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

  • Thread starter toneboy1
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In summary: 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) - 1/2which is about as simple as you can get, with a period...In
  • #1
toneboy1
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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!
 
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  • #2
use sin(2t)=2sin(t)cos(t). then write sin(3t)=sin(2t+t) and expand it. next use
[itex]\cos(2t) = 2 \cos^{2}t - 1[/itex]. you then end up with
[tex]x(t) = \cos t + 2 \cos^{2}t - \frac{1}{2}[/tex]
 
  • #3
samalkhaiat said:
...then write sin(3t)=sin(2t+t) and

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:
  • #4
toneboy1 said:
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!
Once you have a sum of cosines, the period is the lowest common multiple of the periods of the terms.
 
  • #5
toneboy1 said:
That is very clever.

So I got to:

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


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


Thanks

sin(t+2t) = sin(t)cos(2t)+sin(2t)cos(t)
 
  • #6
nasu said:
Once you have a sum of cosines, the period is the lowest common multiple of the periods of the terms.

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
samalkhaiat said:
sin(t+2t) = sin(t)cos(2t)+sin(2t)cos(t)

Isn't that what I did?
 
  • #8
toneboy1 said:
What about if the Cosine is negative? Anyway so that would mean from the answer in my original post the period was 't'?

Thanks
-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
nasu said:
-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.

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
 

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  • #10
What would you expect? What is the period of cos(t)?
 
  • #11
nasu said:
What would you expect? What is the period of cos(t)?

Please enlighten me, 360 deg?
 
  • #12
In radians is 2π or about 6.28.
 
  • #13
nasu said:
In radians is 2π or about 6.28.

so from the result I got, how was 2π≈6.28?
 
  • #14
toneboy1 said:
so from the result I got, how was 2π≈6.28?

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.
Matlab uses radians, I suppose.
 
  • #15
nasu said:
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.
Matlab uses radians, I suppose.

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
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
nasu said:
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.

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
toneboy1 said:
Isn't that what I did?

No, look again at your calculations. you put

[tex]\sin (2t) = \sin (2t) \cos (t) + \cos (2t) \sin (2t)[/tex]

which is wrong.
 
  • #19
samalkhaiat said:
No, look again at your calculations. you put

[tex]\sin (2t) = \sin (2t) \cos (t) + \cos (2t) \sin (2t)[/tex]

which is wrong.

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
 

1. What is the formula for combining cos(t)'s from (sin2t + sin3t)/2*sint?

The formula for combining cos(t)'s from (sin2t + sin3t)/2*sint is cos(t) = (sin2t + sin3t)/2*sint.

2. Can the formula be simplified?

Yes, the formula can be simplified to cos(t) = (sin5t)/2*sint.

3. How can I solve for t?

To solve for t, you can use trigonometric identities and algebraic manipulation to isolate t. You can also use a graphing calculator or software to find the solutions visually.

4. Are there any special cases for this formula?

Yes, there are special cases where the formula may not hold true. For example, when sin2t + sin3t = 0, the formula becomes undefined. Additionally, when sint = 0, the equation simplifies to 0 = 0, making it difficult to solve for t.

5. How is this formula used in real-world applications?

This formula is commonly used in physics and engineering to model oscillatory motion, such as in pendulums or springs. It can also be used in trigonometric identities and manipulations to simplify complex equations.

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