Trigonometrical Equations 3 Question 16
17. Find the smallest positive number $p$ for which the equation $\cos (p \sin x)=\sin (p \cos x)$ has a solution $x \in[0,2 \pi]$.
(1995, 5M)
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Answer:
Correct Answer: 17. Smallest positive value of $p=\frac{\pi}{2 \sqrt{2}}$
Solution:
- Given, $\cos (p \sin x)=\sin (p \cos x), \forall x \in[0,2 \pi]$
$\begin{array}{rlrl}\Rightarrow & & \cos (p \sin x) & =\cos \frac{\pi}{2}-p \cos x \ \Rightarrow & p \sin x & =2 n \pi \pm \frac{\pi}{2}-p \cos x, n \in I\end{array}$
$ \Rightarrow \quad p \sin x+p \cos x=2 n \pi+\pi / 2 $
or $\quad p \sin x-p \cos x=2 n \pi-\pi / 2, n \in I$
$\Rightarrow \quad p(\sin x+\cos x)=2 n \pi+\pi / 2$
or $\quad p(\sin x-\cos x)=2 n \pi-\pi / 2, n \in I$
$\Rightarrow p \sqrt{2}\left(\cos \frac{\pi}{4} \sin x+\sin \frac{\pi}{4} \cos x\right)=2 n \pi+\frac{\pi}{2}$
or $p \sqrt{2} \cos \frac{\pi}{4} \sin x-\sin \frac{\pi}{4} \cos x=2 n \pi-\frac{\pi}{2}, n \in I$
$\Rightarrow \quad p \sqrt{2}[\sin (x+\pi / 4)]=\frac{(4 n+1) \pi}{2}$
or
$ p \sqrt{2}[\sin (x-\pi / 4)]=(4 n-1) \frac{\pi}{2}, n \in I $
Now, $\quad-1 \leq \sin (x \pm \pi / 4) \leq 1$
$\Rightarrow \quad-p \sqrt{2} \leq p \sqrt{2} \sin (x \pm \pi / 4) \leq p \sqrt{2}$
$\Rightarrow \quad-p \sqrt{2} \leq \frac{(4 n+1) \cdot \pi}{2} \leq p \sqrt{2}, n \in I$
or $\quad-p \sqrt{2} \leq \frac{(4 n-1) \pi}{2} \leq p \sqrt{2}, n \in I$
Second inequality is always a subset of first, therefore we have to consider only first.
It is sufficient to consider $n \geq 0$, because for $n>0$, the solution will be same for $n \geq 0$.
If $n \geq 0$
$ -\sqrt{2} p \leq(4 n+1) \pi / 2 $
$ \Rightarrow \quad(4 n+1) \pi / 2 \leq \sqrt{2} p $
For $p$ to be least, $n$ should be least.
$ \begin{array}{ll} \Rightarrow & n=0 \\ \Rightarrow & \sqrt{2} p \geq \pi / 2 \quad \Rightarrow \quad p \geq \frac{\pi}{2 \sqrt{2}} \end{array} $
Therefore, least value of $p=\frac{\pi}{2 \sqrt{2}}$
