# Check if $(2,\pi/2)$ lies on $r=2\cos(2\theta)$

Check if $$(2,\pi/2)$$ lies on $$r=2\cos(2\theta)$$

Plugging $$\theta=\pi/2$$ gives $$r=-2$$
Plugging $$r=2$$ gives $$\cos(2\theta)=1\Rightarrow \theta=n\pi$$
However the graph shows the point lies on the graph. I know this has to do with multiple representations of the same point, but couldn't proceed further. Any help?

• Is $(2,\pi/2)$ in Cartesian or polar coordinates? Sep 7, 2019 at 5:32
• @Ennar polar coordinates Sep 8, 2019 at 6:03

I assume the pair $$(2, \pi/2)$$ is given in polar, rather than cartesian, coordinates.

In this context, the typical convention is that $$r < 0$$ is permitted. In other words: $$r$$ and $$\theta$$ are both permitted to be any real value. As you know, distinct pairs $$(r_1, \theta_1)$$ and $$(r_2, \theta_2)$$ may map to the same point on the plane. So the question of whether $$(2, \pi/2)$$ "lies on the graph" should be interpreted like this: Does there exist some pair $$(r, \theta)$$ of real numbers—equivalent to $$(2, \pi/2)$$ in the sense that it represents the same point on the cartesian plane—which satisfies the equation $$r = 2\cos(2\theta)$$?

The point in question has the following polar representations, and no others: $$(2, \pi/2 + 2\pi n), n \in \mathbb Z$$, and $$(-2, 3\pi/2 + 2\pi n), n \in \mathbb Z$$.

Some of these pairs do satisfy the equation: $$2\cos(2(\pi/2 + 2\pi n)) = -2 \ne 2,$$ $$2\cos(2(3\pi/2 + 2\pi n)) = -2.$$ Therefore, the point lies on the graph.

To elaborate a bit more on what's going on here: The cartesian graph of the polar relation $$r = 2\cos(2\theta)$$ is (by definition) the set of all points $$(x,y) \in \mathbb R^2$$ such that there exists a pair $$(r, \theta) \in \mathbb R^2$$, with $$(x,y) = (r\cos\theta,r\sin\theta)$$, which satisfies $$r = 2\cos(2\theta)$$. This is a bit of a mouthful, but the key thing to understand is this: Although the pair $$(2, \pi/2)$$ does not satisfy our relation, it is true that a pair equivalent to that pair satisfies the relation. So if you plot on the cartesian plane the equation $$y = 2\cos(2\theta)$$, and the point $$(x=\pi/2, y=2)$$, you will see that the point doesn't lie on the graph. But once we replace $$x, y$$ by $$r, \theta$$, multiple points on the graph "collapse" to a single point, so the answer becomes "yes".

To be fair, the question is slightly ambiguous, but I believe this is the usual interpretation in precalculus courses.

• +1. Good explanation with the period. Sep 7, 2019 at 5:10
• Missing $2$ in front of cosines at the last two equations. Better write $r=2\cos...$. Sep 7, 2019 at 5:11
• You are wrong it doesn't exist Sep 7, 2019 at 5:19
• It might seem that it does by looking on the graph but that point is actually (-2,3π/2) Sep 7, 2019 at 5:20
• @farruhota: Thanks, corrected. Sep 7, 2019 at 5:28

The following polar equation depicts a polar rose: $$r(\theta)=a\cos (k\theta +b).$$ So, the given polar equation $$r=2\cos 2\theta$$ has $$a=2,k=2,b=0$$. The property of the polar rose equation is when $$k$$ is even, it will have $$2k$$ petals. Hence, the point $$(r(\theta),\theta)=(2,\frac{\pi}{2})$$ does belong to the curve. It is explained by another property of the polar rose called symmetry by vertical axis: $$r(\pi-\theta)=r(\theta)$$. Indeed: $$r(\pi-\theta)=2\cos2(\pi-\theta)=2\cos(2\pi-2\theta)=2\cos 2\theta=r(\theta).$$

If the coordinates are in Cartesian form, then you are correct. The point does not lie on the graph. If they are in polar form you can refer to Yakov's answer.

If the coordinates are in Cartesian form, the blue point on your graph is at $$r = 2, \theta = \frac{\pi}{2}$$. Converting to Cartesian form we have $$(r \cos \theta, r \sin \theta) = (0,2) \ne (2, \frac{\pi}{2})$$ where $$(2, \frac{\pi}{2})$$ is in Cartesian form.

If they are in polar form, then you can substitute $$r$$ and $$\theta$$ to verify that $$2 = 2 \cos \left(2\cdot\frac{\pi}{2} \right)$$. What you have actually shown is that there are infinitely many points which are on the curve in the range when there is no restriction on $$\theta$$. However, by restricting the domain to $$[0, 2\pi)$$, there is only one solution to the equation.

• Excuse me, what? You converted polar coordinates $(2,\pi/2)$ to Cartesian $(0,2)$ and then claim they are different? Sep 7, 2019 at 5:37