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Bogi V.
matematika - sš

Goniometria

Dobrý deň potrebovala by som poradiť s touto úlohou. Treba nájsť periódu funkcie sinx²

2 answers
Viliam H.
Viliam H. Contact
Ak myslis funkciu sin(x^2) tak tato funkcia nie je periodicka, pretoze s vacsim |x| sa "oscilovanie zrychluje".
Druha moznost je sin^2(x) a to je to iste ako (sin x)^2, tato funkcia je periodicka a jej perioda je pi. Da sa vyjst zo vzorca sin^2(x) = 1/2 * (1 - cos(2x)), z toho je potrebne urcit periodu pre cos(2x) a ta je pi, pretoze cos(x) je 2pi.
Peter K.
Peter K. Contact
Pre objasnenie predchádzajúcej odpovede prikladám obrázok.
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