math-square-journal

[image description: a mathematical formula on a plane white background. The formula reads |sin(x+iy) | = | sin(x) + sin(iy) | /end description]  ----------------------------------------------------------------------------------------------------------- Now, following along with the math in the proof to prove this....  I start with what I know, which is the sin addition formula and the relationship between sin and cos and complex exponentials, and also the sinh and cosh formulas. The latter would not be an obvious choice but it’s in the proof I’m following, and they do jump out as I simplify the algebraic expressions.  the sin addition formula gives: sin(a+b) = sin(a)cos(b) + cos(a)sin(b)  so sin(x+iy) = sin(x)cos(iy)+cos(x)sin(iy) given, sin(x) = (1/2i)*[e^ix - e^-ix] and cos(x) = (1/2)*[e^ix + e^-ix] and given, sinh(x) = 1/2[e^x - e^-x] and cosh = 1/2[e^x + e^-x] sin(x+y) = (1/2i)*[e^ix-e^-ix]*(1/2)*[e^i(iy)+e^-i(iy)] + stuff               = (1/4i)*[e^ix-e^-ix] [e^-y+e^y] + stuff                = sin(x)*cosh(y) + stuff                =  sin(x)*cosh(y) + cos(x)*(1/2i)*[e^i(iy)-e^-i(iy)]               =  sin(x)*cosh(y) + (1/2)[ e^ix + e^-ix]*(1/2i)*[e^-y - e^y)]               =  sin(x)*cosh(y) + (-1)(1/4i)[ e^ix + e^-ix]*[e^y - e^-y)] aside: since (-1)[u]*[w_1 - w_2] = u*[w_2-w_1]                =  sin(x)*cosh(y) + (-1/i)cos(x)*sinh(y) aside: (-1/i) = i so                =  sin(x)*cosh(y) + i*cos(x)*sinh(y) Now, the desired formula, |sin(x+iy) | = | sin(x) + sin(iy) |, involves absolute values so we take the absolute value. The absolute value of a complex number is the square root of the sum of the squares of it’s imaginary and real parts. This is intuitive and also what the proof says, so I’m going with it.  | sin(x+iy) |    = | sin(x)*cosh(y) + i*cos(x)*sinh(y) |                       = \sqrt( [sin^2(x)cosh^2(y) + cos^2(x)sinh^2(y)])  aside: we can use the classic trig identity sin^2 + cos^2 = 1                       = \sqrt(  [sin^2(x)cosh^2(y) + (1 - sin^2(x))sinh^2(y)])                      =  \sqrt( [sin^2(x)cosh^2(y) + sinh^2(y) - sin^2(x))sinh^2(y) ])                      = \sqrt( [sin^2(x)*[cosh^2(y)-sinh^2(y)] + sinh^2(y)]) aside: we can change sinh and cosh back into exponential forms to see that  cosh^2-sinh^2 = 1 so                       =  \sqrt( [sin^2(x) + sinh^2(y)]) Which is a very pretty expression but it’s not quite there. It looks like the next step is to hold that thought and then make the other side of the formula also equal to this same expression. Look at the two of them and say, done.  SO the other side of the expression is  | sin(x) + sin(iy) |  we have that sin(iy) = i*sinsh(y)  SO,