2018.

Osnovne relacije između trigonometrijskih funkcija

TEKST ZADATKA

Ako je tg α+ctg α=3, \text{tg} \, \alpha + \text{ctg} \, \alpha = 3 , odrediti: tg3 α−ctg3 α. \text{tg}^3 \, \alpha - \text{ctg}^3 \, \alpha .

REŠENJE ZADATKA

Prvo ćemo kvadrirati dati izraz tg α+ctg α=3 \text{tg} \, \alpha + \text{ctg} \, \alpha = 3 kako bismo odredili vrednost izraza tg2 α+ctg2 α. \text{tg}^2 \, \alpha + \text{ctg}^2 \, \alpha . Koristimo identitet tg α⋅ctg α=1. \text{tg} \, \alpha \cdot \text{ctg} \, \alpha = 1 .

(tg α+ctg α)2=32tg2 α+2tg αctg α+ctg2 α=9tg2 α+2(1)+ctg2 α=9tg2 α+ctg2 α=7(\text{tg} \, \alpha + \text{ctg} \, \alpha)^2 = 3^2 \\ \text{tg}^2 \, \alpha + 2 \text{tg} \, \alpha \text{ctg} \, \alpha + \text{ctg}^2 \, \alpha = 9 \\ \text{tg}^2 \, \alpha + 2(1) + \text{ctg}^2 \, \alpha = 9 \\ \text{tg}^2 \, \alpha + \text{ctg}^2 \, \alpha = 7

Sada računamo vrednost izraza (tg α−ctg α)2 (\text{tg} \, \alpha - \text{ctg} \, \alpha)^2 koristeći prethodno dobijeni rezultat.

(tg α−ctg α)2=tg2 α−2tg αctg α+ctg2 α(tg α−ctg α)2=(tg2 α+ctg2 α)−2(1)(tg α−ctg α)2=7−2=5(\text{tg} \, \alpha - \text{ctg} \, \alpha)^2 = \text{tg}^2 \, \alpha - 2 \text{tg} \, \alpha \text{ctg} \, \alpha + \text{ctg}^2 \, \alpha \\ (\text{tg} \, \alpha - \text{ctg} \, \alpha)^2 = (\text{tg}^2 \, \alpha + \text{ctg}^2 \, \alpha) - 2(1) \\ (\text{tg} \, \alpha - \text{ctg} \, \alpha)^2 = 7 - 2 = 5

Iz prethodnog koraka sledi da je razlika tangensa i kotangensa koren broja 5. Uzimamo u obzir obe mogućnosti.

tg α−ctg α=±5\text{tg} \, \alpha - \text{ctg} \, \alpha = \pm \sqrt{5}

Koristimo formulu za razliku kubova a3−b3=(a−b)(a2+ab+b2) a^3 - b^3 = (a - b)(a^2 + ab + b^2) da bismo izračunali traženi izraz.

tg3 α−ctg3 α=(tg α−ctg α)(tg2 α+tg αctg α+ctg2 α)\text{tg}^3 \, \alpha - \text{ctg}^3 \, \alpha = (\text{tg} \, \alpha - \text{ctg} \, \alpha)(\text{tg}^2 \, \alpha + \text{tg} \, \alpha \text{ctg} \, \alpha + \text{ctg}^2 \, \alpha)

Zamenjujemo poznate vrednosti tg2 α+ctg2 α=7, \text{tg}^2 \, \alpha + \text{ctg}^2 \, \alpha = 7 , tg αctg α=1 \text{tg} \, \alpha \text{ctg} \, \alpha = 1 i tg α−ctg α=±5 \text{tg} \, \alpha - \text{ctg} \, \alpha = \pm \sqrt{5} u formulu.

tg3 α−ctg3 α=(±5)(7+1)tg3 α−ctg3 α=±85\text{tg}^3 \, \alpha - \text{ctg}^3 \, \alpha = (\pm \sqrt{5})(7 + 1) \\ \text{tg}^3 \, \alpha - \text{ctg}^3 \, \alpha = \pm 8\sqrt{5}

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