# Change of Bases Consisting of Trigonometric Functions

Consider the set of functions $\mathcal{A}=\left\{\sin x, \cos x\right\}$. Denote $V = Span(\mathcal{A})$ the linear space obtained by taking the linear combination of functions in $\mathcal{A}$.

1) Consider another basis $\mathcal{B}=\left\{\sin x-\cos x, \sin x + \cos x\right\}$ of $V$. Find the change of basis matrix $S_{\mathcal{B}\rightarrow\mathcal{A}}$.

2) Given a linear transformation $D: V\rightarrow V,\,\,\,\,\,\,f(x)\mapsto f^{'}(x)$ where $f^{'}$ is the first order derivative of $f$. Find the $\mathcal{A}$-matrix and $\mathcal{B}$-matrix of $D$.

I understand that $S=\bigg[[\sin x - \cos x]_{\mathcal{A}},[\sin x + \cos x]_{\mathcal{A}}\bigg]$, but given $S = \begin{pmatrix} 1/2 & 1/2\\ -1/2 & 1/2\\\end{pmatrix}$, would someone mind illustrating the computation that moves us from one basis to the other? I keep trying, for example, to see how $\frac{1}{2}\sin x - \frac{1}{2}\cos x$ and $\frac{1}{2}\sin x + \frac{1}{2}\cos x$ can be expressed as functions of the original basis, but I'm lost. Should I be taking $\sin (\frac{1}{2}) - \cos (\frac{1}{2})$ and $\sin (\frac{1}{2}) + \cos (\frac{1}{2})$, and seeing how these relate to $\cos (\frac{1}{2})$ and $\sin (\frac{1}{2})$ individually?

Apologies for any previous incoherence or lack of specificity. Thanks again!

For $(1)$ the Matrix $S$ should be
$$\left[ \begin {array}{cc} 1/2&1/2\\ -1/2&1/2 \end {array} \right] .$$
• Why wouldn't it be $S = \begin{pmatrix} 1 & 1\\ -1 & 1\\\end{pmatrix}$? – Benjamin Loya Mar 19 '15 at 17:07