KOPMOPAH написа:Да се намерят сумите:$$S=\sum_{k=0}^{n}\sin(y+kx) \\ C=\sum_{k=0}^{n}\cos(y+kx)$$
[tex]S = \sum_{k=0}^{n }sin(y + kx) = siny + sin(y + x) + sin(y + 2x) + sin(y + 3x) + ..........sin(y + nx) =[/tex]
[tex]= siny + siny.cosx + cosy.sinx + siny.cos2x + cosy.sin2x + siny.cos3x + sin3x.cosy + ........... + siny.cosnx + sinnx.cosy =[/tex]
[tex]= siny(1 + cosx + cos2x + cos3x....+ cosnx) + cosy(sinx + sin2x + sin3x + ......+ sinnx ) =[/tex]
[tex]= siny(1 + \displaystyle\frac{sin\displaystyle\frac{(n + 1)x}{2}.cos\displaystyle\frac{nx}{2}}{sin\displaystyle\frac{x}{2}}) + cosy(\displaystyle\frac{sin\displaystyle\frac{(n + 1)x}{2}.sin\displaystyle\frac{nx}{2}}{sin\displaystyle\frac{x}{2}})[/tex]
[tex]C = \sum_{k=0}^{n }cos(y + kx) = cosy + cos(y + x) + cos(y + 2x ) + cos(y + 3x) + ........ + cos(y + nx) =[/tex]
[tex]= cosy + cosy.cosx - siny.sinx + cosy.cos2x - siny.sin2x + cosy.cos3x - siny.sin3x +.........+ cosy.cosnx - siny.sinnx =[/tex]
[tex]= cosy(1 + cosx + cos2x + cos3x + ......+ cosnx) - siny(sinx + sin2x + sin3x + .......sinnx) =[/tex]
[tex]= cosy( 1 + \displaystyle\frac{sin\displaystyle\frac{(n + 1)x}{2}.cos\displaystyle\frac{nx}{2}}{sin\displaystyle\frac{x}{2}}) - siny(\displaystyle\frac{sin\displaystyle\frac{(n + 1)x}{2}.sin\displaystyle\frac{nx}{2}}{sin\displaystyle\frac{x}{2}})[/tex]
Никой любовен роман не е разплакал толкова много хора,колкото учебникът по математика.
Ако нещо мърда - това е биология,ако мирише -това е химия,ако има сила - това е физика,а ако нищо не разбираш - това е математика