Discussion
What is the maximum value of \(3\sin \theta + 4\cos \theta\)?
Answer: C
For an expression of the form \(a\sin \theta + b\cos \theta\), the maximum value is \(\sqrt{a^2+b^2}\) and the minimum value is \(-\sqrt{a^2+b^2}\).
Here, a = 3 and b = 4.
Maximum value = \(\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\).
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