Question

prove that \[\cos(\cos(\cos......\cos(\theta ))))))...)=1 \]

Answer 1

no...the question is right...

Answer 2

Use the cosine expansion?
\[\cos(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{(2k)!}\]

Answer 3

in the problem i have used infinite times cos(cos.... so can u solve by that way...!!!!!

Answer 4

how can it be infinite if theres a starting point and an ending point? Clearly there is one cos on the outside and one on the very very inside.
If you treat it as an infinite amount im getting 1 = 0, which is why i thought it was finite instead.

Answer 5

this problem will be solved analytically so go for that don't think of such big formulas....
if u have done the cosine problem then do the same using sine and prove that in that case answer will be zero.

Answer 6

how do u find 1=0 can u proveit???

Answer 7

*prove it

Answer 8

have u done the previous problem???

Answer 9

If it was infinite, i was doing this:
\[\cos(\cos(\cos(\ldots \cos(\theta)\ldots)))) = 1 \]
take arccos of both sides:
\[\cos(\cos(\cos(\ldots \cos(\theta)\ldots))) = \cos^{-1}(1)\]
If there were an infinite amount of cosines, then the left hand side is the same as the very beginning of the problem, and it equals 1:
\[1 = \cos^{-1}(1) \iff 1 = 0\]
Thats why i assumed it wasnt infinite, but finite.

Answer 10

i havent done this problem before, but ive seen problems like:
\[x^{x^{x^{x^{\ldots}}}} = 2\]
which uses a similar technique.

Answer 11

abywhos, i dont believe there is an infinite amount of cosines.

Answer 12

anywhos*

Answer 13

does anyone agree/disagree with me? any opinions on the matter? lolol

Answer 14

the problem as stated is not true

Answer 15

right, i think that as well.

Answer 16

okay....wait...in case of sine how would you proceed if i ask to find the value of that function.

Answer 17

why have you taken theta in the range .

Answer 18

what do you mean by

'in case of sine how would you proceed if i ask to find the value of that function'

Answer 19

find the value of sin(sin......sin(a))))...)

Answer 20

0

Answer 21

how ??

Answer 22

any reason...?

Answer 23

show that after the first application of sine the sequence is monotone.

it will be obviously bounded. Monotone bounded sequences converge.

Answer 24

exactly and by the same rule when the arguments become zero then cosine gives 1.

Answer 25

do u agree with me?

Answer 26

no

Answer 27

the infinite regression of cosines give an answer other than 1

Answer 28

why not ??

Answer 29

why don't you try a few values of theta to see for yourself

Answer 30

it will converge to a number that is approx ...

0.73908513321515

Answer 31

the solution to the equation \[\cos(x)=x\]

Answer 32

yes that's true ...i made a mistake...actually i compared with the sine one. you are right also the Gauss iteration formula also gives a solution like that of u

Answer 33

okay its very nice to deal with you. hope we will meet again thanks both of you guys ...good job.