MHB Lowest Positive $p$: Solving $\cos(p\sin \, x) = \sin (p\cos\, x)$

[kaliprasad]

find lowest positive p such that $\cos(p\sin \, x) = \sin (p\cos\, x)$

 

[Opalg]

find lowest positive p such that $\cos(p\sin \, x) = \sin (p\cos\, x)$

[sp]If $\sin\alpha = \cos\beta$ then the possible values for $\beta$ are $\beta = \pm\bigl(\frac\pi2 - \alpha\bigr) + 2k\pi.$ Putting $\alpha = p\cos x$ and $\beta = p\sin x$, we get $p\sin x = \pm\bigl(\frac\pi2 - p\cos x\bigr) + 2k\pi,$ from which $$p = \frac{\pm\pi + 4k\pi}{2(\sin x \pm\cos x)},$$ where $k$ is an integer, and the $\pm$ sign in the numerator must be the same as the one in the denominator.

We want to choose the values of $\pm$ and $k$ that will give the minimal positive value for $p$. The correct choice will depend on $x$, and we may assume that $x$ lies in the interval $[0,2\pi].$

Look at the case of the $+$ signs first, so that $p = \frac{\pi + 4k\pi}{2(\sin x + \cos x)}.$ The denominator is positive when $x\in \bigl[0,\frac34\pi\bigr) \cup \bigl(\frac74\pi,2\pi\bigr]$ and in that case the smallest positive value for the numerator clearly occurs when $k=0$, giving $p = \frac{\pi}{2(\sin x + \cos x)}.$ But when $x\in \bigl(\frac34\pi,\frac74\pi\bigr)$ the denominator is negative, so the numerator must also be negative. The value of $k$ that minimizes $p$ is then $k=-1$, and $p = \frac{-3\pi}{2(\sin x + \cos x)}.$

A similar analysis of the case of the $-$ signs shows that we should take $k=0$ if $x\in \bigl[0,\frac14\pi\bigr) \cup \bigl(\frac54\pi,2\pi\bigr]$, and $k=1$ if $x\in \bigl(\frac14\pi,\frac54\pi\bigr).$

Thus we have four possible candidates for the optimal value of $p$, as follows: $$(1)\qquad p = \tfrac{\pi}{2(\sin x + \cos x)} \text{ for } x\in \bigl[0,\tfrac34\pi\bigr) \cup \bigl(\tfrac74\pi,2\pi\bigr],$$ $$(2)\qquad p = \tfrac{-3\pi}{2(\sin x + \cos x)} \text{ for } x\in \bigl(\tfrac34\pi,\tfrac74\pi\bigr),$$ $$(3)\qquad p = \tfrac{-\pi}{2(\sin x - \cos x)} \text{ for } x\in \bigl[0,\tfrac14\pi\bigr) \cup \bigl(\tfrac54\pi,2\pi\bigr],$$ $$(4)\qquad p = \tfrac{3\pi}{2(\sin x - \cos x)} \text{ for } x\in \bigl(\tfrac14\pi,\tfrac54\pi\bigr).$$

Trying to decide which of those candidates actually gives the best result is frankly a bit of a nightmare. It is useful to find the crossover points, where two of the four formulas are equal. For example, formulas $(1)$ and $(4)$ are equal when $ \tfrac{\pi}{2(\sin x + \cos x)} = \tfrac{3\pi}{2(\sin x - \cos x)}$, which simplifies to $\tan x = -2.$ If I have not made any mistakes then the final values for $p$ come out like this, where $\theta = \arctan2$:

If $0\leqslant x \leqslant \pi-\theta$ then $p = \tfrac{\pi}{2(\sin x + \cos x)}$.

If $\pi-\theta \leqslant x \leqslant \pi$ then $p = \tfrac{3\pi}{2(\sin x - \cos x)}$.

If $\pi \leqslant x \leqslant \pi + \theta$ then $p = \tfrac{-3\pi}{2(\sin x + \cos x)}$.

If $\pi+\theta \leqslant x \leqslant 2\pi$ then $p = \tfrac{-\pi}{2(\sin x - \cos x)}$.

[/sp]

 

[Albert1]

find lowest positive p such that $\cos(p\sin \, x) = \sin (p\cos\, x)$

my solution:

Spoiler

let $0\leq x\leq 2\pi$
$cos(p sin x)=sin(p cos x)$
let $y=pcosx$
we have :$siny=cos(\dfrac{\pi}{2}-y)$---(A)
or $sin y=cos(\dfrac{3\pi}{2}+y)$---(B)
from (A):$psinx=\dfrac{\pi}{2}-y$
$\therefore p=\dfrac{\pi}{2(sinx+cosx)}\geq \dfrac {\pi}{2\sqrt 2 }>0$
from (B):$psinx=\dfrac{3\pi}{2}+y$
$\therefore p=\dfrac{3\pi}{2(sinx-cosx)}\geq \dfrac {3\pi}{2\sqrt 2 }>0$
because we want to find the lowest positive $p>0$
we get $p=\dfrac{\pi}{2\sqrt 2}=\dfrac {\sqrt 2 \pi}{4}$

 

[Opalg]

I agree with Albert's solution if $x$ is allowed to vary. I read the question as implying that $x$ is fixed.

 

[kaliprasad]

My solution

Spoiler

we have
$\cos(p\sin\, x) = \sin (p\cos\,x)= \cos (\frac{\pi}{2} - p\cos\,x)$
hence
$p\sin \,x = \frac{\pi}{2} - p \cos \,x$ (other values shall given -ve / larger p)
hence
$p(\cos \ , x + \sin\, x) = \frac{\pi}{2}$
or $\sqrt{2}p( \sin \frac{\pi}{4} \cos \, x + \cos \frac{\pi}{4} \sin \, x) = \frac{\pi}{2}$
or $\sqrt{2}p( \sin ( x + \frac{\pi}{4}) = \frac{\pi}{2}$
the largest value of $\sin ( x + \frac{\pi}{4})$ is 1 hence smallest positive $p$ is $\frac{\pi}{2\sqrt{2}}$