Question

The maximum value of \( 5\cos \theta + 3\cos \left(\theta + \frac{\pi}{3}\right) + 3 \) is

• A. 5
• B. 11
• C. 10
• D. -1
• E. 2

Hint:

Expression of type \( A\cos\theta + B\sin\theta \) has maximum \( \sqrt{A^2 + B^2} \).

Answer 1

The Correct Option is C

Concept: Use identity: \[ \cos(\theta + \alpha) = \cos\theta \cos\alpha - \sin\theta \sin\alpha \] Then convert into form: \[ A\cos\theta + B\sin\theta \] whose maximum is: \[ \sqrt{A^2 + B^2} \]

Step 1: Expand cosine term.
\[ \cos\left(\theta + \frac{\pi}{3}\right) = \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta \]

Step 2: Substitute.
\[ 5\cos\theta + 3\left(\frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta\right) + 3 \] \[ = \left(5 + \frac{3}{2}\right)\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3 \] \[ = \frac{13}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3 \]

Step 3: Find amplitude.
\[ R = \sqrt{\left(\frac{13}{2}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} = \frac{1}{2}\sqrt{169 + 27} = \frac{1}{2}\sqrt{196} = 7 \]

Step 4: Maximum value.
\[ 7 + 3 = 10 \] Final Answer: \[ 10 \]