Question

prove this trigonometry ratio​

Answer 1

Step-by-step explanation:

We have,

$2 \sin \bigg( \frac{6\pi}{11} \bigg) \cos \bigg( \frac{\pi}{11} \bigg) + \sin \bigg( \frac{16\pi}{11} \bigg) + \sin \bigg( \frac{18\pi}{11} \bigg)$

$= 2 \sin \bigg( \frac{6\pi}{11} \bigg) \cos \bigg( \frac{\pi}{11} \bigg) + 2 \sin \bigg \{ \frac{1}{2} \bigg( \frac{16\pi}{11} + \frac{18\pi}{11} \bigg) \bigg \} \cos\bigg \{ \frac{1}{2} \bigg( \frac{16\pi}{11} - \frac{18\pi}{11} \bigg) \bigg \}$

$= 2 \sin \bigg( \frac{6\pi}{11} \bigg) \cos \bigg( \frac{\pi}{11} \bigg) + 2 \sin \bigg \{ \frac{1}{2} \bigg( \frac{34\pi}{11} \bigg) \bigg \} \cos\bigg \{ \frac{1}{2} \bigg( - \frac{2\pi}{11} \bigg) \bigg \}$

$= 2 \sin \bigg( \frac{6\pi}{11} \bigg) \cos \bigg( \frac{\pi}{11} \bigg) + 2 \sin \bigg( \frac{17\pi}{11} \bigg) \cos \bigg( - \frac{\pi}{11} \bigg)$

$= 2 \sin \bigg( \frac{6\pi}{11} \bigg) \cos \bigg( \frac{\pi}{11} \bigg) + 2 \sin \bigg( \frac{17\pi}{11} \bigg) \cos \bigg( \frac{\pi}{11} \bigg)$

$= 2 \cos \bigg( \frac{\pi}{11} \bigg) \bigg \{ \sin \bigg( \frac{6\pi}{11} \bigg) + \sin \bigg( \frac{17\pi}{11} \bigg) \bigg \}$

$= 2 \cos \bigg( \frac{\pi}{11} \bigg) \bigg [ 2\sin \bigg \{ \ \frac{1}{2} \bigg( \frac{6\pi}{11} + \frac{17\pi}{11} \bigg) \bigg \} \cos \bigg \{ \ \frac{1}{2} \bigg( \frac{6\pi}{11} - \frac{17\pi}{11} \bigg) \bigg \} \bigg]$

$= 2 \cos \bigg( \frac{\pi}{11} \bigg) \bigg [ 2\sin \bigg \{ \ \frac{1}{2} \bigg( \frac{23\pi}{11} \bigg) \bigg \} \cos \bigg \{ \ \frac{1}{2} \bigg( - \frac{11\pi}{11} \bigg) \bigg \} \bigg]$

$= 4 \cos \bigg( \frac{\pi}{11} \bigg) \sin \bigg( \frac{23\pi}{22} \bigg) \cos \bigg \{ \ \frac{1}{2} \bigg( - \pi \bigg) \bigg \}$

$= 4 \cos \bigg( \frac{\pi}{11} \bigg) \sin \bigg( \frac{23\pi}{22} \bigg) \cos \bigg ( - \frac{ \pi}{2} \bigg)$

$= 4 \cos \bigg( \frac{\pi}{11} \bigg) \sin \bigg( \frac{23\pi}{22} \bigg) \cos \bigg (\frac{ \pi}{2} \bigg)$

$= 4 \cos \bigg( \frac{\pi}{11} \bigg) \sin \bigg( \frac{23\pi}{22} \bigg) \times 0$

$= 0$