Let x, y, z be real numbers such that x geq y | vedclass

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(D) Given $x+y+z = \frac{\pi}{2}$ and $x \geq y \geq z \geq \frac{\pi}{12}$.We want to minimize $f(x, y, z) = \cos x \sin y \cos z$.Using the product-

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$\frac{1}{8}$

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(D) Given $x+y+z = \frac{\pi}{2}$ and $x \geq y \geq z \geq \frac{\pi}{12}$.We want to minimize $f(x, y, z) = \cos x \sin y \cos z$.Using the product-to-sum formula: $\cos x \cos z = \frac{1}{2}(\cos(x+z) + \cos(x-z))$.Since $x+z = \frac{\pi}{2} - y$,we have $\cos(x+z) = \sin y$.So,$f = \frac{1}{2}(\sin y + \cos(x-z)) \sin y = \frac{1}{2}(\sin^2 y + \sin y \cos(x-z))$.To minimize this,we consider the boundary conditions. Since $x \geq y \geq z \geq \frac{\pi}{12}$ and $x+y+z = \frac{\pi}{2}$,the minimum occurs when $y$ and $z$ are as small as possible,i.e.,$y = z = \frac{\pi}{12}$.Then $x = \frac{\pi}{2} - (y+z) = \frac{\pi}{2} - \frac{2\pi}{12} = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.Substituting these values: $\cos(\frac{\pi}{3}) \sin(\frac{\pi}{12}) \cos(\frac{\pi}{12}) = \frac{1}{2} \cdot \frac{1}{2} \sin(\frac{\pi}{6}) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$.

Let $x, y, z$ be real numbers such that $x \geq y \geq z \geq \frac{\pi}{12}$. If $x+y+z = \frac{\pi}{2}$,then the minimum value of $\cos x \cdot \sin y \cdot \cos z$ is

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