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For any $z=x+iy$, prove the following: $$|\sin z| \geq |\sin x|$$ $$|\cos z| \geq |\cos x|$$ $\epsilon$-$\delta $ proof is not required.

I don't really know how to proceed. I know in order to remove the absolute values I can square both sides and I have tried proving this statement using the hyperbolic forms and then the exponential forms but I keep running into circles and I am getting no where...

So for the first one all I have is $$ |\sin z| \geq |\sin x| $$ $$|\sin z|^2=\sin^2x +\sinh^2y \geq \sin^2x $$ or $${1\over4} |e^{iz}-e^{-iz} |^2≥{1\over4}|e^{ix}-e^{-ix}|^2$$And at this point I think I am beginning to overthink how to square absolute values.

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  • $\begingroup$ If you're referring to abs. value of real number then; $$|x+y|^2 = (x+y)^2$$ $\endgroup$ – Faraad Armwood Mar 15 '16 at 15:20
  • $\begingroup$ Thank you but it was more towards complex numbers like is $$ |e^{iz}-e^{-iz}|^2 = (e^{iz}-e^{-iz})(e^{-iz}-e^{iz})? $$ $\endgroup$ – xxXx Mar 15 '16 at 15:35
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Note that since $\cosh(y)\ge 1$, we have

$$\begin{align}|\sin(z)|^2&=|\sin(x)\cosh(y)+i\cos(x)\sinh(y)|^2\\\\ &=\sin^2(x)\cosh^2(y)+\cos^2(x)\sinh^2(y)\\\\ &\ge \sin^2(x) \end{align}$$

and

$$\begin{align}|\cos(z)|^2&=|\cos(x)\cosh(y)+i\sin(x)\sinh(y)|^2\\\\ &=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)\\\\ &\ge \cos^2(x) \end{align}$$

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Hint:

use addition formulas for $\sin z=\sin (x+iy)$ and the fact that $$ \cos (iy)=\cosh y \qquad \sin (iy)=i\sinh y $$

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  • $\begingroup$ Yes i'm aware of these identities..as that is what I used to get to the point where I am at now. What I don't understand is how I can prove that a function of y is greater than a function of x $\endgroup$ – xxXx Mar 15 '16 at 15:30
  • $\begingroup$ It seems that you have not used correctly the addition formula. $\endgroup$ – Emilio Novati Mar 15 '16 at 15:41
  • $\begingroup$ If you use the correct formula for addition you find the result show in the answer of @Dr.MV $\endgroup$ – Emilio Novati Mar 15 '16 at 15:44
  • $\begingroup$ Mine is further simplified than @Dr.MV's answer but thank you $\endgroup$ – xxXx Mar 15 '16 at 15:56

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