2448. Adicione formule | Balkan Tutor

Prvo moramo odrediti vrednost $\cos \alpha$ koristeći osnovni trigonometrijski identitet $\sin^2 \alpha + \cos^2 \alpha = 1 .$

\cos^2 \alpha = 1 - \sin^2 \alpha

Zamenjujemo datu vrednost $\sin \alpha = \frac{12}{13}$ u formulu.

\cos^2 \alpha = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}

Kako ugao $\alpha$ pripada drugom kvadrantu $\left(\frac{\pi}{2}, \pi\right) ,$ kosinus je u tom intervalu negativan.

\cos \alpha = -\sqrt{\frac{25}{169}} = -\frac{5}{13}

Sada računamo vrednost $\text{tg} \alpha$ koristeći odnos sinusa i kosinusa.

\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5}

Koristimo adicionu formulu za tangens zbira uglova:

\text{tg}(\alpha + \beta) = \frac{\text{tg} \alpha + \text{tg} \beta}{1 - \text{tg} \alpha \text{tg} \beta}

Zamenjujemo $\beta = \frac{\pi}{4}$ i znamo da je $\text{tg} \frac{\pi}{4} = 1 .$

\text{tg}\left(\frac{\pi}{4} + \alpha\right) = \frac{\text{tg} \frac{\pi}{4} + \text{tg} \alpha}{1 - \text{tg} \frac{\pi}{4} \text{tg} \alpha} = \frac{1 + \text{tg} \alpha}{1 - \text{tg} \alpha}

Uvrštavamo izračunatu vrednost $\text{tg} \alpha = -\frac{12}{5}$ u izraz.

\text{tg}\left(\frac{\pi}{4} + \alpha\right) = \frac{1 + \left(-\frac{12}{5}\right)}{1 - \left(-\frac{12}{5}\right)} = \frac{1 - \frac{12}{5}}{1 + \frac{12}{5}}

Sređujemo dvojni razlomak.

\text{tg}\left(\frac{\pi}{4} + \alpha\right) = \frac{\frac{5 - 12}{5}}{\frac{5 + 12}{5}} = \frac{-\frac{7}{5}}{\frac{17}{5}} = -\frac{7}{17}

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