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[SOLVED] Proof of inequality involving circular and hyperbolic trig. functions

shen07

Member
Aug 14, 2013
54
Hi guys,

Can you help me im 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

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hi guys,

Can you help me im 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 [tex]\displaystyle \begin{align*} \left| \sinh{(y)} \right| = \sqrt{ \sinh^2{(y)}} \end{align*}[/tex], which is clearly [tex]\displaystyle \begin{align*} \leq \sqrt{ 2\sinh^2{(y)} + 1 } \end{align*}[/tex].