First slide

Multiple and sub- multiple Angles

Question

$If a_{n + 1} = \sqrt{\frac{1}{2} (1 + a_{n})}, then \cos (\frac{\sqrt{1 - a_{0}^{2}}}{a_{1} a_{2} \dots \dots \infty}) =$

Difficult

A

1
B

-1
C

$a_{0}$
D

$\frac{1}{a_{0}}$

Solution

We have
$\begin{array}{l} a_{n + 1} = \sqrt{\frac{1 + a_{n}}{2}} \\ Let a_{0} = \cos θ \\ \Rightarrow a_{1} = \sqrt{\frac{1 + \cos θ}{2}} = \cos \frac{θ}{2} \\ \Rightarrow a_{2} = \cos \frac{θ}{4} \\ \Rightarrow a_{3} = \cos \frac{θ}{8} \\ \Rightarrow a_{n + 1} = \cos \frac{θ}{2^{n + 1}} a n d s o o n \end{array}$
We know that $\cos \frac{θ}{2} \cos \frac{θ}{2^{2}} \cos \frac{θ}{2^{3}} .............. \cos \frac{θ}{2^{n - 1}} + ............ \infty = \frac{\sin θ}{θ}$
$\begin{array}{l} ∴ \cos (\frac{\sqrt{1 - a_{0}^{2}}}{a_{1} a_{2} \dots \dots \infty}) = \cos (\frac{\sin θ}{\sin θ} \cdot θ) \\ = \cos θ = a_{0} \end{array}$

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