Questions: Question 16 Use the fact that the trigonometric functions are periodic to find the exact value of the expression. Do not use a calculator sin(11π/3) A. √3/2 B. -√3/2 C. -1 D. -1/2

Solution Steps

The sine function has a period of \( 2\pi \), meaning \( \sin(\theta) = \sin(\theta + 2\pi k) \) for any integer \( k \).

To find \( \sin \frac{11\pi}{3} \), reduce \( \frac{11\pi}{3} \) to an angle between \( 0 \) and \( 2\pi \): \[ \frac{11\pi}{3} - 2\pi = \frac{11\pi}{3} - \frac{6\pi}{3} = \frac{5\pi}{3}. \] Thus, \( \sin \frac{11\pi}{3} = \sin \frac{5\pi}{3} \).

The angle \( \frac{5\pi}{3} \) is in the fourth quadrant, where the sine function is negative. The reference angle is \( \frac{\pi}{3} \), and \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \). Therefore: \[ \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}. \]

Final Answer