Question

If the period of the function \[ f(x) = 2\cos(3x+4) - 3\tan(2x-3) + 5\sin(5x) - 7 \] is \( k \), then \( f(x) \) is periodic with fundamental period \( k \). Find \( k \).

• A. \( \sin \frac{k}{8} = \frac{1}{2} \)
• B. \( \cos \frac{k}{6} = \frac{1}{\sqrt{2}} \)
• C. \( \tan \frac{k}{3} = -\sqrt{3} \)
• D. \( \sec \frac{k}{2} = 2 \)

Hint:

To find the period of a function composed of multiple trigonometric terms, compute the LCM of individual function periods.

Answer 1

The Correct Option is C

If the period of the function \[ f(x) = 2\cos(3x+4) - 3\tan(2x-3) + 5\sin(5x) - 7 \] is \( k \), then \( f(x) \) is periodic with fundamental period \( k \). Find \( k \). The period of \(\cos(ax+b)\) is \(\frac{2\pi}{|a|}\). The period of \(\tan(ax+b)\) is \(\frac{\pi}{|a|}\). The period of \(\sin(ax+b)\) is \(\frac{2\pi}{|a|}\). The period of \(2\cos(3x+4)\) is \(\frac{2\pi}{3}\). The period of \(-3\tan(2x-3)\) is \(\frac{\pi}{2}\). The period of \(5\sin(5x)\) is \(\frac{2\pi}{5}\). The period of \(f(x)\) is the least common multiple (LCM) of the periods of its terms. We need to find the LCM of \(\frac{2\pi}{3}\), \(\frac{\pi}{2}\), and \(\frac{2\pi}{5}\). To find the LCM of fractions, we find the LCM of the numerators and the GCD of the denominators.
LCM of \(2\pi, \pi, 2\pi\) is \(2\pi\).
GCD of \(3, 2, 5\) is \(1\).
So, the LCM of the periods is \(\frac{2\pi}{1} = 2\pi\). Therefore, \(k = 2\pi\). Now we check the given options: (1) \( \sin \frac{k}{8} = \sin \frac{2\pi}{8} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \neq \frac{1}{2} \)
(2) \( \cos \frac{k}{6} = \cos \frac{2\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2} \neq \frac{1}{\sqrt{2}} \)
(3) \( \tan \frac{k}{3} = \tan \frac{2\pi}{3} = -\sqrt{3} \)
(4) \( \sec \frac{k}{2} = \sec \frac{2\pi}{2} = \sec \pi = -1 \neq 2 \)
Only option (3) is correct. Final Answer: The final answer is $\boxed{(3)}$