A trigonometric problem from IIT JEE 1995

 The IIT JEE 1995 subjective maths paper had some nice problems. Already posted an integral, previously.

Here is a trigonometric one.

Find the smallest positive $\rho$ such that the equation in $x$ $$\cos (\rho \sin x) = \sin (\rho \cos x)$$ has a solution in $[0, 2\pi]$.

Scroll down for a solution.

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Let's first try to find some $\rho$. Notice that for $x = \frac{\pi}{4}$  to be a solution we need $$\cos \left(\frac{\rho}{\sqrt{2}}\right) = \sin\left(\frac{\rho}{\sqrt{2}}\right)$$

We can try $$\frac{\pi}{2} - \frac{\rho}{\sqrt{2}} =  \frac{\rho}{\sqrt{2}}$$

giving us $$\rho = \frac{\pi}{2\sqrt{2}}$$

 Thus the answer we seek must $ \leq \frac{\pi}{2\sqrt{2}}$.

Now if $\cos x = \cos y $ the we must have that $x = 2n\pi \pm y$ and so $\cos (\rho \sin x) = \sin (\rho \cos x)$ gives us

$$\rho \sin x = 2n\pi \pm \left(\frac{\pi}{2} - \rho \cos x\right)$$

i.e

$$\rho (\sin x \pm \cos x) = 2n \pi \pm \frac{\pi}{2}$$

i.e $$ \sqrt{2} \rho \sin (x \pm \frac{\pi}{4}) = 2n\pi \pm \frac{\pi}{2}$$

Since $0 < \rho \le \frac{\pi}{2\sqrt{2}}$  and $|\sin \theta| \leq 1$, we must have that $n = 0$ and

$$\sqrt{2} \rho |\sin (x \pm \frac{\pi}{4})| = \frac{\pi}{2}$$

This implies that $ \rho \geq \frac{\pi}{2\sqrt{2}}$.

Thus the least possible value is $\rho = \frac{\pi}{2\sqrt{2}}$.