За любителите на тригонометрията

[KOPMOPAH]

Докажете тъждеството:
$\frac{\sin x+\sin2 x+ \sin3 x+ \sin4 x+ \sin5 x}{\cos x+ \cos2 x+ \cos3 x+ \cos4 x+ \cos5 x}=\tan 3x$

[Davids]

Числителя ще е A, знаменателят - B. Тогава
$A = sinx + sin2x + sin3x + sin4x + sin5x = sin(3x-2x) + sin(3x-x) + sin3x + sin(3x+x) + sin(3x + 2x) = sin3xcos2x \cancel{- cos3xsin2x} + sin3xcosx \cancel{- cos3xsinx} + sin3x + sin3xcosx \cancel{+ cos3xsinx} + sin3xcos2x \cancel{+ cos3xsin2x} = sin3x(2cos2x + 2cosx + 1)$
А за знаменателя аналогично:
$B = cosx + cos2x + cos3x + cos4x + cos5x = cos3xcos2x \cancel{+ sin3xsin2x} + cos3xcosx \cancel{+ sin3xsinx} + cos3x + cos3xcosx \cancel{- sin3xsinx} + cos3xcos2x \cancel{- sin3xsin2x} = sin3x(2cos2x + 2cosx + 1)$
Веднага виждаме, че $\frac{A}{B} = \frac{sin3x(2cos2x + 2cosx + 1)}{sin3x(2cos2x + 2cosx + 1)} = tg3x$

[ева]

[tex]\frac{(sinх+sin5х)+(sin2х+sin4х)+sin3х}{(cosх+cos5х)+(cos2х+cos4х)+cos3х}[/tex]=

=[tex]\frac{2sin3хcos(-2х)+2sin3хcos(-х)+sin3х}{2cos3хcos(-2х)+2cos3хcos(-х)+cos3х}[/tex]=

=[tex]\frac{sin3х(2cos2х+2cosх+1)}{cos3х(2cos2х+2cosх+1)}[/tex]=

=[tex]\frac{sin3х}{cos3х}[/tex]=

=tg3х

[Knowledge Greedy]

[tex]z=cosx+isinx[/tex]
По Моавър [tex]z^k=coskx+isinkx[/tex]
Явно [tex]sinkx=Im z^k[/tex] и [tex]coskx=Re z^k[/tex]
Лявата страна
[tex]L=\frac{Im ( z+z^2+z^3+z^4+z^5) }{Re(z+z^2+z^3+z^4+z^5)}=\frac{Im \frac{z^3\left (z^{\frac{5}{2}}-z^{-\frac{5}{2}} \right )}{z^{\frac{1}{2}}-z^{-\frac{1}{2}} }}{Re \frac{z^3\left (z^{\frac{5}{2}}-z^{-\frac{5}{2}} \right )}{z^{\frac{1}{2}}-z^{-\frac{1}{2}} }}=\frac{Im \frac{(cos3x+isin3x).\cancel{2i}sin\frac{5x}{2}}{\cancel{2i}sin\frac{x}{2}}}{Re \frac{(cos3x+isin3x).\cancel{2i}sin\frac{5x}{2}}{\cancel{2i}sin\frac{x}{2}}}[/tex]
След още съкращаване
[tex]L=\frac{Im(cos3x+isin3x)}{Re(cos3x+isin3x)}=\frac{sin3x}{cos3x}=tg3x[/tex]