Needed a short method

Prove that sin(cosx)=cos(sinx)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.

Note by Tanishq Varshney
4 years, 6 months ago

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1 vote

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

Now f(0)=1>0.f(0) = 1 \gt 0. Also, for x[π2,π]x \in [\frac{\pi}{2}, \pi] we have cos(sin(x))>0\cos(\sin(x)) \gt 0 and sin(cos(x))0,\sin(\cos(x)) \le 0, so f(x)>0f(x) \gt 0 on this interval.

So we now just need to focus on the interval (0,π2).(0, \frac{\pi}{2}). On this interval we have that both cos(x)\cos(x) and sin(x)\sin(x) lie between 00 and 1,1, for which sin(cos(x))<cos(x)<cos(sin(x)),\sin(\cos(x)) \lt \cos(x) \lt \cos(\sin(x)),, i.e., f(x)>0.f(x) \gt 0. Thus f(x)>0f(x) \gt 0 for all xx.

(To expand on this last step, note that for θ\theta in (0,π2)(0, \frac{\pi}{2}) we have 0<sin(θ)<θ,0 \lt \sin(\theta) \lt \theta, and so cos(sin(θ))>cos(θ).\cos(\sin(\theta)) \gt \cos(\theta). )

Brian Charlesworth - 4 years, 6 months ago

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

Calvin Lin Staff - 4 years, 6 months ago

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

Chew-Seong Cheong - 4 years, 6 months ago

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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)?"

Calvin Lin Staff - 4 years, 6 months ago

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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.

Chew-Seong Cheong - 4 years, 6 months ago

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It is easy to see that cos(cos(x))sin(sin(x))=0cos(cos(x))-sin(sin(x))=0 has no solutions. Just replace xx with π2x\frac{\pi} {2}-x

Raghav Vaidyanathan - 4 years, 6 months ago

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

Curtis Clement - 4 years, 6 months ago

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Sketch the rough graph of both the functions

Nikhil Tanwar - 4 years, 6 months ago

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