Attempt \( 1: 10 \) attempts remaining. Fit a trigonometric function of the form \( f(t)=c_{0}+c_{1} \sin (t)+c_{2} \cos (t) \) to the data points \( (0,7),\left(\frac{\pi}{2}, 7\right),(\pi,-7),\left(\frac{3 \pi}{2},-11\right) \), using least squares. \( c_{0}= \) \( c_{1}= \) You may find it helpful to review the video below:

To fit a trigonometric function of the form \( f(t) = c_{0} + c_{1} \sin(t) + c_{2} \cos(t) \) to the given data points, we can use the least squares method to determine the coefficients \( c_{0} \), \( c_{1} \), and \( c_{2} \). First, we will set up a system of equations based on the given data points: 1. \( f(0) = c_{0} + c_{1} \sin(0) + c_{2} \cos(0) = c_{0} + c_{2} = 7 \) 2. \( f\left(\frac{\pi}{2}\right) = c_{0} + c_{1} \sin\left(\frac{\pi}{2}\right) + c_{2} \cos\left(\frac{\pi}{2}\right) = c_{0} + c_{1} = 7 \) 3. \( f(\pi) = c_{0} + c_{1} \sin(\pi) + c_{2} \cos(\pi) = c_{0} - c_{2} = -7 \) 4. \( f\left(\frac{3\pi}{2}\right) = c_{0} + c_{1} \sin\left(\frac{3\pi}{2}\right) + c_{2} \cos\left(\frac{3\pi}{2}\right) = c_{0} - c_{1} = -11 \) Now, let's solve this system of equations step by step. From the first equation, \( c_{0} + c_{2} = 7 \) (1). From the second equation, \( c_{0} + c_{1} = 7 \) (2). From the third equation, \( c_{0} - c_{2} = -7 \) (3). From the fourth equation, \( c_{0} - c_{1} = -11 \) (4). Now, we can rearrange equations (1) and (3): From (1): \( c_{2} = 7 - c_{0} \) (5). From (3): \( c_{0} = -7 + c_{2} \) or \( c_{0} + c_{2} = 7 \). Substituting (5) into (3): \( c_{0} - (7 - c_{0}) = -7 \) This simplifies to: \( 2c_{0} - 7 = -7 \) Solving for \( c_{0} \): \( 2c_{0} = 0 \) So, \( c_{0} = 0 \). Now substituting \( c_{0} = 0 \) back into (1): \( 0 + c_{2} = 7 \) Thus, \( c_{2} = 7 \). Substituting \( c_{0} = 0 \) into (2): \( 0 + c_{1} = 7 \) Thus, \( c_{1} = 7 \). Finally, we have: \( c_{0} = 0 \), \( c_{1} = 7 \), \( c_{2} = 7 \). So, the final coefficients are: \( c_{0} = 0 \) \( c_{1} = 7 \)