Reading Tip: Dear reader, you may choose not to read this article, but if you do choose to read it, please make sure to read it to the end. BoJone takes no responsibility for any consequences caused by “giving up halfway” ^_^.
Let us consider a rotation problem: given a function graph y=f(x), find the analytic expression of the graph obtained after rotating it counterclockwise by an angle \theta around the point (p,q).
First, since it is merely a rotation, the overall shape of the function’s graph has not changed, so it must still be the original function f. We pick an arbitrary point (x,y) on the original function graph, which corresponds to the point (x',y') on the rotated graph. Then the new analytic expression should be y'=f(x'). Furthermore, since the graph is rotated around (p,q), the distance from (x,y) and (x',y') to (p,q) should be equal. Let this distance be r, i.e., \sqrt{(x-p)^2+(y-q)^2}=r. Let the angle formed by the line segment connecting (x,y) and (p,q) with “the line passing through (p,q) parallel to the x-axis” be \alpha. Thus we have: \sin\alpha=\frac{y-q}{r}, \quad \cos\alpha=\frac{x-p}{r}
Subsequently: \begin{aligned} \sin(\alpha +\theta) &= \sin\alpha \cos\theta + \sin\theta \cos\alpha = \frac{y-q}{r} \cos\theta + \frac{x-p}{r}\sin\theta \\ \cos(\alpha +\theta) &= \cos\alpha \cos\theta - \sin\theta \sin\alpha = \frac{x-p}{r} \cos\theta - \frac{y-q}{r}\sin\theta \end{aligned}
Then it is obvious: \begin{aligned} y' &= \left(\frac{y-q}{r} \cos\theta + \frac{x-p}{r}\sin\theta\right)r + q = (y-q)\cos\theta + (x-p)\sin\theta + q \\ x' &= \left(\frac{x-p}{r} \cos\theta - \frac{y-q}{r}\sin\theta\right)r + p = (x-p)\cos\theta - (y-q)\sin\theta + p \end{aligned}
At this point, the problem seems solved. The analytic expression of the new function is: (y-q)\cos\theta + (x-p)\sin\theta + q = f\left[(x-p)\cos\theta - (y-q)\sin\theta + p\right]
In particular, the equation for rotation around the origin is: y \cos\theta + x \sin\theta = f(x \cos\theta - y \sin\theta)
For example: y=6-x, after rotating counterclockwise around (0,0) by 45^\circ, the result is: \begin{aligned} \frac{\sqrt{2}}{2}(x+y) &= 6 - \left[\frac{\sqrt{2}}{2}(x-y)\right] \\ x &= 3\sqrt{2} \end{aligned}
Attention, here comes the problem! It is clearly y=3\sqrt{2}, so how did it become x=3\sqrt{2}? Haha, everyone has fallen into the “trap” just like me! Upon closer scrutiny, there seems to be a problem; yet if you derive it again, it seems correct. Could it be...? Actually, the problem appeared right at the beginning!
At the start, we assumed the function of the new image is f (note the underlined part); this is groundless and incorrect. We already know the function in the original graph is y=f(x), and then we can determine the relationship between (x,y) and (x',y'). What we are looking for is the relationship between x' and y'. The correct approach is: solve for x and y in terms of x' and y', and then substitute them into y=f(x). The result will be the relationship between x' and y'! The original reasoning process was not entirely wrong; we just need to modify the procedure to get the answer:
Let \beta = \alpha + \theta, then we have: \begin{aligned} \sin(\beta-\theta) &= \sin\beta \cos\theta - \sin\theta \cos\beta = \frac{y'-q}{r} \cos\theta - \frac{x'-p}{r}\sin\theta \\ \cos(\beta -\theta) &= \cos\beta \cos\theta + \sin\theta \sin\beta = \frac{x'-p}{r} \cos\theta + \frac{y'-q}{r}\sin\theta \end{aligned}
Then it is obvious: \begin{aligned} y &= \left(\frac{y'-q}{r} \cos\theta - \frac{x'-p}{r}\sin\theta\right)r + q = (y'-q)\cos\theta - (x'-p)\sin\theta + q \\ x &= \left(\frac{x'-p}{r} \cos\theta + \frac{y'-q}{r}\sin\theta\right)r + p = (x'-p)\cos\theta + (y'-q)\sin\theta + p \end{aligned}
Substituting into y=f(x), we have:
Finally, we have obtained the analytic expression for the new function (counterclockwise): (y-q)\cos\theta - (x-p)\sin\theta + q = f\left[(x-p)\cos\theta + (y-q)\sin\theta + p\right]
If it is a clockwise rotation: (y-q)\cos\theta + (x-p)\sin\theta + q = f\left[(x-p)\cos\theta - (y-q)\sin\theta + p\right]
In particular, the equations for rotation around the origin are:
y \cos\theta - x \sin\theta = f(x \cos\theta + y \sin\theta) (counterclockwise)
y \cos\theta + x \sin\theta = f(x \cos\theta - y \sin\theta) (clockwise)
There should be no mistake this time, right? For y=6-x, after rotating counterclockwise around (0,0) by 45^\circ, the result is: \begin{aligned} \frac{\sqrt{2}}{2}(y-x) &= 6 - \left[\frac{\sqrt{2}}{2}(x+y)\right] \\ y &= 3\sqrt{2} \end{aligned} After some twists and turns, the answer finally comes out...
Reflection:
Now readers understand why I emphasized reading this article to the end, right? It’s not that this article is particularly important, but if you stop halfway and copy the formula or method, you will find it always results in errors when you use it later, which would be unfortunate ^_^.
Why did I lead the readers into a “trap” at the beginning? Actually, I believe many people have made this kind of mistake. This article describes BoJone’s entire thinking process on this problem, from initial thought to error and then to correction. The root of the error lies in thoughts like “taking it for granted” or “it should be.” If one cannot break through these shackles, it is difficult to advance in the fields of mathematics and physics. This article serves as a reminder to myself and a hope that readers will not “follow the overturned cart in front.” Remember: there is no “taking for granted” in mathematics and physics!
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