Prove: 1^° |sinh y|≤|sin z|≤cosh y; 2^° |sinh y|≤|cos…

Prove: $\ 1^\circ\ |\sinh y|\le|\sin z|\le\cosh y;\quad 2^\circ\ |\sinh y|\le|\cos z|\le\cosh y;\quad 3^\circ\ |\sinh x|\le|\sinh z|\le\cosh x;\quad 4^\circ\ |\sinh x|\le|\cosh z|\le\cosh x\ $ (with $z=x+iy$).

Part $1^\circ$: bounds on $|\sin z|$

From the previous problem, $|\sin z|^2=\sin^2x+\sinh^2y$. Since $\sin^2x\in[0,1]$:

$$ \sinh^2y\ \le\ |\sin z|^2\ \le\ 1+\sinh^2y=\cosh^2y $$

$$ \boxed{|\sinh y|\le|\sin z|\le\cosh y} $$

Part $2^\circ$: bounds on $|\cos z|$

Likewise $|\cos z|^2=\cos^2x+\sinh^2y$ with $\cos^2x\in[0,1]$:

$$ \boxed{|\sinh y|\le|\cos z|\le\cosh y} $$

Part $3^\circ$: bounds on $|\sinh z|$

Using $\sinh(x+iy)=\sinh x\cos y+i\cosh x\sin y$ and $\cosh^2x=1+\sinh^2x$:

$$ |\sinh z|^2=\sinh^2x\cos^2y+\cosh^2x\sin^2y=\sinh^2x\cos^2y+(1+\sinh^2x)\sin^2y=\sinh^2x+\sin^2y $$

Since $\sin^2y\in[0,1]$:

$$ \sinh^2x\ \le\ |\sinh z|^2\ \le\ 1+\sinh^2x=\cosh^2x \qquad\Longrightarrow\qquad \boxed{|\sinh x|\le|\sinh z|\le\cosh x} $$

Part $4^\circ$: bounds on $|\cosh z|$

Using $\cosh(x+iy)=\cosh x\cos y+i\sinh x\sin y$ (derived the same way as $\sinh$, from $e^z+e^{-z}$):

$$ |\cosh z|^2=\cosh^2x\cos^2y+\sinh^2x\sin^2y=\cosh^2x\cos^2y+(\cosh^2x-1)\sin^2y=\cosh^2x-\sin^2y $$

Since $\sin^2y\in[0,1]$:

$$ \cosh^2x-1\ \le\ |\cosh z|^2\ \le\ \cosh^2x \qquad\Longrightarrow\qquad \boxed{|\sinh x|\le|\cosh z|\le\cosh x} $$

Verification

Numerically at $z=0.6+1.3i$: $|\sinh z|\approx1.1549$ and $|\cosh z|\approx0.6906$, both lying in $[|\sinh(0.6)|,\cosh(0.6)]=[0.6367,1.1855]$ ✓. And $|\sin z|,|\cos z|$ both lie in $[|\sinh(1.3)|,\cosh(1.3)]=[1.6984,1.9709]$ when checked against $x=0.6$ (any $x$ gives values inside that $y$-determined band).