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 using the least squares method, we will follow these steps: 1. **Set up the equations**: We will create a system of equations based on the least squares criterion, which minimizes the sum of the squares of the residuals (the differences between the observed values and the values predicted by the model). 2. **Formulate the system**: For each data point \( (t_i, y_i) \), we can express the residual as: \[ r_i = y_i - (c_0 + c_1 \sin(t_i) + c_2 \cos(t_i)) \] The goal is to minimize the sum of the squares of these residuals: \[ S = \sum_{i=1}^{n} r_i^2 \] 3. **Calculate the necessary sums**: We will compute the following sums based on the data points: - \( n \): number of data points - \( \sum y_i \) - \( \sum \sin(t_i) \) - \( \sum \cos(t_i) \) - \( \sum \sin(t_i)^2 \) - \( \sum \cos(t_i)^2 \) - \( \sum \sin(t_i) \cos(t_i) \) 4. **Set up the normal equations**: The normal equations derived from minimizing \( S \) will be: \[ n c_0 + c_1 \sum \sin(t_i) + c_2 \sum \cos(t_i) = \sum y_i \] \[ c_0 \sum \sin(t_i) + c_1 \sum \sin(t_i)^2 + c_2 \sum \sin(t_i) \cos(t_i) = \sum y_i \sin(t_i) \] \[ c_0 \sum \cos(t_i) + c_1 \sum \sin(t_i) \cos(t_i) + c_2 \sum \cos(t_i)^2 = \sum y_i \cos(t_i) \] 5. **Substitute the data points**: The data points are \( (0, 7), \left(\frac{\pi}{2}, 7\right), (\pi, -7), \left(\frac{3\pi}{2}, -11\right) \). Now, let's calculate the necessary sums: - For \( t_1 = 0 \), \( y_1 = 7 \): - \( \sin(0) = 0 \) - \( \cos(0) = 1 \) - For \( t_2 = \frac{\pi}{2} \), \( y_2 = 7 \): - \( \sin\left(\frac{\pi}{2}\right) = 1 \) - \( \cos\left(\frac{\pi}{2}\right) = 0 \) - For \( t_3 = \pi \), \( y_3 = -7 \): - \( \sin(\pi) = 0 \) - \( \cos(\pi) = -1 \) - For \( t_4 = \frac{3\pi}{2} \), \( y_4 = -11 \): - \( \sin\left(\frac{3\pi}{2}\right) = -1 \) - \( \cos\left(\frac{3\pi}{2}\right) = 0 \) Now we can compute the sums: - \( n = 4 \) - \( \sum y_i = 7 + 7 - 7 - 11 = -4 \) - \( \sum \sin(t_i) = 0 + 1 + 0 - 1 = 0 \) - \( \sum \cos(t_i) = 1 + 0 - 1 + 0 = 0 \) - \( \sum \sin(t_i)^2 = 0^2 + 1^2 + 0^2 + (-1)^2 = 2 \) - \( \sum \cos(t_i)^2 = 1^2 + 0^2 + (-1)^2 + 0^2 = 2 \) - \( \sum \sin(t_i) \cos(t_i) = 0 \cdot 1 + 1 \cdot 0 + 0 \cdot (-1) + (-1) \cdot 0 = 0 \) - \( \sum y_i \sin(t_i) = 7 \cdot 0 + 7 \cdot 1 - 7 \cdot 0 - 11 \cdot (-1) = 7 + 11 = 18 \) - \( \sum y_i \cos(t_i) = 7 \cdot 1 + 7 \cdot 0 - 7 \cdot (-1) + (-11) \cdot 0 = 7 + 7 = 14 \) Now we can set up the normal equations: 1. \( 4c_0 + 0c_1 + 0c_2 = -4 \) 2. \( 0c_0 + 2c_1 + 0c_2 = 18 \) 3. \( 0c_0 + 0c_1 + 2c_2 = 14 \) This simplifies to: 1. \( 4c_0 = -4 \) → \( c_0 = -1 \) 2. \( 2c_1 = 18 \) → \( c_1 = 9 \) 3. \( 2c_2 = 14 \) → \( c_2 = 7 \) Thus, the coefficients are: \[ c_0 = -1, \quad c_1 = 9, \quad c_2 = 7 \] Final results: - \( c_0 = -1 \) - \( c_1 = 9 \)