Question

Let $x,y,z$ be real numbers with $x\ge y\ge z\ge \frac{\pi }{12}$ such that $x+y+z=\frac{\pi }{2}$. If $p=\cos x\sin y\cos z,$ then

• A. the maximum value of $p$ is $\frac{2+\sqrt{3}}{8}$
• B. the minimum value of $p$ is $\frac{1}{8}$
• C. the maximum value of $p$ is attained when $x=y=\frac{5\pi }{24}$ and $z=\frac{\pi }{12}$
• D. the minimum value of $p$ is attained when $x=y=z=\frac{\pi }{6}$

Answer 1

Answer: (A), (B), (C)

the maximum value of $p$ is attained when $x=y=\frac{5\pi }{24}$ and $z=\frac{\pi }{12}$

Since, $\frac{\pi }{2}>x\ge y\ge z\ge \frac{\pi }{12}$
$\Rightarrow \sin (y-z)\ge 0,\sin (x-y)\ge 0$

$p=\cos x\sin y\cos z$
$\Rightarrow p=\frac{1}{2}\cos x\left[\sin \right(y+z)+\sin (y-z\left)\right]$
$\Rightarrow p\ge \frac{1}{2}\cos x\sin (y+z)[\because \sin (y-z)\ge 0]$
$\Rightarrow p\ge \frac{1}{2}{\cos }^{2}x$

$x+y+z=\frac{\pi }{2}$
$x=\frac{\pi }{2}-(y+z)\le \frac{\pi }{2}-2\times \frac{\pi }{12}=\frac{\pi }{3}$
$\Rightarrow x\le \frac{\pi }{3}$
Hence, the minimum value of $p$ is $\frac{1}{2}{\cos }^2\frac{\pi }{3}=\frac{1}{8}$, which is obtained when $x=\frac{\pi }{3}$ and $\sin (y-z)=0$
i.e., $p$ is minimum when $x=\frac{\pi }{3}$ and $y=z=\frac{\pi }{12}$ as $x+y+z=\frac{\pi }{2}$

On the other hand, we also have
$p=\frac{1}{2}\cos z\left[\sin \right(x+y)-\sin (x-y\left)\right]$
$\Rightarrow p\le \frac{1}{2}\cos z\sin (x+y)[\because \sin (x-y)\ge 0]$
$\Rightarrow p\le \frac{1}{2}{\cos }^{2}z$
$\Rightarrow p\le \frac{1}{4}(1+\cos 2z)$
$\Rightarrow p\le \frac{1}{4}(1+\cos \frac{\pi }{6})=\frac{2+\sqrt{3}}{8}$
This maximum value is obtained when$x=y=\frac{5\pi }{24}$ and $z=\frac{\pi }{12}$