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I just want you guys to double check if I'm on the right track.

so to prove $\sin '(z)$ = $ \cos (z)$

I did this: $\sin (z)$ = $\frac {e^{iz} - e^{-iz}}{2i}$

$\sin '(z)$ = $\frac {i(e^{iz} + e^{-iz})}{2i}$ = $\frac {e^{iz} + e^{-iz}}{2}$ = $\cos (z)$

is that correct?

and for $\cos '(z)$ = $-\sin (z)$ I did something similar.

$\cos (z)$ = $\frac {e^{iz} + e^{-iz}}{2}$

$\cos '(z)$ = $\frac {i(e^{iz} - e^{-iz})}{2}$ = $\frac {i(e^{iz} - e^{-iz})}{2}* \frac{i}{i}$ = $\frac {i^2(e^{iz} - e^{-iz})}{2i}$ = $\frac {-e^{iz} + e^{-iz}}{2i}$ = $-\sin (z)$

or using the Cauchy-Riemann formula/equation:

$f '(z_0)= \frac{du}{dx} (x_0, y_0) - i \frac{du}{dy} (x_0,y_0)$

where

$\frac{du}{dx} (x_0,y_0) = \frac{dv}{dy} (x_0,y_0)$ and $\frac{du}{dy} (x_0,y_0) = -\frac{dv}{dx} (x_0,y_0)$

so $\cos z = \cos x\cosh y - i\sin x\sinh y$ using the formula/equations we get:

$\cos '(z) = -\sin x\cosh y - i\cos x\sinh y$. I believe this is the correct way. but please correct me if I'm wrong. Thank you!

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    $\begingroup$ To know if you proof is correct we need to know which definition of sin and cos you use. Why did you use the complex analysis tag? the real analysis (or even the calculus tag) would be better for this question $\endgroup$ Apr 15, 2013 at 14:18
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    $\begingroup$ @DominicMichaelis, I think Tim means they are complex-valued functions. $\endgroup$
    – Easy
    Apr 15, 2013 at 14:21
  • $\begingroup$ @DominicMichaelis do you mean whether we use $\cos (z)$ = $\frac {e^{iz} - e^{-iz}}{2i}$ or whether it equals $\cos x\cosh y - i \sin x\sinh y$ ? $\endgroup$
    – Tim
    Apr 15, 2013 at 14:21
  • $\begingroup$ @Easy Yes I'm referring to the $\mathbb{C}$ derivative of $\cos (z)$. Thank you. $\endgroup$
    – Tim
    Apr 15, 2013 at 14:23
  • $\begingroup$ I understand how to get it both ways now. I realized I can use the Cauchy-Riemann equations to solve it the other way as well. Thanks for the help and insight everyone! $\endgroup$
    – Tim
    Apr 15, 2013 at 14:30

1 Answer 1

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Looks okay (although I am not sure about the Cauchy-Riemann approach). The best way to prove this is by using the exponential definition of $\sin$e or $\cos$ine to verify the derivatives. Just a quick note; For a more rigorous method, you can also prove the result by differentiation from first principles: \begin{equation*} \frac{d}{dx}\sin(x)=\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h} \end{equation*} and apply the double angle formulae. Proceed in a similar fashion for $\cos$ine.

You may also be interested in the following geometric approach to your derivative problem:

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-8-limits-of-sine-and-cosine/MIT18_01SCF10_Ses8d.pdf

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