Respuesta :
Answer
[TeX]f(t)=-0.2+4.1sin(t)+4cos(t) [/TeX]
Step-By-Step Explanation
Given the function [TeX]f(t)=c_0+c_1sin(t)+c_2cos(t) [/TeX].
For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)
[TeX]f(0)=c_0+0c_1+c_2=5.5 [/TeX].
[TeX]f(\pi /2)=c_0+c_1+0c_2=0.5 [/TeX].
[TeX]f(\pi)=c_0+0sin(t)-c_2=-2.5 [/TeX].
[TeX]f(3\pi /2)=c_0-c_1+0c_2=-7.5 [/TeX].
Expressing this as a system of linear equations in matrix form AX=B
[TeX]\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right) [/TeX]
Where
[TeX]A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right) [/TeX],
[tex]B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)[/tex]
[tex]X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)[/tex]
To determine the values of X, we use the expression
[TeX]X=(A^{T}A)^{-1}A^{T}B[/TeX]
[TeX]A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right) [/TeX]
[TeX] (A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right) [/TeX]
[TeX]A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right) [/TeX]
Therefore:
[TeX]X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right) [/TeX]
[tex]X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)=\left(\begin{array}{c} -0.2 \\4.1\\4\end{array}\right)[/tex]
Therefore, the trigonometric function which fits to the given data is:
[tex]f(t)=-0.2+4.1sin(t)+4cos(t)[/tex]
