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Home » A_\textcircle = \pi r^2 = \pi \cdot 25 = 25\pi - Abu Waleed Tea

A_\textcircle = \pi r^2 = \pi \cdot 25 = 25\pi - Abu Waleed Tea

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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2\sin x + 3\cos x = \sqrt{13} \sin(x + \phi)
The range of $ \sqrt{13} \sin(x + \phi) $ is $ [-\sqrt{13}, \sqrt{13}]...
2\sin x + 3\cos x + 4 \in [4 - \sqrt{13}, 4 + \sqrt{13}]
Since $ \sqrt{13} \approx 3.605 $, $ 4 - \sqrt{13} \approx 0.395 > 0 $...
Therefore, the absolute value $ |2\sin x + 3\cos x + 4| $ achieves its...
The minimum occurs when $ 2\sin x + 3\cos x $ is minimized, but since...
\min |2\sin x + 3\cos x + 4| = \min (2\sin x + 3\cos x + 4) = 4 - \sqr...
\min |2\sin x + 3\cos x + 4| = 4 - \sqrt{13} \quad \text{if } 4 - \sqr...
But $ 4 - \sqrt{13} > 0 $ since $ \sqrt{13} < 4 $, so yes.
But wait: is this the minimum? Since the expression inside is always g...

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