Prove: 1^° |sin z|=√(sin^2x+sinh^2y); 2^° |cos…
Prove: $\ 1^\circ\ |\sin z|=\sqrt{\sin^2x+\sinh^2y};\quad 2^\circ\ |\cos z|=\sqrt{\cos^2x+\sinh^2y}\ $ (with $z=x+iy$).
Explanation of Method
Start from the real/imaginary splits of $\sin z$ and $\cos z$ (proved previously), take the modulus, and simplify using $\cosh^2y=1+\sinh^2y$ and $\sin^2x+\cos^2x=1$.
Part $1^\circ$: $|\sin z|=\sqrt{\sin^2x+\sinh^2y}$
From $\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$:
$$ |\sin z|^2=\sin^2x\cosh^2y+\cos^2x\sinh^2y=\sin^2x(1+\sinh^2y)+\cos^2x\sinh^2y $$
$$ =\sin^2x+\sinh^2y(\sin^2x+\cos^2x)=\sin^2x+\sinh^2y $$
$$ \boxed{|\sin z|=\sqrt{\sin^2x+\sinh^2y}} $$
Part $2^\circ$: $|\cos z|=\sqrt{\cos^2x+\sinh^2y}$
From $\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$:
$$ |\cos z|^2=\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(1+\sinh^2y)+\sin^2x\sinh^2y $$
$$ =\cos^2x+\sinh^2y(\cos^2x+\sin^2x)=\cos^2x+\sinh^2y $$
$$ \boxed{|\cos z|=\sqrt{\cos^2x+\sinh^2y}} $$
Special cases
$y=0$: $\ |\sin x|=|\sin x|,\ |\cos x|=|\cos x|$ ✓, both reduce correctly to the real case.
$x=0$ (i.e. $z=iy$): $\ |\sin(iy)|=\sqrt{0+\sinh^2y}=|\sinh y|$, matching $\sin(iy)=i\sinh y$; and $|\cos(iy)|=\sqrt{1+\sinh^2y}=\cosh y$, matching $\cos(iy)=\cosh y$.
A contrast with the real case
$$ |\sin z|^2+|\cos z|^2=(\sin^2x+\cos^2x)+2\sinh^2y=1+2\sinh^2y $$
which equals $1$ only when $y=0$ — unlike the real identity $\sin^2x+\cos^2x=1$, the moduli of complex sine and cosine do not satisfy a Pythagorean relation once $y\neq0$.
Result
These are the two formulas that make the growth of $\sin z,\cos z$ along the imaginary direction explicit — both blow up like $\sinh|y|$ as $y\to\pm\infty$, regardless of $x$.