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> 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,
by NAFV_P 13y ago
> 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 13y agoMy 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 13y agoYeah, 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.