Cos(Sin) = Sin(Cos)?

Here, I've used only two trigonometric identities: $sin(x+\frac{\pi}{2})=cos (x)\quad and \quad sin(\frac{\pi}{2}-x)=cos (x).$ First, simplify the result that you get using the latter identity. Now, if you check the case for which $psinx+pcosx=\pi/2\Rightarrow psin(x+\frac{\pi}{4})=\frac{\pi}{2\sqrt2},$ you can clearly see that it's not possible for p>=1 as it would take a negative x to satisfy the equation. So, we try the next thing, which is $psin(\frac{\pi}{4}-x)=\frac{\pi}{2\sqrt2}$ ; and we can see that even for this equation, p=1 is not going to work. Now, if we try p=2, take the sine inverse and subtract the result from $\pi/4$ , we can see that the x thus obtained is between 0 and $2\pi.$ In addition to this, we can also see that $\frac{\pi}{2\sqrt2}>1$ and therefore, a value of p=1 will never satisfy the set of solutions to the equation for any real x. As 1 is removed and 2 confirmed, the answer is 2.

The easiest way to show this is to plot $f(x) = \cos{(p\sin{x})}-\sin{(p\cos{x})}$ for various values of $p$ . I have done that for $p = 1, 2, 3$ with a spreadsheet. And the graph is as follows:

It is note that $p=1$ (blue line) has no solution. When $p=2$ (red line) and $p=3$ (green line) each has $4$ solutions. Therefore, the smallest $p$ with solutions is $\boxed{2}$ .

Simplifying the equation with Vishnu C's Method would give you the final equation as which further reduces to . Now since RHS is greater than 1 therefore for x to exist p must be greater than 1. So lowest possible Integer will be p=2 .

As the result of a cos and sin can only be the same if

cos(x)=sin(PI/2 - x)

Then here we have:

p cos (x) = PI/2 - p sin(x) , or

PI/2 = p ( cos(x) + sin(x) )

As the maximum of cos(x)+sin(x) is sqrt(2), then to reach PI/2 in above, integer p has to be:

p >= sqrt(2) PI /2 = 1.11 => p = 2 or higher