So I have the the column matrix $X=\begin{pmatrix} a\\ b \end{pmatrix}$ with $a,b$ real numbers and to this matrix I have the following vector associated: $\vec{x}=a\cdot \vec{i}+b\cdot\vec{j}$

I have the matrix: $A(t)=\begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}$ with $t$ real number

$B=\begin{pmatrix} 1\\ 1 \end{pmatrix}$ , $U=A(-\frac{\pi}{12})\cdot B$ and $V=A(\frac{\pi}{6})\cdot B$

I need to find the cosinus of vectors $\vec{u}$ and $\vec{v}$ these vectors being associated to column matrix $U$ and $V$.The response is $\frac{\sqrt{2}}{2}$

So I just did calculations I it's a little bit of work but I find that the matrix A(t) is a rotation matrix with angle t and the cosinus between u and v is pi/4 but at school I didn't learn about rotation matrix.Can you explain me why the angle is pi/4? I mean, if you can, in simple terms how rotation matrix work.


  • 3
    $\begingroup$ $A(t)$ acts on the vector $B$ by rotating it by $t$. So, $A(-\pi/12)$ rotates $B$ by $-\pi/12$, and $A(\pi/6)$ rotates $B$ by $\pi/6$. The difference between these two angles is $\pi/6 - (-\pi/12) = \pi/6 + \pi/12 = 3\pi/12 = \pi/4$. $\endgroup$
    – user169852
    Jul 7 '19 at 8:29
  • $\begingroup$ I agree with @Bungo's comment/explanation. I just wanted to remark that $A(t)$ is a rotation by angle minus $t$ counterclockwise (i.e. $t$ clockwise). $\endgroup$
    – Malkoun
    Jul 7 '19 at 9:03
  • $\begingroup$ “How a rotation matrix works” has been answered in many places in many ways, for example math.stackexchange.com/questions/363652/… $\endgroup$
    – David K
    Jul 7 '19 at 12:32
  • $\begingroup$ So do you want to completely understand what makes that particular formulation of $A(t)$ be a rotation matrix, or do you just want to know how the angle between $u$ and $v$ comes to be $\pi/4$? Did Bungo's comment answer your question? If not, where did it stop making sense? Remember you can edit the question to make it clearer what you really need to know. $\endgroup$
    – David K
    Jul 8 '19 at 1:56
  • $\begingroup$ Bungo's comment was helpful.I understood.Thanks! $\endgroup$
    – DaniVaja
    Jul 9 '19 at 8:44

The important thing is that every rotation $\varphi$ about the origin is a linear transformation, meaning that, for all vectors $v, w$ of the plane and scalar $\lambda$, we have $$\varphi(\lambda v) =\lambda\cdot\varphi(v) \\ \varphi(v+w) =\varphi(v) +\varphi(w)$$ These properties can be easily seen geometrically, in particular, drawing the second one says something like '$\varphi$ takes parallelograms to parallelograms'.
It is also clear that the composition of rotation by angle $t$ with rotation by angle $s$ is the rotation by $s+t$.

Now, by the basics of linear algebra, since $\varphi$ is linear, once a basis $e_1,e_2$ is fixed, it can be expressed as multiplication by the matrix whose columns are $\varphi(e_1),\varphi(e_2)$ (coordinated in basis $e_1,e_2$).
Consequently, composition of linear transformations corresponds to matrix multiplication.

In the given example $A(t)$ is the matrix of rotation by angle $-t$, with respect to the standard basis $i,j$.
Based on the above, it should be straightforward that $$A(t)\cdot A(s)\ =\ A(t+s)$$ which boils down to the trigonometric addition formulas for $\sin$ and $\cos$ (and proves them at once).


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