Question

Find the period of the function. y = -3 cos 1/4x

Answer 1

Normally, the period of \(y=\cos x\) is \(2\pi\). You have \(y=-3\cos\frac{x}{4}\). Does the presence of the \(-3\) in front change the period?

Answer 2

I'm not sure @whpalmer4

Answer 3

No, it doesn't, it only affects the up and down size of the wiggles, not how often they happen.

Here are 3 different cos functions, differing only in the number in front. One (the blue one) is plain old \(\cos x\). The purple one is \(3\cos x\). The olive one is \(-2\cos x\), which is inverted relative to the other two because I multiplied it by a negative number, but you can see, I hope, that the periods are identical for all 3.

Answer 4

Let's try attaching the diagram, Bill!

Answer 5

Any question about that part?

Answer 6

No questions

Answer 7

But 2pi isn't an answer choice for me

Answer 8

pi/4

-3

3pi/4

8π

Answer 9

2pi is the period for y= cos(x)
2pi isn't the period for y=cos(ax) (note a doesn't equal 0)

Answer 10

though you could find the period of y=cos(ax) by doing 2pi/a

Answer 11

I got 4pi but it's not an answer choice

Answer 12

because it isn't the answer

Answer 13

what is your a here according to what I'm calling a?

Answer 14

\[y=-3 \cos(\frac{1}{4}x)=-3 \cos(a x)\]
it should be easy to identify what I'm calling a

Answer 15

1/4

Answer 16

So what is 2pi/(1/4) ?

Answer 17

\[\frac{2\pi}{\frac{1}{4}}\]

Answer 18

8pi

Answer 19

great

Answer 20

to find the period of sin or cos you do 2pi/(what's in front of x)
once you have the function like this: (usually they call the thing in front of b instead a)
y=acos(bx+c)
y=asin(bx+c)

Answer 21

okay thank you (:

Answer 22

Sorry, I got distracted, looks like @myininaya handled it for you.

What I was going to say is that you can think of \(2\pi\) as the conversion factor between real-world numbers and trig-world numbers.

If you want a cos function that has 1 cycle per unit on your x-axis, you use \(\cos(2\pi x)\).

If you want a cos function that has 4 cycles per unit on your x-axis, you use \(\cos( 4*2\pi x) = \cos(8\pi x)\).

Similarly, if you want a cos function that has 1/3 of a cycle per unit on your x-axis, you would use \(\cos(\frac{1}{3}*2\pi x) = \cos(\frac{2}{3}\pi x)\).

Multiplying by a number whose magnitude is greater than 1 increases the frequency of the wiggles (and thus shortens the period), and multiplying by a number whose magnitude is less than 1 decreases the frequency of the wiggles (and thus lengthens the period).

Answer 23

To get one complete cycle, the argument of the \(\cos x\) function has to go from \(0\) to \(2\pi\), or go through any other range of \(2\pi\). Each time the quantity you are using for the argument to \(\cos x\) goes through \(2\pi\) radians, your graph makes a complete cycle. Finding period and frequency is just a matter of doing proportions with \(2\pi\).

If you remember the business of scaling and transforming functions, like \(y = f(x)\) becomes \(y = f(x)+a\) if you shift it by \(a\) units along the y-axis, or \(y = f(x)\) becomes \(y = f(x-a)\) if you shift it by \(a\) units along the x-axis, we're doing the same thing here, except we are using the transformation \(y = f(ax)\) which compresses or expands along the x-axis by a factor of \(a\).