I need to applied the T- student testing to the parameters $\beta_0, \beta_1, \beta_2,\beta_3,\beta_4$ which model is:
$$ y = \beta_0 + \beta_1 t + \beta_2 Cos(\beta_3 t+\beta_4)$$
to do that, I need the design matrix X, and then $(X'X)^{-1}$ to get $c_{jj}$ diagonal elements to find $Se(B_i)$
I have tried to take the matrix like:
\begin{equation} \beta_0 \quad \beta_1 \quad \; \beta_2 \quad \quad \beta_3 \quad \beta_4 \end{equation}
\begin{equation} \begin{bmatrix} 1 & t_1 & Cos(t_1)& t_1 & 1\\ 1 & t_2 & Cos(t_2)& t_2 & 1\\ \vdots & \vdots & \vdots& \vdots& \vdots\\ 1 & t_n & Cos(t_3)& t_n & 1 \end{bmatrix} \end{equation}
or even;
\begin{equation} \beta_0 \quad \beta_1 \qquad \quad \beta_2 \qquad \quad \beta_3 \quad \beta_4 \end{equation}
\begin{equation} \begin{bmatrix} 1 & t_1 & Cos(\beta_3 t_1 + \beta_4)& t_1 & 1\\ 1 & t_2 & Cos(\beta_3 t_2 + \beta_4)& t_2 & 1\\ \vdots & \vdots & \vdots& \vdots& \vdots\\ 1 & t_n & Cos(\beta_3 t_3 + \beta_4)& t_n & 1 \end{bmatrix} \end{equation}
but with that form of the matrix is not possible to calculate $$(X'X)^{-1}$$