# $4\sin^2\frac{\theta}{2}.S=(n+1)\sin n\theta-n \sin (n+1)\theta$, and $4\sin^2\frac{\theta}{2}.C=-1+(n+1)\cos n\theta-n \cos (n+1)\theta$

If $S\equiv \sin\theta+2\sin2\theta+3\sin3\theta+......+n\sin n\theta$ and $C\equiv \cos\theta+2\cos2\theta+3\cos3\theta+......+n\cos n\theta$,prove that

$4\sin^2\frac{\theta}{2}.S=(n+1)\sin n\theta-n \sin (n+1)\theta$,

and $4\sin^2\frac{\theta}{2}.C=-1+(n+1)\cos n\theta-n \cos (n+1)\theta$

My try:$4\sin^2\frac{\theta}{2}.S$
$=2(1-\cos \theta)[\sin\theta+2\sin2\theta+3\sin3\theta+......+n\sin n\theta]$
$=2[\sin\theta+2\sin2\theta+3\sin3\theta+......+n\sin n\theta-\cos\theta\sin\theta-2\cos\theta\sin2\theta+.....-n\cos\theta\sin n\theta]$

HINT:

$$2r(1-\cos y)\cdot\sin ry=2r\sin ry-r(2\sin ry\cos y)$$ $$=2r\sin ry-r\{\sin(r+1)y-\sin(r-1)y\}$$

$$=-r\{\sin(r+1)y-\sin ry\}+r\{\sin ry-\sin(r-1)y\}$$

Do you know about Telescoping Series?

Set $r=1,2,3,\cdots,n-1,n$ and add

Similarly, $$2r(1-\cos y)\cdot\cos ry=2r\cos ry-r(2\cos ry\cos y)=?$$

HINT:

Using Euler's formula, $$S_n=C+iS=\sum_{r=1}^n re^{ir\theta}$$

Now using Arithmetic-Geometric Series, $$S_n=\dfrac{1-(n+1)e^{in\theta}+ne^{i(n+1)\theta}}{(1-e^{i\theta})^2}$$

Simplify and equate the real & the imaginary parts.