Let $R_\theta(v)$ be the result of rotating a point $v = (x, y)$ around the origin through an angle of $\theta$.
Then $R_\theta(v) = (x\cos\theta-y\sin\theta, x\sin\theta + y\cos\theta)$.
Now given diagram like this:
(From Calculus/Spivak)
Show that:
$$x'=\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y$$ $$y'=-\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y$$
Intuitively that makes sense, but I haven't been able to express the sides $x',y'$ using some expression.
If we start with some point $v =(x,y)$ and rotate it by $45^\circ$, then:
$$R_\theta(v) = (x\cos\theta-y\sin\theta, x\sin\theta + y\cos\theta) \\ = (x\frac{1}{\sqrt{2}}-y\frac{1}{\sqrt{2}}, x\frac{1}{\sqrt{2}} + y\frac{1}{\sqrt{2}})$$
And if we refer to $x',y'$ as vectors, then:
$$v = x' + y'$$
But I can't bring my thoughts together. How can I express the sides of the rectangle in terms of $x, y$?
One thing is to keep the point fixed and rotate the the axes, another is to rotate the point keeping the axes fixed.
Clearly you pass from one to the other by inverting the sign of the angle.
The matrix $$ {\bf R}_{\,\,z} (\alpha) = \left( {\matrix{ {\cos \alpha } & { - \sin \alpha } \cr {\sin \alpha } & {\cos \alpha } \cr } } \right) $$ when applied to a vertical vector (point or axis), returns the vector rotated by $\alpha$ in the old (i.e. fixed) coordinates.
While $$ {\bf R}_{\,\,z} ( - \alpha) = {\bf R}_{\,\,z} ^{\,T} (\alpha) = {\bf R}_{\,\,z} ^{\, - \,{\bf 1}} (\alpha) $$ when applied to a vertical vector gives that vector (fixed) in the new coordinates rotated by $+\alpha$
(or the vector rotated by $-\alpha$, keeping reference fixed).
And that is what you were asked to prove.
Told in other terms, without using the concept of matrix, if we indicate by $\bf i,\, \bf j$ and $\bf {i'},\, \bf {j'}$ the unit reference vectors, and if (as it is usually done) you start with determining the new vectors referenced to the old, which upon a rotation of $+\alpha$ gives the system on the LHS $$ \left\{ \matrix{ {\bf i'} = \cos \alpha \,{\bf i} + \sin \alpha {\kern 1pt} {\bf j} \hfill \cr {\bf j'} = - \sin \alpha \,{\bf i} + \cos \alpha {\kern 1pt} {\bf j} \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{ {\bf i} = \cos \alpha \,{\bf i'} - \sin \alpha {\kern 1pt} {\bf j'} \hfill \cr {\kern 1pt} {\bf j} = \sin \alpha \,{\bf i'} + \cos \alpha {\kern 1pt} {\bf j'} \hfill \cr} \right. $$ and which, when solved for $\bf i,\, \bf j$ provides that on the RHS.
Now the same vector $\bf v$ ( a "fixed" vector) will be expressed in the two references as $$ {\bf v} = x\,{\bf i} + y{\kern 1pt} {\bf j} = x'{\bf i'} + y'{\bf j'} $$ Then $$ \eqalign{ & {\bf v} = x'{\bf i'} + y'{\bf j'} = x\,{\bf i} + y{\kern 1pt} {\bf j} = \cr & = x\left( {\cos \alpha \,{\bf i'} - \sin \alpha {\kern 1pt} {\bf j'}} \right) + y\left( {\sin \alpha \,{\bf i'} + \cos \alpha {\kern 1pt} {\bf j'}} \right)\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ x' = x\cos \alpha + y\sin \alpha \hfill \cr y' = - x\sin \alpha + y\cos \alpha \hfill \cr} \right. \cr} $$
That is: $$ \left( {\matrix{ {x'} \cr {y'} \cr } } \right) = \left( {\matrix{ {\cos \alpha } & {\sin \alpha } \cr { - \sin \alpha } & {\cos \alpha } \cr } } \right)\left( {\matrix{ x \cr y \cr } } \right) = {\bf R}_{\,\,z} ( - \alpha)\left( {\matrix{ x \cr y \cr } } \right) $$ as told above (in matrix terms).