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Question Number 198558 by hardmath last updated on 22/Oct/23

Commented by mr W last updated on 22/Oct/23

$$hardmath=shrinava?$$

Answered by mr W last updated on 22/Oct/23

$$\cos \frac{\pi }{6}=\frac{1}{2}\sqrt{3}$$$$2{\cos }^2\frac{\pi }{12}-1=\frac{1}{2}\sqrt{3}$$$$\Rightarrow \cos \frac{\pi }{12}=\frac{1}{2}\sqrt{2+\sqrt{3}}$$$$similarly$$$$\Rightarrow \cos \frac{\pi }{24}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{3}}}$$$$\Rightarrow \cos \frac{\pi }{48}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}$$$$\Rightarrow \cos \frac{\pi }{96}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$$$$\Rightarrow \cos \frac{\pi }{192}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}}$$$$i.e.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}}=2\cos \frac{\pi }{192}$$

Commented by hardmath last updated on 28/Oct/23

$$\mathrm{thank}\mathrm{you}\mathrm{professor}$$

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