Question

$\cos\left(\dfrac\pi7\right)\cdot\cos\left(\dfrac{2\pi}7\right)\cdot\cos\left(\dfrac{4\pi}7\right)=-\dfrac18$ olduğunu gösteriniz.

Answer 1

$A=\cos\left(\dfrac\pi7\right)\cdot\cos\left(\dfrac{2\pi}7\right)\cdot\cos\left(\dfrac{4\pi}7\right)$ olsun.

 

$A=\cos\left(\dfrac\pi7\right)\cdot\cos\left(\dfrac{2\pi}7\right)\cdot\cos\left(\dfrac{4\pi}7\right)$

$$\Rightarrow$$

$2\cdot\sin\left(\dfrac{\pi}{7}\right) \cdot A  =2\cdot \sin\left(\dfrac{\pi}{7}\right) \cdot \cos\left(\dfrac\pi7\right)\cdot\cos\left(\dfrac{2\pi}7\right)\cdot\cos\left(\dfrac{4\pi}7\right)$

$$\Rightarrow$$

$2\cdot \sin\left(\dfrac{\pi}{7}\right) \cdot A = \sin\left(\dfrac {2\pi}{7}\right)\cdot\cos\left(\dfrac{2\pi}7\right)\cdot\cos\left(\dfrac{4\pi}7\right)$

$$\Rightarrow$$

$4\cdot \sin\left(\dfrac{\pi}{7}\right) \cdot A = 2\cdot \sin\left(\dfrac {2\pi}{7}\right)\cdot\cos\left(\dfrac{2\pi}7\right)\cdot\cos\left(\dfrac{4\pi}7\right)$

$$\Rightarrow$$

$8\cdot \sin\left(\dfrac{\pi}{7}\right) \cdot A =2\cdot \sin\left(\dfrac {4\pi}{7}\right)\cdot\cos\left(\dfrac{4\pi}7\right)$

$$\Rightarrow$$

$8\cdot \sin\left(\dfrac{\pi}{7}\right) \cdot A =\sin\left(\dfrac {8\pi}{7}\right)=\sin\left(\pi +\dfrac {\pi}{7}\right)=-\sin\frac{\pi}{7}$

$$\Rightarrow$$

$A =-\dfrac{1}{8}.$