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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@Brian Charlesworth @Tanishq Varshney

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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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Sketch the rough graph of both the functions

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