Question

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.

r = 5 cos 3θ

Answer 1

lol

Answer 2

read it http://www.most.gov.mm/techuni/media/EM_02011_chap5b.pdf

Answer 3

testing for y-symmetry
let's make r = 5 cos 3θ and set \(\theta = \pi - \theta\)
so
\(\bf r = 5 cos 3\theta\\
\theta = (\pi - \theta)\\
r = 5cos(3(\pi - \theta)) \implies r = 5cos(3\pi- 3\theta)
\)

Answer 4

our resultant equation, DOESN'T RESEMBLE the original equation, therefore, no y-symmetry
now let's check for x-symmery
by setting \(\theta = - \theta\)

Answer 5

hold on, lemme check really quick and expand the cosine above

Answer 6

$$\bf
r = 5 cos 3\theta\\
\theta = (\pi - \theta)\\
r = 5cos(3(\pi - \theta)) \implies r = 5cos(3\pi- 3\theta)\\
cos(3\pi- 3\theta) = cos(3\pi)cos(3\theta)+sin(3\pi)sin(3\theta)\\
cos(3\pi- 3\theta) =-cos(3\theta)\\
\text{thus then}\\
r = 5cos(3\pi- 3\theta) = -5cos(3\theta)
$$
so our resultant equation, DOESN'T RESEMBLE the original equation, therefore, no y-symmetry

Answer 7

now let's test for x-symmetry by setting \(\theta = -\theta\)

Answer 8

$$\bf
r = 5 cos 3\theta\\
\theta = ( - \theta)\\
r = 5cos(3(- \theta)) \implies r = 5cos(-3\theta)\\
\color{blue}{cos(-x) = cos(x)} \text{ thus then}\\
r = 5cos(-3\theta) = r = 5 cos 3\theta
$$
so our resultant equation DOES RESEMBLE the original one, thus there's an x-symmetry

Answer 9

so now let's test for the origin
by setting "r" to -r
so
$$\bf
r = 5 cos 3\theta\\
r = -r\\
-r = 5 cos 3\theta \implies r = -5 cos 3\theta
$$

so our resultant DOESN'T RESEMBLE the original function, thus no dice on the origin symmetry

Answer 10

thank you soooooo much :)

Answer 11

yw

Answer 12

I'm not understanding why \(\theta = \pi - \theta\) ...

Answer 13

to check for symmetry along the y-axis, or the so-called \(\bf\large \frac{\pi}{2}\) line
you set the \(\theta = \pi - \theta\)
and then factor and simplify to see if you get the original equation

Answer 14

for the test one can use either \(\bf \theta = \pi - \theta\) or you set "r" and \(\theta\) both to \(\bf -r\ and \ -\theta\)

Answer 15

and \(\bf \large \pi-a\) on the 2nd quadrant, will show symmetry over the y-axis