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Trignometry

Blazing goIITian

Joined:	10 Feb 2008
Post:	303

1 Apr 2008 21:38:13 IST

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379

good question

None

The value of

[A].
[B].
[C].
[D].

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Comments (5)

ram kumar

Blazing goIITian

Joined: 14 Aug 2007

Posts: 407

1 Apr 2008 22:58:38 IST

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put n=2.......

the answer is [C]........

ram kumar

Blazing goIITian

Joined: 14 Aug 2007

Posts: 407

2 Apr 2008 10:00:45 IST

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here is one method that i have found out......

LET

C;=;cosx+nC_1cos2x+nC_2cos3x+.........+nC_ncos(n+1)x

S;=;sinx+nC_1sin2x+nC_2sin3x+.........+nC_nsin(n+1)x

now

C+iS;=;(cosx+isinx)+nC_1(cos2x+isin2x)+....+cos(n+1)x+isin(n+1)x

using euler's formula... $e^{ix};=;cosx+isinx$

$C+iS;=;e^{ix}+nC_1e^{i2x}+nC_2e^{i3x}+....+e^{i(n+1)x}$

$C+iS;=;e^{ix}(1+nC_1e^{ix}+nC_2e^{i2x}+....+e^{inx})$

$C+iS\;=\;e^{ix}(1+e^{ix})^n$

see hsbhatt sir's post after this........

the answer is [C].......

sandeep ramesh

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2 Apr 2008 10:01:55 IST

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another plausible methos is grouping the first and last terms, as they have the same coefficients also :)

I still am yet to work out fully though

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

2 Apr 2008 11:02:01 IST

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small error in Ramkumar's working:

$C+iS = e^{ix} (1+\binom{n}{1}e^{ix} + \binom{n}{2}e^{2ix} + ...+e^{nix}) \\ \\ = e^{ix} (1+e^{ix})^n$

Now $1+e^{ix} = 1+\cos x + i\sin x = 2 \cos\frac{x}{2} (\cos\frac{x}{2} + i\sin \frac{x}{2})$

Which is how you get that the given expression equals

$2^n \cos^n \frac{x}{2} \cos (\frac{n+2} {2}) x$

ram kumar

Blazing goIITian

Joined: 14 Aug 2007

Posts: 407

2 Apr 2008 12:19:11 IST

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oh!!!!! i am sorry... anyway thanks hsbhatt sir for correcting me............

i have edited my post now.......

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