Complex Trig, DE

  • Thread starter jlatshaw
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  • #1
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Homework Statement


Hello,
I am in differential equations currently and I have a homework question regarding simplifying

sin( Pi t)/4

into

.5 * i E^(-.25 i pi t) - .5 * i E^(.25 i pi t)


Homework Equations


I think they might be using Euler's Identity, but I am unsure.
E^(a + ib)t = E^(at) (cos[bt] + i sin[bt])​​
 

Answers and Replies

  • #2
LCKurtz
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What happens if you subtract these two equations:$$
e^{i\theta} = \cos\theta + i\sin\theta$$ $$
e^{-i\theta} = \cos\theta - i\sin\theta$$
 
  • #3
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I get:

2 i Sin(theta)

Even still, how does that get me closer to my end?

Thank you.
-James
 
  • #4
LCKurtz
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You didn't write the whole equation. When you add/subtract two equations you get another equation.
 
  • #5
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Ok,

E^(i theta) - E^(-i theta) = 2 i sin(theta)

I'm still a little bit confused as to where I can go from this.
I like the way the right hand side of the equation is looking, but I don't know what to do with the imaginary component in 2 i sin(theta)

Thanks,
-James
 
  • #6
LCKurtz
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Look at what you have and what you are trying to get and what value of ##\theta## you need.
 
  • #7
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Ok, so I have
.5 * i E^(-.25 i pi t) - .5 * i E^(.25 i pi t)

knowing:
E^(i theta) = cos(theta) + i sin(theta)
E^(-i theta) = cos(theta) - i sin(theta)
so
.5 i E^(i theta) = .5 i (cos(theta) + i sin(theta)) = .5 (i cos(theta) - sin(theta))
.5 i E^(i -theta) = .5 i (cos(theta) - i sin(theta)) = .5 (i cos(theta) + sin(theta))

I guess we can let theta = .25 pi t
Can I say,
-.5 i E^(i theta) + .5 i E^(i -theta)
= -.5 (i cos(theta) - sin(theta)) + .5 (i cos(theta) + sin(theta))
= sin(theta)
= sin(.25 pi t)

But where does the 1/4 come into play?
 
  • #8
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So sorry! made a big bobo!

I want to show:
.5 * i E^(-.25 i pi t) - .5 * i E^(.25 i pi t) = Sin(pi t/4)
Using the above does just that.

Thanks so much!!! :)
 
  • #9
LCKurtz
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You are making this way too hard. You already have this:

Ok,

E^(i theta) - E^(-i theta) = 2 i sin(theta)

Now, instead of doing everything all over like you have here:

Ok, so I have
.5 * i E^(-.25 i pi t) - .5 * i E^(.25 i pi t)

knowing:
E^(i theta) = cos(theta) + i sin(theta)
E^(-i theta) = cos(theta) - i sin(theta)
so
.5 i E^(i theta) = .5 i (cos(theta) + i sin(theta)) = .5 (i cos(theta) - sin(theta))
.5 i E^(i -theta) = .5 i (cos(theta) - i sin(theta)) = .5 (i cos(theta) + sin(theta))

I guess we can let theta = .25 pi t
Can I say,
-.5 i E^(i theta) + .5 i E^(i -theta)
= -.5 (i cos(theta) - sin(theta)) + .5 (i cos(theta) + sin(theta))
= sin(theta)
= sin(.25 pi t)

But where does the 1/4 come into play?

Just put your ##\theta = \frac{\pi t}{4}## in that equation you already have at the top of this post:$$
e^{i\theta}-e^{-i\theta} = 2i\sin\theta$$Lose the decimals, do the substitution, and solve that equation for the sine term.
 

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