trigno

[twin yacht]

n= 1 to infinity
multiplicationsigma (1/1-tan^2(2^-n) ) = tank ; k =?
helpp. ping me plz. thanks.

[lone mason]

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[copper bear]

so you are calculating

$\tan^{-1}\qty(\prod_{n=1}^{\infty} \frac{1}{1-\tan^2(2^{-n})})$

[daring lake]

cute rizzly bear (won't eat you)

[twin yacht]

yes

[twin yacht]

did u get it?

[copper bear]

mm no

i found a slight simplification
nothing that solves it yet

my simplification is to write tan in terms of sine and cosine, simplify, and use trig identity

and then you can simplify over the whole product

[twin yacht]

yeah i did that too

[copper bear]

(i also used a calculator so i know the answer)

i just don't know how to justify it

ok i think i know

[twin yacht]

howw

[copper bear]

i gtg sorry

[twin yacht]

okay : (

[copper bear]

well im back
did you figure it out

[twin yacht]

no can u tell plz

[copper bear]

i will give you this

$\frac{1}{1 - \tan^2(2^{-n})} = \frac{\cos^2(2^{-n})}{\cos(2*2^{-n})}$

[daring lake]

cute rizzly bear (won't eat you)

[copper bear]

so then you write out a few terms of the infinite product

you can cancel some things out

[twin yacht]

and cancel out

yeah

[copper bear]

$\frac{1}{\cos\left(1\right)}\prod_{n=1}^{\infty}\cos\left(2^{-n}\right)$

[twin yacht]

(cos1/2 cos1/4 cos1/8..... ) /cos1

[daring lake]

cute rizzly bear (won't eat you)

[copper bear]

o ok you already had that

[twin yacht]

yeah then what

i'm stuck there

[copper bear]

recall that sin(2x) = 2sin(x)cos(x)

im not 100% sure on the logic here bear with me

sin(2x) = 4sin(x/2)cos(x/2)cos(x)

= 8sin(x/4)cos(x/4)cos(x/2)cos(x)

this is awfully reminiscent of the infinite product

[twin yacht]

what next? i tried a lot but ntg works.