kind of implied it... anyway I don't know off the top of my head to to do it any other way either, but perhaps you are right in that i need to think more broadly. can you give me a hint towards another direction?
Tx
4. The direction I'd choose is, think about an argand diagram

arg Z = arctan y/x, where Z = x + iy is the complex number
5. Despite being in a chapter about Maclaurin, you're not supposed to use Maclaurin to prove the result.

You probably already know that pi/4 = arctan(1); the problem is that the Maclaurin series for pi/4 converges very slowly, so although the result is true, it's not a terribly good way of calculating pi.

So the point of a question like this is that with the formula pi/4 = arctan(1/2) + arctan(1/3) the Maclaurin series for arctan(1/2), arctan(1/3) converge relatively quickly, so it's actually quite a practical way of calculating pi/4.

In other words, it comes in the Maclaurin chapter because you can use the formula and Maclaurin series to calculate pi, but actually proving the formula has nothing at all to do with Maclaurin series.

An even nicer formula for calculating pi is

You can deduce it by calculating (1+5i)^4 and dividing by (1-i) (having noted that (1+5i)^4 is very close to a multiple of (1-i)) and then calculating arguments.
6. Alternatively:
Spoiler:
Show
and just use tan-compound-angle formulae.
7. (Original post by GHOSH-5)
Alternatively:
Spoiler:
Show
and just use tan-compound-angle formulae.
You seem to know way too much maths for your age
8. thanks all, very much, your comments are as always very! useful
9. I would do the following:
and . Then . Now see if you can get anywhere with that:

Spoiler:
Show
then .

Hope this helps w/o need for Maclaurin - which incidentally I'm looking forward to doing this year in Maths
10. (Original post by GHOSH-5)
Alternatively:
Spoiler:
Show
and just use tan-compound-angle formulae.
Which leads to a very useful identity:

11. (Original post by GHOSH-5)
Alternatively:
Spoiler:
Show
and just use tan-compound-angle formulae.
Very nice
12. GHOSH5's way is a nice way of dealing with arc trig functions in general.

The nice thing about arctan's is you can interchange it with arg z.

The reason why that's nice is because the argument behaves similar to logarithms arg A + arg B = arg AB. So you can manipulate argument forms much easier than arc functions.

so
arctan 1/2 = arg (2 + i) (check on an argand diagram for yourself for proof)
arctan 1/3 = arg(3 + i)

arg (2 + i) + arg(3 + i) = arg(2 + i)(3 + i) = arg (5 + 5i) = arctan 1 = pi/4

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