Animated Solution for Mathematics - Straight Lines: Let $0<\alpha <\pi /2$ be fixed angle. If $P=(\cos \theta ,\sin \theta )$ and $Q=(\cos (\alpha -\theta ),\sin (\alpha -\theta ))$, then $Q$ is obtained from $P$ by

Welcome, future engineer! Today, we are going to peel back the layers of a seemingly simple coordinate geometry problem. At first glance, you see two points, $P$ and $Q$, defined by trigonometric functions.

It looks like a standard coordinate problem, but beneath the surface lies a beautiful, elegant symmetry. Let's embark on this journey to understand the transformation that maps $P$ to $Q$.

First, let's visualize our workspace. We are in a standard two-dimensional Cartesian plane.

We are given $P=(\cos \theta ,\sin \theta )$ and $Q=(\cos (\alpha -\theta ),\sin (\alpha -\theta ))$. If you recall your basic trigonometry, any point of the form $(\cos \varphi ,\sin \varphi )$ lies on the unit circle centered at the origin.

Both $P$ and $Q$ are dancing on the circumference of this circle. The vector $OP$ makes an angle $\theta$ with the positive $x$-axis, and the vector $OQ$ makes an angle $\alpha -\theta$ with the same axis.

Your first instinct might be to assume this is a rotation. It is a very common trap!

If we were to rotate $P$ to $Q$, the angle of rotation would be the difference between their angular positions. Let's calculate that:

Notice something? The rotation angle depends on $\theta$. If you move $P$, the angle of rotation changes. A true rotation must be a rigid motion with a fixed angle. Since our 'rotation' depends on the position of $P$, we can confidently discard the rotation hypothesis.

If it is not a rotation, what could it be? Let's consider a reflection.

Imagine a mirror line passing through the origin. When you reflect a point across a line, the line acts as an angle bisector between the original vector and its image.

If $Q$ is the reflection of $P$, then the line of reflection must be the line that bisects the angle between $OP$ and $OQ$. To find this line, we simply need to find the average of the two angles.

Let's perform the calculation. The angle of $P$ is $\theta$, and the angle of $Q$ is $\alpha -\theta$. The angle of the bisector line is the average of these two:

Watch closely as we simplify this. The $\theta$ and $-\theta$ in the numerator cancel each other out perfectly! We are left with:

This is the magic of the problem. The variable $\theta$ has vanished! This tells us that the line of reflection is fixed at an angle of $\frac{\alpha }{2}$ with the positive $x$-axis, regardless of where $P$ is located.

Now, we just need to express this line in terms of its slope. A line passing through the origin with an angle of inclination $\varphi$ has a slope $m=\tan \varphi$.

Here, our angle is $\frac{\alpha }{2}$, so the slope of our reflection line is $m=\tan (\frac{\alpha }{2})$. Thus, $Q$ is obtained by reflecting $P$ across the line $y=x\tan (\frac{\alpha }{2})$.

You have just solved a complex geometric transformation by finding the hidden symmetry. Keep this mindset—look for the constant in the variables, and the solution will reveal itself!