proof by induction, trigonometric identity help:

**stalker123** · August 8th 2010, 12:25 AM

first up, sorry if this is the wrong forum to post but i was unsure where to post it under.

i am trying to prove that:

$ 1 + 2* sum( cos(ku), k=1 to n) = sin( (2n+1)u/2) / sin(u/2): $
(ment to read sum( cos(ku), k=1 to n))

my induction assumption is that:
$ 1 + 2( cos(u) +.....+ cos(ku) ) = sin( ku + u/2)/sin(u/2) $

and so i am trying to reduce:
$ sin(ku + u/2)/sin(u/2) + 2cos( ku + u) $
to become
$ sin( ku + 3u/2) / sin(u/2) $

this is where i am stuck. any help would be appreciated greatly

**Also sprach Zarathustra** · August 8th 2010, 01:10 AM

A solution without using induction!

$cos(u) + cos(2u) + cos(3u) + ... + cos(nu) $

Let $T = \Sigma_{k = 1}^ {n}[ cos(ku) ]$

Multiplying by $2 sin(\frac{u}{ 2})$ :

$2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n}[2 sin(\frac{u}{ 2})cos(ku) ] $

Now, using the identity $2 sin(\frac{u}{ 2}) cos(ku) = sin( [ \frac{(2k + 1) }{ 2} ]u ) - sin( [ \frac{(2k - 1)}{ 2} ]u )$ :

$2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n}[ sin( [ \frac{(2k + 1)}{ 2 }]u ) - sin( [\frac{(2k - 1)} { 2} ]u ) ]$

Splitting the R.H.S into two separate summations, we will get:

$2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n} [ sin( [\frac{ (2k + 1)} { 2} ]u ) ] - \Sigma_{k = 1}^ {n} sin( [ \frac{(2k - 1)} { 2} ]u ) ]$

Or:

$2T sin(\frac{u}{ 2})) = \Sigma _{k = 2}^ {n } [ sin( [ \frac{(2k - 1)} { 2} ]u ) ] - \Sigma_{k = 1}^ {n} [ sin( [\frac{ (2k - 1)} { 2} ]u ) ] $

The above is telescopic series:

$2T sin(\frac{u}{ 2}) $
$= sin( [ \frac{(2n + 1)} { 2} ]u ) - sin(\frac{u}{ 2}) $
$= sin [\frac{ (2n + 1)} { 2} ]u ) - sin(\frac{u}{ 2}) $
$= 2 sin(\frac{nu }{ 2}) cos( \frac{[ n + 1 ]u} { 2} ) $

Hence:
$T =\frac{ sin(\frac{nu}{ 2}) cos( \frac{[ n + 1 ]u} { 2} ) }{sin(\frac{u}{ 2})}$

**Prove It** · August 8th 2010, 01:45 AM

First of all, let's write this in a form that is readable...

You're asked to prove that

$1 + 2\sum_{k = 1}^{n}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 1)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$ .

Base step $n = 1$ :

$LHS = 1 + 2\cos{u}$ .

$RHS = \frac{\sin{\left(\frac{3u}{2}\right)}}{\sin{\left( \frac{u}{2}\right)}}$

$= \frac{\sqrt{\frac{1 - \cos{3u}}{2}}}{\sqrt{\frac{1 - \cos{u}}{2}}}$

$= \sqrt{\frac{\frac{1 - \cos{3u}}{2}}{\frac{1 - \cos{u}}{2}}}$

$= \sqrt{\frac{1 - \cos{3u}}{1 - \cos{u}}}$

$= \sqrt{\frac{1 - (4\cos^3{u} - 3\cos{u})}{1 - \cos{u}}}$

$= \sqrt{\frac{1 + 3\cos{u} - 4\cos^3{u}}{1 - \cos{u}}}$

$= \sqrt{\frac{(1 - \cos{u})(2\cos{u} + 1)^2}{1 - \cos{u}}}$

$= \sqrt{(2\cos{u} + 1)^2}$

$= 2\cos{u} + 1$

$= LHS$ .

Inductive step - Assume the statement is true for $n$ .

So we assume $1 + 2\sum_{k = 1}^{n}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 1)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$ .

Now we need to show that the statement is true for $n + 1$ , i.e. that
$1 + 2\sum_{k = 1}^{n+1}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 3)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$ .

$LHS = 1 + 2\sum_{k = 1}^{n+1}\cos{(ku)}$ .

$RHS = \frac{\sin{\left[\frac{(2n + 3)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$

$= \frac{\sin{\left[\frac{(2n+1)u}{2} + \frac{2u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$

$= \frac{\sin{\left[\frac{(2n+1)u}{2} + u\right]}}{\sin{\left(\frac{u}{2}\right)}}$

$= \frac{\sin{\left[\frac{(2n+1)u}{2}\right]}\cos{(u)}+\cos{\left[\frac{(2n+1)u}{2}\right]}\sin{(u)}}{\sin{\left(\frac{u}{2}\right)}}$

Now I'm a little stuck too, will keep thinking about it...

**Vlasev** · August 8th 2010, 02:32 AM

Also sprach, you are looking for $1+2T$ , not just $T$ .

**Vlasev** · August 8th 2010, 02:43 AM

BTW, this sum is known as the Dirichlet Kernel Clicky!

EDIT: Ahh, double post lol. I should have just edited!

**Archie Meade** · August 8th 2010, 06:06 AM

Originally Posted by stalker123

first up, sorry if this is the wrong forum to post but i was unsure where to post it under.

i am trying to prove that:

$ 1 + 2* sum( cos(ku), k=1 to n) = sin( (2n+1)u/2) / sin(u/2): $
(ment to read sum( cos(ku), k=1 to n))

my induction assumption is that:
$ 1 + 2( cos(u) +.....+ cos(ku) ) = sin( ku + u/2)/sin(u/2) $

and so i am trying to reduce:
$ sin(ku + u/2)/sin(u/2) + 2cos( ku + u) $
to become
$ sin( ku + 3u/2) / sin(u/2) $

this is where i am stuck. any help would be appreciated greatly

P(n)

$1+2\displaystyle\huge\sum_{k=1}^ncos(ku)=\frac{sin \left[(2n+1)\frac{u}{2}\right]}{sin\left(\frac{u}{2}\right)}$

P(n+1)

$1+2\displaystyle\huge\sum_{k=1}^{n+1}cos(ku)=\frac {sin\left[(2[n+1]+1)\frac{u}{2}\right]}{sin\left(\frac{u}{2}\right)}\ ?$

Proof

$1+2\displaystyle\huge\sum_{k=1}^{n+1}cos(ku)=1+2\s um_{k=1}^ncosku+2cos[(n+1)u]$

If P(n) is valid, this will equal

$\displaystyle\huge\frac{sin\left[(2n+1)\frac{u}{2}\right]}{sin\frac{u}{2}}+2cos[(n+1)u]$

Hence, the question is

$\displaystyle\huge\frac{sin\left([2(n+1)+1]\frac{u}{2}\right)}{sin\frac{u}{2}}=\frac{sin\left ([2n+1]\frac{u}{2}\right)+\left(sin\frac{u}{2}\right)2cos[(n+1)u]}{sin\frac{u}{2}}\ ?$

$sin\left([2n+3]\frac{u}{2}\right)=sin\left([2n+1]\frac{u}{2}\right)+2cos\left([n+1]u\right)sin\frac{u}{2}\ ?$

Expanding the rightmost term

$2cos\left([n+1]u\right)sin\frac{u}{2}=sin\left([n+1]u+\frac{u}{2}\right)-sin\left([n+1]u-\frac{u}{2}\right)$

$=sin\left([2n+3]\frac{u}{2}\right)-sin\left([2n+1]\frac{u}{2}\right)$

The inter-term relationship is proven,
hence the trail of gunpowder has been placed all the way to that bank.
"What bank?"
That bank at infinity!

All that's required is to prove the equality is true for n=1..

$1+2cosu=\displaystyle\huge\frac{sin\frac{3u}{2}}{s in\frac{u}{2}}=\frac{sin\left(u+\frac{u}{2}\right) }{sin\frac{u}{2}}$

and since

$sinu=2sin\frac{u}{2}cos\frac{u}{2}$

and using

$sin(A+B)=sinAcosB+cosAsinB$

the result follows

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