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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
4 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. · 3 weeks, 6 days ago

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. · 3 weeks, 5 days ago

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. · 3 weeks, 5 days ago

Yes, I found the with the citation of 'Russia' in the book 'The Art And Craft Of Problem Solving'. · 3 weeks, 5 days ago

What have you done? What have you tried? Where are you stuck? Staff · 3 weeks, 6 days ago

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 · 3 weeks, 6 days ago

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? · 3 weeks, 6 days ago

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. Staff · 3 weeks, 5 days ago