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Daniel za post:
Vivienne
Reputacija: 4.55%
od Daniel » Ponedeljak, 05. Jul 2021, 15:52
Kao što je Vivienne lepo primetila, transformacije koje su upotrebljene u prethodnim postupcima važe samo uz određene uslove. Da bi se smelo primeniti [inlmath]\frac{1+\text{tg }\frac{\alpha}{2}}{1-\text{tg }\frac{\alpha}{2}}=\frac{1+\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}}{1-\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}}[/inlmath], morali bismo imati neki uslov da je [inlmath]\text{tg }\frac{\alpha}{2}\ge0[/inlmath], a mi taj uslov nemamo. (Važi da je [inlmath]\text{tg}^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{1+\cos\alpha}[/inlmath], ali odatle imamo [inlmath]\text{tg }\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}[/inlmath], tj. bez određenih uslova ne znamo kada upotrebiti [inlmath]+[/inlmath] a kada [inlmath]-[/inlmath]).
Ja bih zato to radio na sledeći način:
[dispmath]\text{tg}\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\left(\frac{1-\sin\alpha}{\cos\alpha}\right)=1\\
\frac{1+\text{tg }\frac{\alpha}{2}}{1-\text{tg }\frac{\alpha}{2}}\cdot\frac{\overbrace{\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}}^1-\overbrace{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}^{\sin\alpha}}{\underbrace{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}_{\cos\alpha}}=1\\
\frac{\frac{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}{\cancel{\cos\frac{\alpha}{2}}}}{\frac{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}{\cancel{\cos\frac{\alpha}{2}}}}\cdot\frac{\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)^\bcancel2}{(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2})\bcancel{(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})}}=1\\
\frac{\cancel{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}}{\bcancel{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}}\cdot\frac{\bcancel{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}}{\cancel{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}}=1\\
1=1[/dispmath]
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