The period of the function is \( \frac{2\pi}{b} \). Since \( f\left(\frac{\pi}{2b}\right) = 4 \), it indicates a quarter-period shift, consistent with \( b \) being arbitrary for simplicity. Assume one full cycle completes in \( \frac{2\pi}{b} = 2\pi \), giving \( b = 1 \).

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