Zadatak 0971

Priprema za prijemni za "tehničke" fakultete (ETF, MatF, MašF, Fon, SF, TMF,...)

Vrednost izraza $$\frac{\cos \frac{7\pi }{6}\cdot \cos \frac{7\pi }{3}\cdot \tan \frac{7\pi }{4} }{\cot \frac{10\pi }{3}\cdot \cos \frac{7\pi }{4}\cdot \sin \frac{8\pi }{3}}$$ je:

a) \(\frac{\sqrt{6}}{3}\);          b) \(\frac{\sqrt{6}}{2}\);          c) \(-\frac{\sqrt{6}}{3}\);          d) \(\frac{\sqrt{6}}{6}\);          e) \(-\frac{\sqrt{6}}{2}\).

Rešenje:


$$\frac{\cos \frac{7\pi }{6}\cdot \cos \frac{7\pi }{3}\cdot \tan \frac{7\pi }{4} }{\cot \frac{10\pi }{3}\cdot \cos \frac{7\pi }{4}\cdot \sin \frac{8\pi }{3}}$$

$$=\frac{-\cos \frac{\pi }{6}\cdot \cos \frac{\pi }{3}\cdot (-\tan \frac{\pi }{4}) }{\cot \frac{\pi }{3}\cdot \cos \frac{\pi }{4}\cdot \sin \frac{\pi }{3}}$$

$$=\frac{\frac{\sqrt{3}}{2}\cdot \frac{1}{2}\cdot 1}{\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}}=\frac{\sqrt{6}}{2}$$

Tačan odgovor je b).