Prove that \sqrt [44] {\tan 1^{\circ} \tan 2^{\circ}...\tan 44^{\circ}} < \sqrt 2 -1 <\frac{\tan 1^{\circ} + \tan 2^{\circ}+...+\tan 44^{\circ}}{44}
1sir first and last toh can be... but how abt √2 - 1??
i have been working on this problem since it was given but to no effect... how can i proceed??
1sir i got that...
990 = 22.5 + 22.5 + 22.5 + ....(44 times)
Also a1a2a3...an is maximum when a1=a2=a3=...=an
So tan1°tan2°tan3°...tan44° < tan22.5°tan22.5°....(44times)
or tan1°tan2°tan3°...tan44° < (tan22.5)44
or tan1°tan2°tan3°...tan44° < (√2 - 1)44
or ( tan1°tan2°tan3°...tan44°)144 < √2 - 1
Similarly the right inequality can be proved!!
1A Simple Observation -
( tan θ ) x tan ( 45 - θ ) = 2
Does That Help ? I Think It Does !
341There is such an air of mystery about this that I am tempted to say, "The game is afoot Watson!"
But hey, why cant someone make an honest attempt like Euclid did?
@Ricky: how on earth did you come up with that identity \tan \theta \times \tan \left(\frac{\pi}{4} - \theta \right) = 2?
Putting θ = 0 send that identity packing.
Sorry if I am sounding nasty, but I want things to get going.
62The RHS can be proved simply by Jenson's Inequality....
Still thinking on the Left hand side..[12]
341yesssssir! think we need to start a course on intellectual honesty?
1Well, that was a good try by Euclid. I thought that we have to use the AM-GM Inequality(after looking at RHS and LHS of the Inequality).
Now I have to learn Jenson's Inequality so as to get RHS...
341This part a_1a_2...a_n is max when a_1=a_2=...=a_n
needs some attendant conditions like they have all to be +ve and a constraint that a_1+a_2+...+a_n is a constant.
So, its not clear how your steps come about. (Though the final solution will look pretty similar to this!)
62Cracked the LHS as well... dont know why this same approach wasnt working yesterday :P
Take ln (tan x) and apply Jensen's [1] :D
62:)
now for the rest.. specially not knowing Jensen's (which is my most favorite inequality)
Read these two.. there is Jensen's inequality...
http://www.targetiit.com/iit-jee-forum/posts/max-or-min-16090.html
http://www.targetiit.com/iit-jee-forum/posts/solve-using-graph-13875.html
For Detailed theory use:
http://www.math.ust.hk/excalibur/v5_n4.pdf
1My corrected simple observation -
For proper " x " , we must have
( 1 + tan x ) [ 1 + tan ( 45 - x ) ] = 2
So apply AM - GM on the numbers , 1 + tan 1 ; 1 + tan 2 , .................1 + tan 44 .
We directly get the result of the extreme right .
Nowhere near Nishant sir's Brilliance , though ..............
62good work on that one ricky :)
even i was surprised to see your first post ;)