Pn = cos(x/2).cos(x/22). cos(x/23).................cos(x/2n)
Multiplying and diving by 2sin(x/2n)
Pn = 2/2. sin(x/2n). cos(x/2n).cos(x/2n-1)....... cos(x/22).cos(x/2) . cosec(x/2n)
similarly, if u multiply 2, n times, then by the formula 2sinx.cox= sin2x, it'll finally become
= 1/2n sinx. cosec(x/2n)
Now if u carefully observe the summation inside the limit, then u'll see that it is obtained when the log of Pn is differenciated. Hence, if u take log of Pn
log [ cos(x/2).cos(x/22).................cos(x/2n)] = log (1/2n .sinx. cosec (x/2n))
log(cos(x/2)) +logcos(x/22))+....................log cos(x/2n)) = logsinx - log 2n.sin(x/2n)
differenciating both sides,
-1/2tan(x/2) -1/22. tan(x/22) - 1/23 .tan(x/23) -..............-1/2n.tan(x/2n)
= cotx - 1/2n. cos(x/2n)/sin(x/2n)
Therefore
[n]
[ infinity] S = [n][infinity] -[ cotx - 1/2n. cos(x/2n)/sin(x/2n) ] where S is the sum that is obtained after differenciatiion.
Hence limit L = -cotx + [n][infinity] (1/x). (x/2n). cos(x/2n) / sin(x/2n)
Now as n tends to infinity, hence 2n will tend towards 0.
By using
[x][0] sinx/x=1 we get,
L = -cotx +1/x
Hope this helps,
cheers!!!
Moderation Team
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198
infinity r=1]
n (1/2r)tan(x/2r)=(1/x)-cotx
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