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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: keyan on August 09, 2012, 05:32:24 pm

Title: Complex number question
Post by: keyan on August 09, 2012, 05:32:24 pm: Hi guys a bit stuck on this question here.

sin(3x) = 3cos^2(x)*sin(x) - sin^3(x) (x = theta just using x to be clear)

The question is a carry on and says HENCE show that just above. The question before is z3 = cis(x) explain why z3^n = cis(nx).

It's been a long time since complex numbers so i'm stumped atm with this.
Edit: It's a follow up question after proving de moivres theorem so it somehow relates so i don't think you can just use trig identities to solve to clarify.
Title: Re: Complex number question
Post by: paulsterio on August 09, 2012, 05:48:33 pm: This isn't even a complex numbers question, it's just trig identities.

$ sin(3x) = sin(2x+x) = sin(2x)cos(x) + sin(x)cos(2x) $

$ =2sin(x)cos(x)cos(x) + sin(x)(cos^2(x)-sin^2(x)) $

$ =2sin(x)cos^2(x)+sin(x)cos^2(x)-sin^3(x) $

$ =3cos^2(x)sin(x)-sin^3(x) $
Title: Re: Complex number question
Post by: Keyansep on August 09, 2012, 06:06:38 pm: It's basis is off De Moivres theorem though? It says hence so i thought to get the full marks you would need to somehow apply that?
Title: Re: Complex number question
Post by: Lasercookie on August 09, 2012, 06:33:37 pm: I came up with this after playing around with things a bit. Doing it purely through identities is quicker and easier though.

$z_3 = cis(x)$

$(z_3)^3 = cis(3x)$

$cis(x)^3 = (cos(x) + isin(x))^3 = cos(3x) + isin(3x)$

Binomial expansion thingie: 1 3 3 1

$= cos^3(x)i^0sin^0(x) + 3cos^2(x)isin(x) + 3cos(x)i^2sin^2(x) + cos^0(x)i^3sin^3(x)$

$= cos^3(x) + 3cos^2(x)isin(x) + 3cos(x)i^2sin^2(x) - isin^3(x)$

$= cos^3(x) + 3cos^2(x)isin(x) - 3cos(x)sin^2(x) - isin^3(x)$

$cis(x)^3 = cos(3x) + isin(3x) = cos^3(x) - 3cos(x)sin^2(x) + (3cos^2(x)sin(x) - sin^3(x))i$

Equate the real and imaginary coefficients

Real: $cos(3x) = cos^3(x) - 3cos(x)sin^2(x)$

Imaginary: $sin(3x) = 3cos^2(x)sin(x) - sin^3(x)$
Title: Re: Complex number question
Post by: Jenny_2108 on August 09, 2012, 06:58:55 pm: Quote from: laseredd on August 09, 2012, 06:33:37 pm
I came up with this after playing around with things a bit. Doing it purely through identities is quicker and easier though.

$z_3 = cis(x)$

$(z_3)^3 = cis(3x)$

$cis(x)^3 = (cos(x) + isin(x))^3 = cos(3x) + isin(3x)$

Binomial expansion thingie: 1 3 3 1

$= cos^3(x)i^0sin^0(x) + 3cos^2(x)isin(x) + 3cos(x)i^2sin^2(x) + cos^0(x)i^3sin^3(x)$

$= cos^3(x) + 3cos^2(x)isin(x) + 3cos(x)i^2sin^2(x) - isin^3(x)$

$= cos^3(x) + 3cos^2(x)isin(x) - 3cos(x)sin^2(x) - isin^3(x)$

$cis(x)^3 = cos(3x) + isin(3x) = cos^3(x) - 3cos(x)sin^2(x) + (3cos^2(x)sin(x) - sin^3(x))i$

Equate the real and imaginary coefficients

Real: $cos(3x) = cos^3(x) - 3cos(x)sin^2(x)$

Imaginary: $sin(3x) = 3cos^2(x)sin(x) - sin^3(x)$

This method is interesting
I think we can apply Pascal triangle as well to get the coefficient quicker
Title: Re: Complex number question
Post by: Lasercookie on August 09, 2012, 07:04:37 pm: Quote from: Jenny_2108 on August 09, 2012, 06:58:55 pm
I think we can apply Pascal triangle as well to get the coefficient quicker
Would you be able to show me what you mean? I used that triangle to get the coefficients of 1 3 3 1 out for the binomial expansion, is there more to that method than just that? (I hope I haven't been doing things the long way around for all this time)
Title: Re: Complex number question
Post by: paulsterio on August 09, 2012, 07:08:32 pm: For cubics, you should just remember that it's 1, 3, 3, 1 and for quartics you should just remember that it's 1, 4, 6, 4, 1.

Apart from that, use the triangle :P
Title: Re: Complex number question
Post by: Jenny_2108 on August 09, 2012, 07:36:47 pm: Quote from: laseredd on August 09, 2012, 07:04:37 pm
Quote from: Jenny_2108 on August 09, 2012, 06:58:55 pm
I think we can apply Pascal triangle as well to get the coefficient quicker
Would you be able to show me what you mean? I used that triangle to get the coefficients of 1 3 3 1 out for the binomial expansion, is there more to that method than just that? (I hope I haven't been doing things the long way around for all this time)

Click here to see the Pascal triangle
http://daugerresearch.com/vault/PascalsTriangle.gif

Quote from: paulsterio on August 09, 2012, 07:08:32 pm
For cubics, you should just remember that it's 1, 3, 3, 1 and for quartics you should just remember that it's 1, 4, 6, 4, 1.

Apart from that, use the triangle :P

Yeah, as Paul's just explained. You get the next row by finding the sum of two numbers of the above row. The start and end numbers are always 1