Prove that \(sin(cosx)=cos(sinx)\) has no real solution.

The method i came across was this

LHS is not equal to RHS.

If u have got any other method do post it.

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## Comments

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TopNewestLook at the function \(f(x) = \cos(\sin(x)) - \sin(\cos(x)).\) This function is even and periodic with period \(2\pi.\) so we just need to focus on the interval \([0, \pi].\)

Now \(f(0) = 1 \gt 0.\) Also, for \(x \in [\frac{\pi}{2}, \pi]\) we have \(\cos(\sin(x)) \gt 0\) and \(\sin(\cos(x)) \le 0,\) so \(f(x) \gt 0\) on this interval.

So we now just need to focus on the interval \((0, \frac{\pi}{2}).\) On this interval we have that both \(\cos(x)\) and \(\sin(x)\) lie between \(0\) and \(1,\) for which \(\sin(\cos(x)) \lt \cos(x) \lt \cos(\sin(x)),\), i.e., \(f(x) \gt 0.\) Thus \(f(x) \gt 0\) for all \(x\).

(To expand on this last step, note that for \(\theta\) in \((0, \frac{\pi}{2})\) we have \(0 \lt \sin(\theta) \lt \theta,\) and so \(\cos(\sin(\theta)) \gt \cos(\theta).\) )

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Yes! I recall when I first saw bounding those values by \( \cos x \), which seemed really ingenious to me.

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The easiest way for me is to plot the graph for the full range of \(x \in [0,2\pi)\). The graph is as follows. We note that the \(\sin{(\cos{x})} \ne \cos{(\sin{x})}\) always.

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Well, part of the question would then be "How do you know that's how the graphs should be drawn (Other than relying on a calculator / graphing capabilities)?"

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Do you mean how to do graphs with Excel spreadsheet? If so, maybe I should write a wiki on it. There are others requested for it. I just don't like to do it.

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It is easy to see that \(cos(cos(x))-sin(sin(x))=0\) has no solutions. Just replace \(x\) with \(\frac{\pi} {2}-x\)

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It's convincing that \[cos(cos(x)) \ne sin(sin(x)) \] but how can you prove that statement?

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@Brian Charlesworth @Tanishq Varshney

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Sketch the rough graph of both the functions

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