Find the minimum value of p for which cos (p sin x) = sin (p cos x)

1.	Find the minimum value of p for which cos (p sin x) = sin (p cos x) has solution in [0, 2π].
Answer» Given, cos (p sinx) = sin (p cosx) ⇒ sin (\(\frac{\pi}{2}\)− p sinx ) = sin (p cosx) ⇒ \(\frac{\pi}{2}\) − p sinx = πcosx ⇒ p (sinx + cosx) = \(\frac{\pi}{2}\) ⇒ \(\sqrt{2 }\) p (sinx \(\frac{1}{\sqrt{2}}\) + cosx \(\frac{1}{\sqrt{2}}\) ) =\(\frac{\pi}{2}\) ⇒ \(\sqrt{2 }\) p [sinx cos\(\frac{\pi}{4 }\)+ cosx sin \(\frac{\pi}{4 }\) ] = \(\frac{\pi}{2}\) ⇒ \(\sqrt{2 }\) p [sin (x + \(\frac{\pi}{4 }\) )] = \(\frac{\pi}{2}\) ⇒ sin (x + \(\frac{\pi}{4 }\)) = \(\frac{\pi}{2\sqrt{2}p}\) = \(\frac{\pi}{(2)\frac{1}{2^p}}\) Since, sine lies between – 1 and 1 Therefore, − 1 ≤ \(\frac{\pi}{(2)\frac{1}{2^p}}\) ≤ 1 Hence, the minimum value of p is \(\frac{\pi}{(2)\frac{1}{2}}\) [∴ Positive value of p is to taken]

Find the minimum value of p for which cos (p sin x) = sin (p cos x) has solution in [0, 2π].

Answer»

Given, cos (p sinx) = sin (p cosx)

⇒ sin (\(\frac{\pi}{2}\)− p sinx ) = sin (p cosx)

⇒ \(\frac{\pi}{2}\) − p sinx = πcosx

⇒ p (sinx + cosx) = \(\frac{\pi}{2}\)

⇒ \(\sqrt{2 }\) p (sinx \(\frac{1}{\sqrt{2}}\) + cosx \(\frac{1}{\sqrt{2}}\) ) =\(\frac{\pi}{2}\)

⇒ \(\sqrt{2 }\) p [sinx cos\(\frac{\pi}{4 }\)+ cosx sin \(\frac{\pi}{4 }\) ] = \(\frac{\pi}{2}\)

⇒ \(\sqrt{2 }\) p [sin (x + \(\frac{\pi}{4 }\) )] = \(\frac{\pi}{2}\)

⇒ sin (x + \(\frac{\pi}{4 }\)) = \(\frac{\pi}{2\sqrt{2}p}\) = \(\frac{\pi}{(2)\frac{1}{2^p}}\)

Since, sine lies between – 1 and 1

Therefore, − 1 ≤ \(\frac{\pi}{(2)\frac{1}{2^p}}\) ≤ 1

Hence, the minimum value of p is \(\frac{\pi}{(2)\frac{1}{2}}\)

[∴ Positive value of p is to taken]