# Combining Cos(t)'s from (sin2t + sin3t)/2*sint

1. Sep 8, 2012

### toneboy1

The question was given: x(t) = (sin2t + sin3t)/2*sint

determine the period. So I converted numerator and denominator into the exponential forms and ended up getting (after flipping, timesing and cancelling)
(exp(j3t)-exp(-j)-exp(j)+exp(-3j)+exp(j4)-exp(-2j)-exp(j2)+exp(-4j)) / 2

which in turn turns into

x(t) = cos(3t) - cos(t) - cos(2t) + cos(4t) if I'm not mistaken

so is there a way of simplifying this into maybe one Cos or something so I can find the period?

Or have I done this question the wrong way?

Thanks heaps!

2. Sep 9, 2012

### samalkhaiat

use sin(2t)=2sin(t)cos(t). then write sin(3t)=sin(2t+t) and expand it. next use
$\cos(2t) = 2 \cos^{2}t - 1$. you then end up with
$$x(t) = \cos t + 2 \cos^{2}t - \frac{1}{2}$$

3. Sep 10, 2012

### toneboy1

That is very clever.

So I got to:

(2.sin(t).cos(t) + sin(2t).cos(t)+cos(2t).sin(2t)) / 2sin(t)

[2.sin(t).cos(t)+2sin(t).cos(t).cos(t)+(cos^2(t)-sin^2(t))*(2.sin(t).cos(t)) ]
/ 2sint

(then) although I have ended up with cost+cos^2 t + cos^3 t - sin^2(t).cost

Have I done something wrong or am I not finished simplifying it? (I can't think of what else to do).

Thanks

Last edited: Sep 10, 2012
4. Sep 10, 2012

### nasu

Once you have a sum of cosines, the period is the lowest common multiple of the periods of the terms.

5. Sep 10, 2012

### samalkhaiat

sin(t+2t) = sin(t)cos(2t)+sin(2t)cos(t)

6. Sep 10, 2012

### toneboy1

What about if the Cosine is negative? Anyway so that would mean from the answer in my original post the period was 't'?

Thanks

7. Sep 10, 2012

### toneboy1

Isn't that what I did?

8. Sep 11, 2012

### nasu

-cos(x)=cos(x+π)
That component will have a phase of π, in respect to the other components. Does not change the period, though.

9. Sep 11, 2012

### toneboy1

Ah, I see. So it was '1'. When I plotted the sum of: x(t) = cos(3t) - cos(t) - cos(2t) + cos(4t) in matlab it generated a period of 6.275 or something, how can this be?

Observe the picture attached

#### Attached Files:

• ###### PF.png
File size:
13.6 KB
Views:
170
10. Sep 11, 2012

### nasu

What would you expect? What is the period of cos(t)?

11. Sep 11, 2012

### toneboy1

12. Sep 11, 2012

### nasu

13. Sep 11, 2012

### toneboy1

so from the result I got, how was 2π≈6.28?

14. Sep 11, 2012

### nasu

I am not sure what are you asking. The value of 2π is not a consequence of your result but the other way around.
π or "PI" is 3.1415...
so 2π= 6.28....

Now the period of a function like sin(x) or cos (x) is 2π. This means that sin(x+2π) = sin(x).
In these expressions x is a non-dimensional variable.

You can also write sin(x+360) = sin(x) but then your x should be in hexadecimal degrees too.

15. Sep 11, 2012

### toneboy1

AH! Because I was plotting it like it was in Time (seconds)

Playing Devil's advocate, would it be fair to make 't' as seconds and plot the thing and say it took 6.28 seconds for a period?

16. Sep 11, 2012

### nasu

In physics we usually make the arguments of the functions non-dimensional.
A physical quantity periodic in time will be represented by something like
sin(2∏t/T) so t is in seconds and T is the "physical" period, in seconds.
This has the property

sin[2∏(t+T)/T)]=sin(2∏t/T) so the period is T.

17. Sep 11, 2012

### toneboy1

what do you mean by 'non-dimensional' exactly?

Could I plead the case that since that T is 2π, it is 6.28 seconds?

How did the 'sin[2∏(t+T)/T)]' come about?

18. Sep 11, 2012

### samalkhaiat

No, look again at your calculations. you put

$$\sin (2t) = \sin (2t) \cos (t) + \cos (2t) \sin (2t)$$

which is wrong.

19. Sep 11, 2012

### toneboy1

You're right, thanks, would you be happy with:

(2.sin(t).cos(t) + sin(2t).cos(t)+cos(2t).sin(t)) / 2sin(t)

[2.sin(t).cos(t)+2sin(t).cos(t).cos(t)+(cos^2(t)-sin^2(t))*sin(t)) ]
/ 2sint

∴ 2cost + 2cos^2(t) +Cos^2(t)-(1-cos^2(t)) / 2

∴ 2Cos^2(t) + Cos(t) - 1/2

which is about as simple as you can get, with a period 2π?

Thanks