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# Trigonometry Question

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So I have narrowed down a possible aproach to this by using the Double Angle and Half Angle Formulas, but whenever I try to apply them I get a wrong answer.

Here is the problem:

If $$a_0 = \sin^2 \left( \frac{\pi}{45} \right)$$ and $$a_{n + 1} = 4a_n (1 - a_n)$$
for $$n \ge 0$$ find the smallest positive integer $$n$$ such that $$a_n = a_0$$.

Jul 23, 2020

#1
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Are you sure you don't mean $$\sin^2(\frac{\pi}{5})$$ ?

Jul 24, 2020
#3
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No, I ment $$\frac{\pi}{45}$$

That is what has been throwing me off.

Guest Jul 24, 2020
#4
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Hmm, surprisingly, it does work for pi/45:

Alan  Jul 24, 2020
#5
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Huh, thats a nice way to do it.

Thanks Alan!

Guest Jul 25, 2020
#2
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I'm going to assume you mean the angle to be $$\frac{\pi}{5}$$.

Jul 24, 2020