3. Zadatak

Prvi način.

Uočimo da je $1+\cos\dfrac\pi7=2\cos^2\dfrac\pi{14}$ i $\sin\dfrac{6\pi}7=\sin\dfrac\pi7=2\sin\dfrac\pi{14}\cos\dfrac\pi{14}$. Stoga je

$$\begin{aligned} 1+\cos\frac\pi7+i\sin\frac{6\pi}7 &=2\cos^2\frac\pi{14}+2i\sin\frac\pi{14}\cos\frac\pi{14}\\ &=2\cos\frac\pi{14}\left(\cos\frac\pi{14}+i\sin\frac\pi{14}\right). \end{aligned}$$

Potenciranjem konačno dobivamo

$$\begin{aligned} \left(1+\cos\frac\pi7+i\sin\frac{6\pi}7\right)^{14} &=\left(2\cos\frac\pi{14}\left(\cos\frac\pi{14}+i\sin\frac\pi{14}\right)\right)^{14}\\ &=2^{14}\left(\cos\frac\pi{14}\right)^{14}\left(\cos\frac\pi{14}+i\sin\frac\pi{14}\right)^{14}\\ &=2^{14}\left(\cos\frac\pi{14}\right)^{14}(\cos\pi+i\sin\pi)\\ &=-2^{14}\left(\cos\frac\pi{14}\right)^{14}. \end{aligned}$$

Drugi način.

Broj $1+\cos\dfrac\pi7+i\sin\dfrac{6\pi}7$ transformirajmo koristeći formule pretvorbe zbroja u umnožak:

$$\begin{aligned} 1+\cos\frac\pi7+i\sin\frac{6\pi}7 &=\cos0+i\sin0+\cos\frac\pi7+i\sin\frac\pi7\\ &=\left(\cos0+\cos\frac\pi7\right)+i\left(\sin0+\sin\frac\pi7\right)\\ &=2\cos\frac\pi{14}\cos\frac\pi{14}+2i\sin\frac\pi{14}\sin\frac\pi{14}\\ &=2\cos\frac\pi{14}\left(\cos\frac\pi{14}+i\sin\frac\pi{14}\right). \end{aligned}$$

Dalje nastavljamo kao u prvom rješenju.