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Can you help me im stuck:

**By finding the real and imaginary parts of z prove that,**

$$|\sinh(y)|\le|\sin(z)|\le|\cosh(y)|$$

$$|\sinh(y)|\le|\sin(z)|\le|\cosh(y)|$$

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.