User:Paul Wormer/scratchbook

The following equations hold:

\begin{align} \cos(\alpha+\beta) &= \cos\alpha\cos\beta - \sin\alpha\sin\beta\\ \sin(\alpha+\beta) &= \sin\alpha\cos\beta+\cos\alpha\sin\beta \end{align} $$

Proof

The proof is easiest in matrix notation. Because we do not expect readers to know this notation we give the proof in long-hand notation and only indicate briefly the equivalent matrix form.

The singly primed (green) unit vectors are given in terms of the unprimed (black) unit vectors in long-hand and matrix notation, respectively, by

\begin{align} \mathbf{e}_{x'} &= \cos\alpha\,\mathbf{e}_{x} + \sin\alpha\, \mathbf{e}_{y'} \\ \mathbf{e}_{y'} &= -\sin\alpha\,\mathbf{e}_{x} + \cos\alpha\, \mathbf{e}_{y'} \end{align} \quad \Longleftrightarrow\quad \begin{pmatrix}\mathbf{e}_{x'}\\ \mathbf{e}_{y'}\end{pmatrix}= \begin{pmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{pmatrix} \begin{pmatrix}\mathbf{e}_{x}\\ \mathbf{e}_{y}\end{pmatrix} $$ The doubly primed (red) unit vector along the x&quot;-axis is given in terms of the singly primed (green) unit vectors  by

\mathbf{e}_{x''} = \cos\beta\,\mathbf{e}_{x'} + \sin\beta\, \mathbf{e}_{y'} \quad \Longleftrightarrow\quad\mathbf{e}_{x''} = (\cos\beta,\,\,\sin\beta) \begin{pmatrix}\mathbf{e}_{x'}\\ \mathbf{e}_{y'}\end{pmatrix} $$

Substitution of the expression for the singly primed unit vectors (green) in terms of the unprimed vectors (black) gives (in matrix notation this expression would be obtained from the multiplication of a row vector with a square matrix):

\mathbf{e}_{x''} = (\cos\alpha\cos\beta - \sin\alpha\sin\beta)\,\mathbf{e}_{x} + (\sin\alpha\cos\beta+\cos\alpha\sin\beta)\,\mathbf{e}_{y} $$

If we put this next to the expression of the doubly primed (red) vector ex &quot; in terms of the unprimed vectors (black):

\mathbf{e}_{x''} = \cos(\alpha+\beta)\,\mathbf{e}_{x} + \sin(\alpha+\beta)\, \mathbf{e}_{y} $$ we find the equations to be proved. This conclusion is allowed because ex and ey are linearly independent.