Passage
For x ε (0 , Π/4),
Let Sn = \sum_{r=1}^{2n}{sin(sin^{-1}x^{3r-2}})
Cn = \sum_{r=1}^{2n}{cos(cos^{-1}x^{3r-1}})
Tn = \sum_{r=1}^{2n}{tan(tan^{-1}x^{3r}})
1) The correct order of Sn, Cn, Tn is
2) The value of \lim_{n\rightarrow infinity } ( S_{n} + C_{n} + T_{n} ) is equal to
(a) 11 - x (b) x1 - x
(c) 11 + x (d) x1 + x
3) the value of x for which Sn = Cn + Tn is
(a) sinÎ 5 (b) 2sinÎ 5
(c) sinÎ 10 (d) 2sinÎ 10
23sin (sin-1x ) will be x only
same for cos and tan
so there is nothing difficult here ( or m i missing something ? )
for Sn, tr=x3r-2, tr+1=x3r+1
tr+1=x3tr
so Sn=x+x4+x7+x10+..... =x(1-x6n)1-x3
for Cn, similarly tr+1=x3tr
so Cn=x2+x5+x8+x11+..... =x2(1-x6n)1-x3
similarly
Tn= x3(1-x6n)1-x3
now since x ε (0 , Π/4)
i.e x > 0 and x < 1
so x > x2 > x 3, so Sn > Cn > Tn
so Sn + Cn + Tn = (x+x2+x3)(1-x6n)1-x3 = x(1+x+x2)(1-x6n)1-x3 =x(1-x6n)1-x
so lim n ->∞ Sn + Cn + Tn = x1-x
1shouldnt da limit be zero instead of ∞? the standard limit that qwerty has used applies for n→0 and nt ∞.
23@ aditya i applied n→∞ only
fraction tends to zero when multiplied to itself infinite times