If A + B = frac{pi}{2}, the maximum value of | vedclass

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(A) Let $f(A) = \cos A \cos B = \cos A \cos \left( \frac{\pi}{2} - A \right) = \cos A \sin A = \frac{1}{2} \sin 2A.$ Since the maximum value of $\sin

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$\frac{1}{2}$

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(A) Let $f(A) = \cos A \cos B = \cos A \cos \left( \frac{\pi}{2} - A ight) = \cos A \sin A = \frac{1}{2} \sin 2A.$ Since the maximum value of $\sin 	heta$ is $1,$ the maximum value of $\frac{1}{2} \sin 2A$ occurs when $\sin 2A = 1.$ Thus,the maximum value is $\frac{1}{2} 	imes 1 = \frac{1}{2}.$ Alternatively,using derivatives: $f'(A) = \cos 2A.$ Setting $f'(A) = 0$ gives $2A = \frac{\pi}{2},$ so $A = \frac{\pi}{4}.$ Since $f''(A) = -2 \sin 2A,$ at $A = \frac{\pi}{4},$ $f''(\frac{\pi}{4}) = -2 \sin \frac{\pi}{2} = -2 The maximum value is $\cos \frac{\pi}{4} \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}} = \frac{1}{2}.$

If $A + B = \frac{\pi}{2},$ the maximum value of $\cos A \cos B$ is

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