# Trigonometric Identity

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• Aug 29th 2008, 09:56 AM
PersaGell
Trigonometric Identity
Hi, I need help on this problem:

Use the identity $2\sin{\frac{x}{2}}\cos{kx} =
\sin{(2k+1)}\frac{x}{2} - \sin{(2k-1)}\frac{x}{2}$
and the telescoping property of finite sums to prove that if $x \neq 2m\pi$ (m an integer), we have $\sum_{k=1}^{n} \cos{kx} = \frac{\sin{\frac{nx}{2}}\cos{\frac{1}{2}(n+1)x}}{\ sin{\frac{1}{2}x}}$

Here is what I have so far:

Taking $k= 1,2,...,n$
$2\sin{\frac{x}{2}}\cos{kx}= 2\sin{\frac{x}{2}}\sum_{k=1}^{n}\cos{kx}$ =

$[\sin{\frac{3x}{2}}-\sin{\frac{x}{2}} ] +$ $[\sin{\frac{5x}{2}}-\sin{\frac{3x}{2}}]+ ... + [\sin{\frac{(2n-1)x}{2}}-\sin{\frac{(2n-3)x}{2}}] +$ $[\sin{\frac{(2n+1)x}{2}} -\sin{\frac{(2n-1)x}{2}}]$

$= \sin{\frac{(2n+1)x}{2}} - \sin{\frac{x}{2}}$ = $\sin{(n+ \frac{1}{2})x } - \sin{\frac{x}{2}}$

This reduces to $\sin{nx}\cos{\frac{x}{2}} + \cos{nx}\sin{\frac{x}{2}} -\sin{\frac{x}{2}}$

Now I'm stuck.
• Aug 29th 2008, 10:07 AM
Moo
Hi !
Quote:

Originally Posted by PersaGell
Hi, I need help on this problem:

Use the identity $2\sin{\frac{x}{2}}\cos{kx} =
\sin{(2k+1)}\frac{x}{2} - \sin{(2k-1)}\frac{x}{2}$
and the telescoping property of finite sums to prove that if $x \neq 2m\pi$ (m an integer), we have $\sum_{k=1}^{n} \cos{kx} = \frac{\sin{\frac{nx}{2}}\cos{\frac{1}{2}(n+1)x}}{\ sin{\frac{1}{2}x}}$

Here is what I have so far:

Taking $k= 1,2,...,n$
${\color{red}\sum_{k=1}^n} 2\sin{\frac{x}{2}}\cos{kx}= 2\sin{\frac{x}{2}}\sum_{k=1}^{n}\cos{kx}$ =

$[\sin{\frac{3x}{2}}-\sin{\frac{x}{2}} ] +$ $[\sin{\frac{5x}{2}}-\sin{\frac{3x}{2}}]+ ... + [\sin{\frac{(2n-1)x}{2}}-\sin{\frac{(2n-3)x}{2}}] +$ $[\sin{\frac{(2n+1)x}{2}} -\sin{\frac{(2n-1)x}{2}}]$

$= \sin{\frac{(2n+1)x}{2}} - \sin{\frac{x}{2}}$

(There's the red part that was missing :))

Once you're here, try to think "how can I transform a difference of sines into a product of a sine and a cosine ?" and then use this formula (which is quite the same as the identity you're given at the very beginning) :

$\sin(a)-\sin(b)=2 \cos \left(\frac{a+b}{2} \right) \cdot \sin \left(\frac{a-b}{2}\right)$

And you're done :)

(this formula is available, as well as many others, here : Trigonometry)
• Aug 29th 2008, 01:35 PM
ThePerfectHacker
It is even easier.
Use complex numbers and geometric series.
• Aug 29th 2008, 05:01 PM
Shyam
Your last line is wrong, which is
$
\sin{nx}\cos{\frac{x}{2}} + \cos{nx}\sin{\frac{x}{2}} -\sin{\frac{x}{2}}
$

See here,
$
=\sin{(n+ \frac{1}{2})x } - \sin{\frac{x}{2}}
$

$
=2\cos{(\frac{n+1}{2})x }\sin{\frac{nx}{2}}
$