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!
Best Answer
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