If the function f(x) = frac{4sqrt{2}(sin 3x + | vedclass

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(A) Step $1$: Simplify the numerator using $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$.$\sin 3x + \sin x = 2 \sin 2x \cos x$.Numerator

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$(2)^2$

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(A) Step $1$: Simplify the numerator using $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$.$\sin 3x + \sin x = 2 \sin 2x \cos x$.Numerator $= 4\sqrt{2} (2 \sin 2x \cos x) = 8\sqrt{2} \sin 2x \cos x$.Step $2$: Simplify the denominator using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$.$2 \sin 2x \sin \frac{3x}{2} = \cos(2x - \frac{3x}{2}) - \cos(2x + \frac{3x}{2}) = \cos \frac{x}{2} - \cos \frac{7x}{2}$.Denominator $= \cos \frac{x}{2} - \cos \frac{7x}{2} + \cos \frac{5x}{2} - \cos \frac{3x}{2} = (\cos \frac{x}{2} - \cos \frac{3x}{2}) + (\cos \frac{5x}{2} - \cos \frac{7x}{2})$.$= 2 \sin x \sin \frac{x}{2} + 2 \sin 3x \sin \frac{x}{2} = 2 \sin \frac{x}{2} (\sin x + \sin 3x) = 2 \sin \frac{x}{2} (2 \sin 2x \cos x) = 4 \sin \frac{x}{2} \sin 2x \cos x$.Step $3$: Calculate $f(x) = \frac{8\sqrt{2} \sin 2x \cos x}{4 \sin \frac{x}{2} \sin 2x \cos x} = \frac{2\sqrt{2}}{\sin \frac{x}{2}}$.Step $4$: For continuity at $x = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = \lim_{x 	o \frac{\pi}{2}} \frac{2\sqrt{2}}{\sin \frac{x}{2}} = \frac{2\sqrt{2}}{\sin \frac{\pi}{4}} = \frac{2\sqrt{2}}{1/\sqrt{2}} = 4$.

If the function $f(x) = \frac{4\sqrt{2}(\sin 3x + \sin x)}{2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2} - \cos \frac{3x}{2}}$ for $x \neq \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$, then the value of $f(\frac{\pi}{2})$ is equal to

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