Question

prove that equations cos(sin x)=sin(cos x) does not possess real roots

Answer 1

We're going to show that
cos (sinx) - sin (cosx) > 0

<=> cos (sinx) - cos ( π/2 - cosx ) > 0

<=> 2 sin [ (π/4) + (1/2). (sinx - cosx) ].sin [ (π/4) - (1/2). (sinx - cosx) ] > 0 -- (**)

Answer 2

Now if we can show the two terms in sin are positive we're done.

Answer 3

Note first by the trig identity from the last problem that
I sinx - cosx I = I √2 sin (x-π/4) I ≤ √2 < π/2

Answer 4

Thus π/2 < ( sinx - cosx ) < π/2

=> - π/4 < ( sinx - cosx )/2 < π/4

So 0 < π/4 + (1/2) ( sinx - cosx ) < π/2

And therefore sin [ (π/4) + (1/2). (sinx - cosx) ] > 0

This is the first sin term above

Answer 5

You can go through a similar procedure for the second term.

Hence
sin [ (π/4) + (1/2). (sinx - cosx) ].sin [ (π/4) - (1/2). (sinx - cosx) ] > 0

and therefore
cos (sinx) - sin (cosx) > 0

Thus there are no real roots of cos(sin x) = sin(cos x)

Answer 6

***
I'm not crazy about this proof, as it seems a bit clunky. I'll let you know if I think of something better.

Answer 7

any better answer??

Answer 8

Right now, no ...
This is a high school question?

Answer 9

yaa

Answer 10

Tricky. You've 17, yes?

Answer 11

yes

Answer 12

That's what I thought. Anyway, if I think of something better, I'll let you know, but for now, this is what I've got.

Answer 13

wt abt first method u told using graph?????

Answer 14

No, it doesn't work the way I hoped it would. I can see it intuitively, but I can't write down a good proof.

Answer 15

For the record, if you haven't looked at a graph of the function, here it is:
http://www.wolframalpha.com/input/?i=cos%28sin+x%29+-+sin%28cos+x%29

Answer 16

and this:
http://www.wolframalpha.com/input/?i=graph+of+cos%28sin+x%29+and+sin%28cos+x%29

Answer 17

Ok ... now I'm going downstairs to the Starbucks in the building and finally getting that coffee!