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Needed a short method

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

LHS is not equal to RHS.

If u have got any other method do post it.

Note by Tanishq Varshney
1 year, 9 months ago

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Look 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 $$\cos(\sin(\theta)) \gt \cos(\theta).$$ ) · 1 year, 9 months ago

Yes! I recall when I first saw bounding those values by $$\cos x$$, which seemed really ingenious to me. Staff · 1 year, 9 months ago

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.

· 1 year, 9 months 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)?" Staff · 1 year, 9 months 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. · 1 year, 9 months ago

It is easy to see that $$cos(cos(x))-sin(sin(x))=0$$ has no solutions. Just replace $$x$$ with $$\frac{\pi} {2}-x$$ · 1 year, 9 months ago

It's convincing that $cos(cos(x)) \ne sin(sin(x))$ but how can you prove that statement? · 1 year, 9 months ago