MHB Proof of inequality involving circular and hyperbolic trig. functions

[shen07]

Hi guys,

Can you help me I am stuck:

By finding the real and imaginary parts of z prove that,
$$|\sinh(y)|\le|\sin(z)|\le|\cosh(y)|$$

i have tried the following:

Let $$z=x+iy$$,
then $$\sin(z)=sin(x+iy)=\sin(x)\cosh(y)+i\sinh(y)\cos(x)$$

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

And now i am stuck here.

 

[Prove It]

Hi guys,

Can you help me I am stuck:

By finding the real and imaginary parts of z prove that,
$$|\sinh(y)|\le|\sin(z)|\le|\cosh(y)|$$

i have tried the following:

Let $$z=x+iy$$,
then $$\sin(z)=sin(x+iy)=\sin(x)\cosh(y)+i\sinh(y)\cos(x)$$

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

And now i am stuck here.

Well notice that \displaystyle \begin{align*} \left| \sinh{(y)} \right| = \sqrt{ \sinh^2{(y)}} \end{align*}, which is clearly \displaystyle \begin{align*} \leq \sqrt{ 2\sinh^2{(y)} + 1 } \end{align*}.