# What goes wrong in my parametric equation for it to produce only a partial rotation of the graph of $f(x)=\cos(x)$?

As a pastime, I'm trying to rotate the graph of $$f(x)=\cos(x)$$ about the point $$O=(0,0)$$, and I chose, arbitrarily, a rotation of $$\pi /4$$ radians.

My question is : how comes the parametric equation I came up with only works for the part of the graph that is to the right of $$O=(0,0)$$?

I reasoned as follow : rotating the graph is tentamount to rotating each point $$P=(x, f(x))$$ by $$\pi/4$$ rd. on a circle ( one for each point) centered at $$(0,0)$$ and of radius $$OP = \sqrt {x^2+f(x)^2}$$.

So I defined the following auxiliary functions :

(1) Let $$OP= d(x)= \sqrt {x^2 + f(x)^2}$$

(2) Let $$T(x)=$$ slope of $$OP = \cos(x)/x$$

(3) Let $$A(x)$$= angle of $$OP$$ with the $$X-$$axis $$= \arctan (T(x))$$.

Inasmuch as each point $$P$$ is supposed to turn on a circle of radius $$d(x)$$ and of center $$O$$, trigonometry can be used which gives ( with $$P'$$ denoting the image of $$P$$ under the expected $$\pi/4$$ counterclockwise rotation) :

• $$X$$ coordinate of $$P'$$ : $$d(x)\cos(A(x)+\pi/4)$$

• $$Y$$ coordinate of $$P'$$ : $$d(x)\sin(A(x)+\pi/4)$$.

Hence the parametric equation :

$$\bigg$$

However, this parametric equation produces the desired result only for positive values of $$t$$ ( or of $$x$$).

Probably some feature of the $$\tan(x)$$ function could explain this, but I cannot identify which one it is.

here : a link to my attenpt using Desmos https://www.desmos.com/calculator/mjj6mnr55t

Below, an image for $$x\gt 0$$

Here , an image of what happens when $$x\lt 0$$ , point P is in green , its "image " under the parametric function is in dark grey.

• i think it will work when you use a signed distance ie replace d(t) with d(t)sgn (t) Nov 24, 2021 at 15:09
• @CalvinKhor.- I'm going to try this, thanks for this hint! Nov 24, 2021 at 15:12
• Alternatively, plot the points $\left(\dfrac{x-f(x)}{\sqrt2},\dfrac{x+f(x)}{\sqrt2}\right)$. No need for polar coordinates.
– user995027
Nov 24, 2021 at 15:24
• And @SteveDaou's suggestion is just a special case of plotting $\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} x \\ f(x) \end{pmatrix}$. Nov 24, 2021 at 15:26
• @prets gave a more general answer.
– user995027
Nov 24, 2021 at 15:44

By OP's request, a short explanation of the matrix formula: The matrix $$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$ is the $$2 \times 2$$ rotation matrix by $$\theta$$ radians, as can be observed by noting that the first column is the anticlockwise rotation of the unit vector $$(1, 0)$$ by $$\theta$$ radians, and the second column is the same transformation of $$(0, 1)$$. (Draw some triangles with hypotenuse $$1$$ to convince yourself of this if you haven't played this game before!)
Hence if you take any point/vector $$(x, y)$$ in the plane and multiply it by this matrix, you get precisely the same point back, only rotated $$\theta$$ radians anticlockwise.
For our purposes, the points we are interested in are the points $$(x, f(x))$$ on the graph, so $$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} x \\ f(x) \end{pmatrix} = \begin{pmatrix} x \cos(\theta) - f(x) \sin(\theta) \\ x \sin(\theta) + f(x) \cos(\theta) \end{pmatrix}$$ is exactly the rotation (by $$\theta$$ radians anticlockwise) of the graph in question.
Specialising at $$\theta = \pi/4$$ then gives the parametrisation $$\Bigl( \frac{x - f(x)}{\sqrt{2}}, \frac{x + f(x)}{\sqrt{2}} \Bigr)$$ since $$\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$$.
• In complex numbers, $(x+if(x))\,\text{cis}(\theta)$.