|sin z|^2 + |cos z|^2= sinh y^2 + coshy^2 ,can you prove that?

Question

z is a complex number.

Answer 1

|sin(z)|^2 + |cos(z)|^2

Where |f(z)|^2 is understood to mean f(z)*~f(z), where ~f(z) is the complex congugate of f(z): ~f(x + iy) = f(x - iy)

Writing the real and imaginary parts of the arguments explicitly, we have:

= |sin(x+iy)|^2 + |cos(x+iy)|^2

Using the sum of angles formulas:

= |(sin(x)cos(iy) + cos(x)sin(iy)|^2 +
|(cos(x)cos(iy) - sin(x)sin(iy)|^2

Expanding in terms of complex congugates:

= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(-iy) + cos(x)sin(-iy)) +
((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(-iy) - sin(x)sin(-iy))

Noting that cos(-x) = cos(x) and sin(-x) = -sin(x):

= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(iy) - cos(x)sin(iy)) +
((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(iy) + sin(x)sin(iy))


Multiplying this out (both terms are the product of a sum and difference of two terms, which yield the difference of squares):

= (sin(x)cos(iy))^2 - (cos(x)sin(iy))^2 + (cos(x)cos(iy))^2 - (sin(x)sin(iy))^2

Factoring:

= ((sin(x))^2 + (cos(x))^2)*cos(iy) - ((sin(x))^2 + (cos(x))^2)*sin(iy)

but ((sin(x))^2 + (cos(x))^2) = 1, so

= (cos(iy))^2 - (sin(iy))^2

From Euler's formula (exp(ix) = cos(x) + i*sin(x)), we can show that:

cos(iy) = (exp(-y) + exp(y))/2 = cosh(y)
and
sin(iy) = (exp(-y) - exp(y))/(2*i) = i*sinh(y)

Substituting these identities in for cos(iy) and sin(iy), we have:

= (cosh(y))^2 - (i*sinh(y))^2

= (cosh(y))^2 + (sinh(y))^2
QED

Answer 2

No, I haven't learned complex trig functions yet.