Question

Let the range of the function \[ f(x) = 6 + 16 \cos x \cdot \cos\left(\frac{\pi}{3} - x\right) \cdot \cos\left(\frac{\pi}{3} + x\right) \cdot \sin 3x \cdot \cos 6x, \quad x \in R \text{ be } [\alpha, \beta]. \] Then the distance of the point \((\alpha, \beta)\) from the line \(3x + 4y + 12 = 0\) is:

• A. 11
• B. 8
• C. 10
• D. 9

Hint:

To simplify expressions involving trigonometric functions, remember to use the product-to-sum and sum-to-product formulas, as well as double and triple angle identities. Recognizing these patterns helps in efficiently solving the problem.

Answer 1

The Correct Option is A

\[f(x) = 6 + 16 \cos x \cdot \cos\left(\frac{\pi}{3} - x\right) \cdot \cos\left(\frac{\pi}{3} + x\right) \cdot \sin 3x \cdot \cos 6x \] \[ = 6 + 16 \cos x \cdot \left(\cos^2\left(\frac{\pi}{3}\right) - \sin^2 x\right) \sin 3x \cos 6x \] \[ = 6 + 16 \cos x \cdot \left(\frac{1}{4} - \sin^2 x\right) \sin 3x \cos 6x \] \[ = 6 + 4 \cos x \cdot \left(1 - 4 \sin^2 x\right) \sin 3x \cos 6x \] \[ = 6 + 4 \cos x \cdot \left(1 - 4 \sin^2 x\right) \sin 3x \cos 6x \] Since \(\sin 3x = 3 \sin x - 4 \sin^3 x = \sin x(3 - 4 \sin^2 x)\), then \(1-4\sin^2 x = \frac{3\sin x - \sin 3x}{\sin x} - 4 \sin^2 x \) \[ = 6 + 4 \cos x \left(\cos 2x - \sin^2x\right) \sin 3x \cdot \cos 6x \] Using the identity \( \cos A \cos(A - B) \cos(A + B) = \frac{1}{4} \cos 3B \), we have: \[ \cos x \cos\left(\frac{\pi}{3} - x\right) \cos\left(\frac{\pi}{3} + x\right) = \frac{1}{4} \cos 3x \] So, \[ f(x) = 6 + 16 \cdot \frac{1}{4} \cos 3x \cdot \sin 3x \cdot \cos 6x = 6 + 4 \cos 3x \sin 3x \cos 6x = 6 + 2 \sin 6x \cos 6x = 6 + \sin 12x \] Since \( -1 \le \sin 12x \le 1 \), \[ 5 \le f(x) \le 7 \] So \( [\alpha, \beta] = [5, 7] \) The distance of the point \( (5, 7) \) from the line \( 3x + 4y + 12 = 0 \) is \[ \frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}} = \frac{|15 + 28 + 12|}{5} = \frac{55}{5} = 11 \]