Complex number question

[keyan]

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.

[paulsterio]

This isn't even a complex numbers question, it's just trig identities.

[Keyansep]

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?

[Lasercookie]

I came up with this after playing around with things a bit. Doing it purely through identities is quicker and easier though.







Binomial expansion thingie: 1 3 3 1









Equate the real and imaginary coefficients

Real:

Imaginary:

[Jenny_2108]

I came up with this after playing around with things a bit. Doing it purely through identities is quicker and easier though.







Binomial expansion thingie: 1 3 3 1









Equate the real and imaginary coefficients

Real:

Imaginary:

This method is interesting
I think we can apply Pascal triangle as well to get the coefficient quicker

[Lasercookie]

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)

[paulsterio]

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

[Jenny_2108]

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

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

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