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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

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
–
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.

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.

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.

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## Comments

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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

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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?

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

$\begin{aligned} x & \approx & -4.71 \pm 1.21 i \\ x & \approx & 0.76 \pm 0.61i \end{aligned}$

Its graph looks like this:

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

$\cos(\cos(\cos(\cos x))) = \sin(\sin(\sin(\sin x))) \\ \implies \sqrt{1- \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}}} = \sqrt{1- \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}}} \\ \implies \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} \\ \implies \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = 0 \\ \implies \bigg[ \cos \left( \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} + \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} \right) \bigg] \cdot \bigg[ \cos \left( \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} \right) \bigg] = 0 \qquad \qquad \small \color{#3D99F6}{\text{Using} \ \cos^2 A - \sin^2 B = \cos(A+B) \cdot \cos(A-B)}$

Therefore, the trivial solutions are of form:

$1. \ \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} + \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \dfrac{(2n+1) \pi}{2} \qquad \small \color{#3D99F6}{\text{Where} \ n \in \mathbb{N}}$

$2. \ \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \dfrac{(2m+1) \pi}{2} \qquad \small \color{#3D99F6}{\text{Where} \ m \in \mathbb{N}}$

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.

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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.

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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.Log in to reply

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