Question

If the period of the function \( f(x) = \frac{\tan 5x \cos 3x}{\sin 6x} \) is \( \alpha \), then find \( f \left( \frac{\alpha}{8} \right) \):

• A. \( \frac{1}{2} \)
• B. \( -1 \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( -\frac{1}{\sqrt{2}} \)

Hint:

To find the period of a function with multiple trigonometric terms, calculate the period of each term and then find the least common multiple (LCM) of the periods.

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = \frac{\tan 5x \cos 3x}{\sin 6x} \] and we are asked to find \( f \left( \frac{\alpha}{8} \right) \), where \( \alpha \) is the period of the function. 
Step 1: To find the period \( \alpha \) of the function, we need to find the periods of the individual trigonometric functions in the numerator and denominator:
- The period of \( \tan(5x) \) is \( \frac{\pi}{5} \),
- The period of \( \cos(3x) \) is \( \frac{2\pi}{3} \),
- The period of \( \sin(6x) \) is \( \frac{2\pi}{6} = \frac{\pi}{3} \).
The period of the function is the least common multiple (LCM) of these three periods. To find the LCM, we consider the denominators of the periods: - \( \frac{\pi}{5}, \frac{2\pi}{3}, \frac{\pi}{3} \).
The LCM of 5, 3, and 3 is 15. Therefore, the period \( \alpha \) of the function is: \[ \alpha = \frac{2\pi}{15}. \]
Step 2: Now that we know \( \alpha = \frac{2\pi}{15} \), we can find \( f \left( \frac{\alpha}{8} \right) \). We substitute \( \frac{\alpha}{8} \) into the function: \[ f \left( \frac{\alpha}{8} \right) = \frac{\tan \left( 5 \times \frac{\alpha}{8} \right) \cos \left( 3 \times \frac{\alpha}{8} \right)}{\sin \left( 6 \times \frac{\alpha}{8} \right)}. \] Substitute \( \alpha = \frac{2\pi}{15} \) into the equation: \[ f \left( \frac{\alpha}{8} \right) = \frac{\tan \left( 5 \times \frac{2\pi}{15 \times 8} \right) \cos \left( 3 \times \frac{2\pi}{15 \times 8} \right)}{\sin \left( 6 \times \frac{2\pi}{15 \times 8} \right)}. \] After evaluating the trigonometric functions, we find that: \[ f \left( \frac{\alpha}{8} \right) = \frac{1}{\sqrt{2}}. \]