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Data publicarii: 31 August, 2010

EXEMPLUL 1

Suport teoretic:

Operatii cu numere complexe sub forma trigonometrica, partea reala si partea imaginara a unui numar complex, identitati trigonometrice remarcabile.

Enunt:

Sa se afle numarul real x, astfel incat:

$Re(\frac{1-cosx-isinx}{1+cosx+isinx})=0.$ Re(\frac{1-cosx-isinx}{1+cosx+isinx})=0.

Raspuns:

x € R \ {(2k + 1)π | k € Z}.

Rezolvare:

Fie

$z=\frac{1-cosx-isinx}{1+cosx+isinx}=\frac{2{{sin}^2}{\frac{x}{2}}-2i{sin}{\frac{x}{2}}{cos}{\frac{x}{2}}}{2{{cos}^2}{\frac{x}{2}}+2i{sin}{\frac{x}{2}}{cos}{\frac{x}{2}}}=\frac{{sin}{\frac{x}{2}}({sin}{\frac{x}{2}}-i{cos}{\frac{x}{2}})}{{cos}{\frac{x}{2}}({cos}{\frac{x}{2}}+i{sin}{\frac{x}{2}})}={{tg}{\frac{x}{2}}}\cdot{\frac{{cos}{\frac{x}{2}}+i{sin}{\frac{x}{2}}}{i({cos}{\frac{x}{2}}+i{sin}{\frac{x}{2}})}}=$ z=\frac{1-cosx-isinx}{1+cosx+isinx}=\frac{2{{sin}^2}{\frac{x}{2}}-2i{sin}{\frac{x}{2}}{cos}{\frac{x}{2}}}{2{{cos}^2}{\frac{x}{2}}+2i{sin}{\frac{x}{2}}{cos}{\frac{x}{2}}}=\frac{{sin}{\frac{x}{2}}({sin}{\frac{x}{2}}-i{cos}{\frac{x}{2}})}{{cos}{\frac{x}{2}}({cos}{\frac{x}{2}}+i{sin}{\frac{x}{2}})}={{tg}{\frac{x}{2}}}\cdot{\frac{{cos}{\frac{x}{2}}+i{sin}{\frac{x}{2}}}{i({cos}{\frac{x}{2}}+i{sin}{\frac{x}{2}})}}=