\( g ( t ) = 8 \sin t \cos t - 3 \sec t , a = \frac { 11 \pi } { 4 } \)

To analyze the function \( g(t) = 8 \sin t \cos t - 3 \sec t \) at the point \( a = \frac{11\pi}{4} \), we can start by simplifying the expression and then evaluating it at the given value of \( t \). First, we can use the double angle identity for sine: \[ \sin(2t) = 2 \sin t \cos t \] Thus, we can rewrite \( 8 \sin t \cos t \) as: \[ 8 \sin t \cos t = 4 \sin(2t) \] Now, substituting this back into the function \( g(t) \): \[ g(t) = 4 \sin(2t) - 3 \sec t \] Next, we need to evaluate \( g\left(\frac{11\pi}{4}\right) \). First, we simplify \( \frac{11\pi}{4} \): \[ \frac{11\pi}{4} = 2\pi + \frac{3\pi}{4} \] This means that \( \frac{11\pi}{4} \) is equivalent to \( \frac{3\pi}{4} \) in terms of sine and cosine, since sine and cosine are periodic with a period of \( 2\pi \). Now, we can find \( \sin\left(\frac{3\pi}{4}\right) \) and \( \cos\left(\frac{3\pi}{4}\right) \): \[ \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \] Next, we calculate \( \sec\left(\frac{3\pi}{4}\right) \): \[ \sec\left(\frac{3\pi}{4}\right) = \frac{1}{\cos\left(\frac{3\pi}{4}\right)} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2} \] Now we can evaluate \( g\left(\frac{11\pi}{4}\right) \): 1. Calculate \( 2t \) when \( t = \frac{11\pi}{4} \): \[ 2t = 2 \cdot \frac{11\pi}{4} = \frac{22\pi}{4} = \frac{11\pi}{2} \] 2. Find \( \sin\left(\frac{11\pi}{2}\right) \): Since \( \frac{11\pi}{2} = 5\pi + \frac{\pi}{2} \), we can reduce it to \( \frac{\pi}{2} \): \[ \sin\left(\frac{11\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \] 3. Now substitute back into \( g(t) \): \[ g\left(\frac{11\pi}{4}\right) = 4 \sin\left(\frac{11\pi}{2}\right) - 3 \sec\left(\frac{3\pi}{4}\right) \] Substituting the values we found: \[ g\left(\frac{11\pi}{4}\right) = 4 \cdot 1 - 3 \cdot (-\sqrt{2}) = 4 + 3\sqrt{2} \] Thus, the final result is: \[ \boxed{4 + 3\sqrt{2}} \]