Question

If \( 5\sin\theta + 3\cos\left(\theta + \frac{\pi}{3}\right) + 3 \) lies between \( \alpha \) and \( \beta \) (including \( \alpha, \beta \) also), then \( (\alpha-\beta)(\alpha+\beta-6) = \)

• A. \( 28-5\sqrt{3} \)
• B. 0
• C. 3
• D. \( 28+5\sqrt{3} \)

Hint:

Always check the expression to be evaluated before diving into complex calculations. Often, symmetry (like \( \alpha+\beta = 2C \)) simplifies the problem drastically.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We simplify the trigonometric expression into the form \( R\sin(\theta + \phi) + k \). The range of this function is \( [k-R, k+R] \).
Step 2: Detailed Explanation:

Let \( f(\theta) = 5\sin\theta + 3\cos(\theta + 60^\circ) + 3 \). Expand the cosine term: \[ \cos(\theta + 60^\circ) = \cos\theta \cos 60^\circ - \sin\theta \sin 60^\circ = \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta \] Substitute back: \[ f(\theta) = 5\sin\theta + 3\left( \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta \right) + 3 \] \[ f(\theta) = \left( 5 - \frac{3\sqrt{3}}{2} \right)\sin\theta + \frac{3}{2}\cos\theta + 3 \] This is of the form \( A\sin\theta + B\cos\theta + C \). The minimum value is \( \alpha = C - \sqrt{A^2+B^2} \). The maximum value is \( \beta = C + \sqrt{A^2+B^2} \). Here \( C = 3 \). Let \( R = \sqrt{A^2+B^2} \). So, \( \alpha = 3 - R \) and \( \beta = 3 + R \). We need to calculate \( (\alpha-\beta)(\alpha+\beta-6) \). Calculate terms: 1. \( \alpha - \beta = (3-R) - (3+R) = -2R \). 2. \( \alpha + \beta - 6 = (3-R) + (3+R) - 6 = 6 - 6 = 0 \). The product is: \[ (-2R) \times 0 = 0 \] Thus, calculating the actual value of \( R \) is unnecessary.
Step 4: Final Answer:

The value is 0.