I found a brilliant math conclusion

Revision en1, by Indialbedo, 2022-04-24 14:40:19

$$$\sum_{i=1}^n\sin(ix)=\dfrac{\cos(x/2)-\cos((2n+1)x/2)}{2\sin(x/2)}$$$ $$$\sum_{i=1}^n\sin(ix)=\dfrac{\sin(\frac{n+1}2x)\sin(\frac n 2x)}{\sin(\frac x 2)}$$$

Now I give my proof.

notice that $$$\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$$$

$$$\sum_{i=1}^n\sin(ix)=\dfrac{\sin(x/2)(\sin(x)+\sin(2x)+\sin(3x)+\ldots)}{\sin(x/2)}$$$ $$$=\dfrac{\cos(x/2)-\cos(3x/2)+\cos(3x/2)-\cos(5x/2)\ldots+\cos((2n+1)x/2)}{2\sin(x/2)}$$$ $$$=\dfrac{\cos(x/2)-\cos((2n+1)x/2)}{2\sin(x/2)}$$$

inverse $$$\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$$$ to be $$$\cos(a)-\cos(b)=-2\sin(\frac{a+b}2)\sin(\frac{a-b}2)$$$

$$$\text{LHS}=\dfrac{\sin(\frac{n+1}2x)\sin(\frac n 2x)}{\sin(\frac x 2)}$$$

maybe a nice idea! What do u think?