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Prove that, cos $\frac{3\pi }{13}$ + cos $\frac{5\pi }{13}$- 2 cos $\frac{9\pi }{13}$ cos $\frac{12\pi }{13}$=0
Solution:
L.H.S= cos $\frac{3\pi }{13}$+ cos $\frac{5\pi }{13}$ – 2 cos $\frac{9\pi }{13}$ cos $\frac{12\pi }{13}$
= 2 cos $\frac{\frac{3\pi }{13} + \frac{5\pi }{13}}{2}$ cos$\frac{\frac{3\pi }{13} – \frac{5\pi }{13}}{2}$ – 2 cos $\frac{9\pi }{13}$cos $\frac{12\pi }{13}$
= 2 cos $\frac{4\pi }{13}$ cos $\left( – \frac{\pi }{13}\right)$ – 2 cos $\frac{9\pi }{13}$ cos $\frac{12\pi }{13}$
= 2 cos $\frac{4\pi }{13}$ cos $\frac{\pi }{13}$ – 2 cos $\left(\pi – \frac{4\pi }{13}\right)$ cos $\left(\pi – \frac{\pi }{13}\right)$
= 2 cos $\frac{4\pi }{13}$ cos $\frac{\pi }{13}$ – 2 cos $\frac{4\pi }{13}$ cos $\frac{\pi }{13}$
= 0 =R.H.S [Proved]
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