On: Algebraic and calculus concepts may be better way ...

[donatj]

When I started Calculus in high school everything made so much more sense! Trigonometry was just memorizing equations, not having any real idea why the equations were what they were.

With Calculus on the other hand, I could come up with the equations I used in prior classes. I instantly wondered why they didn't start with this?! Trig/Pre Calc was just a waste of my time.

[NAFV_P → donatj]

> 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.

[malisper → NAFV_P]

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)

[NAFV_P → malisper]

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.