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Period of f(θ) = cos 3θ

Hello,

I wonder if anyone can help me make sense of the attached passage which is from the book Engineering Mathematics, by K.A. Stroud.

To begin with, when he writes "Here we see that, for example..." and so on - what is his reasoning here?

Thank you in advance!

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Reply 1

TeeEm

Original post by TanPiPerTwo

period = 2pi/n always

Reply 2

TanPiPerTwo

Thanks! That is true but I am trying to understand his reasoning - but cannot follow...

Reply 3

A.McCall

I am not an expert I just do Mathematics A2 for further maths AS

But when you have the graph sin(theta) it's period is 2Pi in radians the amount of time it takes to make one sinusoidal wave. This means that when you calculate (theta) your answers will be 0,2Pi, 4Pi, 6Pi etc...

But when you have the graph sin(4 theta) it's period is a scale factor of 1/4 the original sin wave. (even I don't quite think that is a good explanation)

So here is some maths :smile:

When you have sin(theta) to find theta you use the sin^-1 function on the calculator for sin(theta) = 0 it would give you 0, 2Pi, 4Pi

For sin(4 theta) you use the sin^-1 function on the calculator for sin(4 theta) = 0 would give you 0, 2Pi, 4Pi but as the answer for 4 (theta) so all your value then have to be divided by 4 to get the period for 4 (theta)

This is just a C3 and C4 student saying this. The only experience I have is alot of trigonometry work so not sure how much help I am but I tried.

Reply 4

TeeEm

Original post by TanPiPerTwo

Thanks! That is true but I am trying to understand his reasoning - but cannot follow...

it is do do with the transformation f(ax)

f(2x) stretches horizontally by 1/2, so there are 2 cycles from 0 to 2pi, so period = 2pi/2
f(3x) stretches horizontally by 1/3, so there are 3 cycles from 0 to 2pi, so period = 2pi/3
f(4x) stretches horizontally by 1/4, so there are 4 cycles from 0 to 2pi, so period = 2pi/4

etc

Reply 5

TanPiPerTwo

But when writes:

"Here we see that, for example:

f(θ) = cos 3θ =1 when 3θ = 0, 2π, 4π etc..."

How am I meant to "see" this? Where did this information come from?

Sorry, being stupid here, I know...

Reply 6

Smaug123

Original post by TanPiPerTwo

\cos(3 \theta) = 1

. Let

u=3\theta

to get

\cos(u) = 1

. This has solutions

u=0, 2 \pi, 4 \pi, \dots

- are you happy with that bit?

Reply 7

A.McCall

it the cosine wave it starts at 1 on the y axis at the point 0 on the x axis then the full wave takes till 2 Pi on the x axis to get back to the point 1 on the y axis.

Just google sin waves and cosine waves it will help you understand alot better.

You are not being stupid just misguided.

Reply 8

TenOfThem

Original post by TanPiPerTwo

You are expected to know that Cos of 0, 2pi, 4pi, etc is 1

Reply 9

TanPiPerTwo

Thanks all.

So because f(θ) = cos θ = 1 when θ = 0, 2π, 4π etc., it follows that f(θ) = cos 3θ = 1 when 3θ = 0, 2π etc.?

Reply 10

TenOfThem

Original post by TanPiPerTwo

Thanks all.

So because f(θ) = cos θ = 1 when θ = 0, 2π, 4π etc., it follows that f(θ) = cos 3θ = 1 when 3θ = 0, 2π etc.?

yes

Reply 11

TanPiPerTwo

But when he goes on to say:

"So that f(θ + 2π/3) = cos 3(θ + 2π/3)..."

I've again no idea where this comes from.

(edited 9 years ago)

Reply 12

Smaug123

Original post by TanPiPerTwo

But when he goes on to say:

"So that f(θ + 2π/3) = cos 3(θ + 2π/3)..."

I've again no idea where this comes from.

That's identically what you get if you evaluate

f

\theta + \frac{2}{3} \pi

. Let

\theta = \theta + \frac{2}{3} \pi

in the definition of

f

Reply 13

TanPiPerTwo

OK but why is he taking that step?

Reply 14

Smaug123

Original post by TanPiPerTwo

OK but why is he taking that step?

It's verifying that, as expected, the period divides

\frac{2}{3} \pi

.

We know that it is actually

\frac{2}{3} \pi

, but this particular line of working just checks that there is a repetition every 2/3pi, not that this is the shortest distance between repetitions. That was done earlier, in finding when the function was 1 and showing that it happened every 2/3 pi: that showed that the period was at least 2/3 pi. This line, however, shows that the period is at most 2/3 pi. Hence the two lines together show that the period is 2/3 pi.

Reply 15

Phichi

Original post by A.McCall

So here is some maths :smile:

Some bad maths!!!!

Reply 16

Smaug123

Original post by A.McCall

When you have sin(theta) to find theta you use the sin^-1 function on the calculator for sin(theta) = 0 it would give you 0, 2Pi, 4Pi

Answers along the lines of "it's true for this mathematical reason" are preferred to answers along the lines of "use this button on your calculator". Note that

\sin^{-1} (\sin(0)) = 0

and not

2 \pi

. Taking inverse-sin of a sin expression will give you one of the possible solutions; the others will differ from it by multiples of 2pi and by subtracting it from

\pi

Reply 17

A.McCall

Original post by Smaug123

Answers along the lines of "it's true for this mathematical reason" are preferred to answers along the lines of "use this button on your calculator". Note that

\sin^{-1} (\sin(0)) = 0

and not

2 \pi

. Taking inverse-sin of a sin expression will give you one of the possible solutions; the others will differ from it by multiples of 2pi and by subtracting it from

\pi

Yeah I get that because you would have to take the solutions in range you have and also you would get the value Pi.

I just think of it as a graph and I can't explain that very well...

I get it stop bullying me :'(