lagrange's trignometric identity

**flower3** · July 5th 2009, 09:39 PM

use the identity $1+ z+ {z^2} + {z^3}+ {z^4}+...+ {z^n} = \frac{1-{z^ (n+1)} }{1-z} , z \neq 0$ to prove lagrange's trignometric identity :
$1+ cos\ominus + cos2\ominus+...+ cosn\ominus = \frac {1}{2} + \frac {sin (n+\frac {\ominus}{2})}{2sin(\frac{\ominus}{2})} , 0< \ominus < 2 \pi$

**Moo** · July 5th 2009, 11:28 PM

Hello,

We know that the cosine is the real part of $e^{i\theta}$

So by some changes, $\cos n\theta=\Re((e^{i\theta})^n)$

We then have :
$F=1+\cos\theta+\cos 2\theta+\dots+\cos n\theta=\Re\left(1+e^{i\theta}+(e^{i\theta})^2+\do ts+(e^{i\theta})^n\right)$

Which is, by the identity you're given, equal to :

$F=\Re\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\Re\left(\frac{1-\cos((n+1)\theta)-i\sin((n+1)\theta)}{1-\cos\theta-i\sin\theta}\right)$

Now, since you want to take the real part of this fraction, transform the denominator into a real number : multiply the numerator and the denominator by the complex conjugate of $1-\cos\theta-i\sin\theta$ , which is $1-\cos\theta+i\sin\theta$

By using the identity $(a+ib)(a-ib)=a^2+b^2$ , we get in the denominator : $(1-\cos\theta)^2+\sin^2\theta=1+\cos^2\theta-2\cos\theta+\sin^2\theta=2-2\cos\theta=2(1-\cos\theta)$

And in the numerator, using basic algebraic operations and rearranging, we get :
$1-\cos\theta-\cos((n+1)\theta)+{\color{red}\cos((n+1)\theta)\co s\theta+\sin((n+1)\theta)\sin\theta}$ $+i \left[\sin\theta-\sin((n+1)\theta)-\sin\theta\cos((n+1)\theta)+\cos\theta\sin((n+1)\t heta)\right]$

Since we only want the real part, we may only keep the first line.

And then the red part is exactly $\cos((n+1)\theta-\theta)=\cos n\theta$

So finally, we have :

$F=\frac{1-\cos\theta-\cos((n+1)\theta)+\cos(n\theta)}{2(1-\cos\theta)}=\frac 12+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2(1-\cos\theta)}$

By the half angle formula, we have $1-\cos\theta=2\sin^2 \tfrac \theta 2$

And by the identity $\cos a-\cos b=-2\sin\tfrac{a+b}{2}\sin\tfrac{a-b}{2}=2\sin\tfrac{a+b}{2}\sin\tfrac{b-a}{2}$ (because the sine function is odd)

So $\cos(n\theta)-\cos((n+1)\theta)=2\sin\left(\tfrac{2n+1}{2} \cdot\theta\right)\sin\tfrac\theta 2=2\sin\left(n\theta+\tfrac \theta 2\right)\sin\tfrac\theta 2$

And we can simplify... :

$F=\frac 12+\frac{\sin\left(n\theta+\tfrac \theta 2\right)}{2\sin\tfrac \theta 2}$

Note : Some steps have been left to your understanding...

**simplependulum** · July 6th 2009, 01:57 AM

$\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}} = \frac{ 1-e^{i(n+1)\theta}}{(e^{i \frac{\theta}{2}}) ( e^{- i \frac{\theta}{2}} - e^{i \frac{\theta}{2}})}$

$= \frac{ e^{-i \frac{\theta}{2}} - e^{i(n+\frac{1}{2})\theta}}{-2i \sin{\frac{\theta}{2}}}$

The series is equal to the real part of it , eg

$\frac{1}{2} + \frac{ \sin{(n+\frac{1}{2})\theta}}{2\sin{\frac{\theta}{2 }}}$

**flower3** · July 7th 2009, 10:11 AM

thanks for moo"i understand your proof

" and simplependulum

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