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Trig in trig in trig in trig

Can anyone please explain how I can solve the following problem:

Solve for \(x\):

\[ \cos(\cos(\cos(\cos x))) = \sin(\sin(\sin(\sin x))) \]

Note by Nilabha Saha
9 months, 3 weeks ago

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Well, according to Wolfram alpha, the solution(s) are:

\(\begin{eqnarray} x & \approx & -4.71 \pm 1.21 i \\ x & \approx & 0.76 \pm 0.61i \end{eqnarray}\)

Its graph looks like this:


I couldn't properly solve the problem, but I managed to get to this point.

\(\cos(\cos(\cos(\cos x))) = \sin(\sin(\sin(\sin x))) \\ \implies \sqrt{1- \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}}} = \sqrt{1- \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}}} \\ \implies \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} \\ \implies \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = 0 \\ \implies \bigg[ \cos \left( \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} + \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} \right) \bigg] \cdot \bigg[ \cos \left( \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} \right) \bigg] = 0 \qquad \qquad \small \color{blue}{\text{Using} \ \cos^2 A - \sin^2 B = \cos(A+B) \cdot \cos(A-B)}\)

Therefore, the trivial solutions are of form:

\(1. \ \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} + \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \dfrac{(2n+1) \pi}{2} \qquad \small \color{blue}{\text{Where} \ n \in \mathbb{N}}\)

\(2. \ \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \dfrac{(2m+1) \pi}{2} \qquad \small \color{blue}{\text{Where} \ m \in \mathbb{N}}\)

The two pairs of answers which are complex will be of the forms as discussed above.


I don't know of a better way to approach this problem. Tapas Mazumdar · 9 months, 3 weeks ago

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@Tapas Mazumdar Is there any other way? Perhaps, since we know that sin(x) = cos(x) for x = \frac{{pi}{4} and \frac{5 * {pi}}{4}. Or that sin({pi} - x) = cos(x). Do we get any lead like that? I couldn't find any, but you might. Thanks for helping. Nilabha Saha · 9 months, 3 weeks ago

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@Nilabha Saha Well, you might get principal solutions like that. But you have to think for general cases while solving problems like these, for example \(\sin x = \cos x\) has the general solution of \(n \pi + \dfrac{\pi}{4}\), where \(n \in \mathbb{N}\). As for the identity \(\sin^k \left( \dfrac{\pi}{2} - x \right) = \cos^k x\), I haven't still tried that approach. I would check it out too, and let you know.

I found it online that this is a problem from the All Russian Olympiad 1995, Grade 10, Problem 1. Tapas Mazumdar · 9 months, 3 weeks ago

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@Tapas Mazumdar Yes, I found the with the citation of 'Russia' in the book 'The Art And Craft Of Problem Solving'. Nilabha Saha · 9 months, 3 weeks ago

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What have you done? What have you tried? Where are you stuck? Calvin Lin Staff · 9 months, 3 weeks ago

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@Calvin Lin This is a problem which apparently came in one of Russia's elite math examinations (I have no idea which one) Source: The Art And Craft Of Problem Solving Nilabha Saha · 9 months, 3 weeks ago

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@Nilabha Saha However, I graphed these two functions and found that they never intersect, implying there are no solutions. But, how do we theoretically prove that this has no solutions within the real numbers? Nilabha Saha · 9 months, 3 weeks ago

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@Nilabha Saha Yup, the idea is to show that these 2 graphs never intersect (over the reals). It's somewhat tricky to prove that though. AFAIK there isn't a "non-calculus" approach. Calvin Lin Staff · 9 months, 3 weeks ago

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@Calvin Lin Can you guide me on the calculus approach? Nilabha Saha · 9 months, 3 weeks ago

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