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.

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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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 \(\cos(\sin(\theta)) \gt \cos(\theta).\) ) – Brian Charlesworth · 2 years, 1 month ago

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– Calvin Lin Staff · 2 years, 1 month ago

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

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.

– Chew-Seong Cheong · 2 years, 1 month agoLog in to reply

– Calvin Lin Staff · 2 years, 1 month ago

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)?"Log in to reply

– Chew-Seong Cheong · 2 years, 1 month ago

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.Log in to reply

It is easy to see that \(cos(cos(x))-sin(sin(x))=0\) has no solutions. Just replace \(x\) with \(\frac{\pi} {2}-x\) – Raghav Vaidyanathan · 2 years, 1 month ago

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– Curtis Clement · 2 years, 1 month ago

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

@Brian Charlesworth @Tanishq Varshney – Raghav Vaidyanathan · 2 years, 1 month ago

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Sketch the rough graph of both the functions – Nikhil Tanwar · 2 years, 1 month ago

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