###################################
Formula for rotating a vector in 2D
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Let's say we have a point $(x_1, y_1)$.  The point also defines the vector $(x_1, y_1)$.

The vector $(x_1, y_1)$ has length $L$.

We rotate this vector anticlockwise around the origin by $\beta$ degrees.

The rotated vector has coordinates $(x_2, y_2)$

The rotated vector must also have length $L$.

*******
Theorem
*******

.. math::

    x_2 = \cos \beta x_1 - \sin \beta y_1 \\
    y_2 = \sin \beta x_1 + \cos \beta y_1

See: `wikipedia on rotation matrices`_.

*************
Preliminaries
*************

Call the angle between $(x_1, y_1)$ and the x-axis : $\alpha$.  Then:

.. math::
    :label: x_1_y_1

    x_1 = L \cos(\alpha) \\
    y_1 = L \sin(\alpha)

We rotate $(x_1, y_1)$ by angle $\beta$ to get $(x_2, y_2)$.  So the angle
between $(x_2, y_2)$ and the x-axis is $\alpha + \beta$:

.. math::
    :label: x_2_y_2

    x_2 = L \cos(\alpha + \beta) \\
    y_2 = L \sin(\alpha + \beta)

***************************
Proof by the angle sum rule
***************************

If you are happy with :doc:`angle_sum` proof, then we are most of the way
there.

The angle sum rule gives us:

.. math::

    \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\
    \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

So, substituting from equations :eq:`x_1_y_1`, :eq:`x_2_y_2`:

.. math::

    L \cos(\alpha + \beta) =
    L \cos \alpha \cos \beta - L \sin \alpha \sin \beta \implies \\
    x_2 = x_1 \cos \beta - y_1 \sin \beta \\

We do the matching substitutions into $\sin(\alpha + \beta)$ to get $y_2$.

$\blacksquare$

*********************************************
Proof by long-hand variant of angle sum proof
*********************************************

This section doesn't assume the angle sum rule, but uses a version of the
angle-sum proof to prove the rotation formulae.

.. image:: images/rotation_2d.png

We can see from the picture that:

.. math::

    x_2 = r - u

    y_2 = t + s

We are going to use some basic trigonometry to get the lengths of $r, u, t,
s$.

Because the angles in a triangle sum to 180 degrees, $\phi$ on the picture is
$90 - \alpha$ and therefore the angle between lines $q, t$ is also $\alpha$.

Remembering the definitions of $\cos$ and $\sin$:

.. math::

    \cos\theta = \frac{A}{H} \implies A = \cos \theta H

    \sin\theta = \frac{O}{H} \implies O = \sin \theta H

Thus:

.. math::

    x_1 = \cos \alpha L

    y_1 = \sin \alpha L

    p = \cos \beta L

    q = \sin \beta L

    r = \cos \alpha p = \cos \alpha \cos \beta L = \cos \beta x_1

    s = \sin \alpha p = \sin \alpha \cos \beta L = \cos \beta y_1

    t = \cos \alpha q = \cos \alpha \sin \beta L = \sin \beta x_1

    u = \sin \alpha q = \sin \alpha \sin \beta L = \sin \beta y_1

So:

.. math::

    x_2 = r - u = \cos \beta x_1 - \sin \beta y_1

    y_2 = t + s = \sin \beta x_1 + \cos \beta y_1

$\blacksquare$.

.. include:: links_names.inc