Hacker News new | comments | show | ask | jobs | submit login

> Trigonometry was just memorizing equations, not having any real idea why the equations were what they were.

The old proofs of those equations are geometric, they are a bit harder to get your head around than learning complex algebra.

Taking Euler's identity:

  e^(ix)=cos(x)+i*sin(x)

  > e^(ikx)=cos(kx)+i*sin(kx)=(cos(x)+i*sin(x))^k
Expand the right side of the equation, then split into real and imaginary parts. That is a quick way to determine sin(kx) or cos(kx) if sin(x) or cos(x) are already known and k is a positive integer.

EDIT: bloomin' asterisks.




My favorite is a way to prove the sum of angles formula for sine and cosine.

  cos(x1+x2) + isin(x1+x2) = e^((x1+x2)i)
                           = e^(i*x1) * e^(i*x2)
                           = (cos(x1) + isin(x1)) * (cos(x2) + isin(x2))
                           = cos(x1)cos(x2) - sin(x1)sin(x2)
                             + i*(sin(x1)cos(x2) + sin(x2)cos(x1))
Since the imaginary parts have to be equal and the real parts have to be equal:

  sin(x1+x2) = sin(x1)cos(x2) + sin(x2)cos(x1)
  cos(x1+x2) = cos(x1)cos(x2) - sin(x1)sin(x2)


Yeah, that's a more generic form of what I said.

ok, hold on...

  e^(i*x)= cos(x)+i*sin(x)
  e^(-i*x)= cos(-x)+i*sin(-x)= cos(x)-i*sin(x)
  cos(x)= (1/2) * (e^(i*x)+e^(-i*x))
  sin(x)= (-i/2) * (e^(i*x)-e^(-i*x))
  cos(i*x)= (1/2) * (e^(i*i*x)+e^(-i*i*x))=(1/2) * (e^x+e^(-x))
  sin(i*x)= (-i/2) * (e^(x)-e^(-x))
That's the amusing part.




Guidelines | FAQ | Support | API | Security | Lists | Bookmarklet | DMCA | Apply to YC | Contact

Search: