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Question Number 157706 by Oberon last updated on 26/Oct/21

Answered by mr W last updated on 26/Oct/21

$$2a_n=\sqrt{\frac{2a_{n-1}+1}{2}}$$$$let{b}_n=2a_n$$$$b_n=\sqrt{\frac{b_n+1}{2}}$$$$2b_n^2-1=b_{n-1}$$$$say{b}_n=\cos \frac{\theta }{2^n}$$$$2b_n-1=2{\cos }^2\frac{\theta }{2^n}-1=\cos \frac{\theta }{2^{n-1}}=b_{n-1}✓$$$$b_1=\cos \frac{\theta }{2}=2a_1=2\times \frac{1}{2}\sqrt{\frac{1}{2}}$$$$\cos \frac{\theta }{2}=\sqrt{\frac{1}{2}}$$$$\Rightarrow \frac{\theta }{2}=\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{2}$$$$\Rightarrow b_n=\cos \frac{\pi }{2^{n+1}}$$$$\Rightarrow a_n=\frac{b_n}{2}$$$$\Rightarrow a_n=\frac{1}{2}\cos \frac{\pi }{2^{n+1}}$$

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