Question

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

Answer 1

Ok. 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 ...........(1)

Answer 2

actually those => should be <=>

I.e., if equation (1) is true then the top equation is true.

Answer 3

Now, If we could prove that both the factors on the left hand side of (1) are positive then we're done.

Note first

| sinx - cosx | = I √2 sin (x-π/4) I ≤ √2 < π/2

We have - π/2 < ( sinx - cosx ) < π/2

=> - π/4 < ( sinx - cosx )/2 < π/4

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

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

Similarly we can prove that

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

Hence (1) is true and we now the result.

Answer 4

I'm not crazy about this proof as it seems a bit clunky, but that's what it is for now. If I think of something better, I'll let you know.