1my previous ans was not correct.. dats y i had hidden it... :(
it will be a g.p. n will get the sum of it n take the imaginary part of it..
This is a very standard problem.
Many of you must have solved this one. This is not for those of you who get the 99.5%ile+ in FIITJEE Mathematics tests ;)
But for many other mortals who fear things. Pls try to solve this. It is not that tough. Only a small trick is involved.
If you dont know how to prove this one.. Pls try to make sure that you do. It is much simpler than you can think :)
We know or have heard of this sum
sin.x + sin 2x + sin 3x ............... + sin nx.
How will you solve
sin kx + sin 2kx + sin 3kx .............. + sin nkx ?
(If you look closely both are exactly the same. x<=>kx!!)
Hint Think of "Imaginary Part of eix
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22 Answers
1sum = eix ( 1 - einx) / ( 1 - eix )
multiply and divide by ( 1 - e-ix )
= (eix) ( 1 - e-ix ) (1 - enix) / 2 - 2cosx
= (eix - 1)(1 - enix) / 2 - 2cosx
= (eix - e(n+1)ix - 1 + enix) / 2 - 2cosx
Imginary part =
sin(x) - sin(n+1)x + sin(nx) / 2 - 2cosx..
Is it correct ?
62Sin x= Imaginary(eix)
Sin 2x= Imaginary(ei2x)
Sin rx = Imaginary(eirx)
Sin nx = Imaginary(einx)
Take the sum
Sinx+sin2x+......... sin nx = Imaginary (eix+ ei2x+ ......... einx)
Note that the RHS is a Geometric Progression
Take the sum oft the GP it will be something something.. (A complex number...)
Then for that complex number find the Imaginary part! (Done!)
Could someone ppost the complete solution? (Last 2 steps?!)
162not exactly ram kumar
it is different from what you have written
hint Sin rx = Imaginary(eirx)
Sum that we need is sum of all such terms..
take the sum inside and then apply imaginary
62yup but that was long back.. and not a part of the question of the day.. :)
and then we had hardly what 3 active users ? ;)
62But then the good thing is that you will learn this forever :)
And hopefully appreciate it as well :)
62Srinath and sky(i am not fully sure of what she means) but yeah they two probably know it :)
I am a bit surprised unique and Anirudha.. bcos i really felt that this was a very well known problem!
1i got this result as a conequence of trignometric derivation... but how can we use complex numbers here ?
1there we get the eiα form for every term , they are in gp . then take Sn no?
1it's in the derivation of the formula for sinα + sin(α+β) + .......+sin(α+nβ) no ?
substituting in that formula we get it no?
62Well unique it may be that.. i dont remember the sum...
But if you could fit in some complex number this dirty looking problem would look absolutely marvellous. You will start to love how closely mathematics is intermingled
And how closely everything meets together.. Be it vectors, coordinate geometry, complex numbers.. and even trigonometry! (This is one awesome example)
Anirudh.. I thought that this was a well known problem... But the trigonometric proof of it is very dirty...
It really really becomes beautiful when you conceal the dirt in a beautiful envelope of complex (I wud rather cut it short before I get too philosophical ... ;)
11What is sinx + sin 2x + sin 3x ............... + sin nx? I didn't know that it was a special sum :(