If cos A sin left( A - frac{pi}{6} right) is | vedclass

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(A) Let $f(A) = \cos A \sin \left( A - \frac{\pi}{6} \right)$.Using the identity $2 \sin x \cos y = \sin(x+y) + \sin(x-y)$,we have:$f(A) = \frac{1}{2}

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$\frac{\pi}{3}$

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(A) Let $f(A) = \cos A \sin \left( A - \frac{\pi}{6} \right)$.Using the identity $2 \sin x \cos y = \sin(x+y) + \sin(x-y)$,we have:$f(A) = \frac{1}{2} \left[ \sin \left( A + A - \frac{\pi}{6} \right) + \sin \left( A - (A - \frac{\pi}{6}) \right) \right]$$f(A) = \frac{1}{2} \left[ \sin \left( 2A - \frac{\pi}{6} \right) + \sin \left( \frac{\pi}{6} \right) \right]$$f(A) = \frac{1}{2} \sin \left( 2A - \frac{\pi}{6} \right) + \frac{1}{4}$.For $f(A)$ to be maximum,$\sin \left( 2A - \frac{\pi}{6} \right)$ must be maximum,which is $1$.So,$2A - \frac{\pi}{6} = \frac{\pi}{2}$.$2A = \frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.$A = \frac{\pi}{3}$.

If $\cos A \sin \left( A - \frac{\pi}{6} \right)$ is maximum,then the value of $A$ is equal to

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