Question 3.12: Prove that: (a) d/dz sin z = cos z, (b) d/dz cos z = - sin z...
Prove that: (a) \frac{d}{d z} \sin z=\cos z (b) \frac{d}{d z} \cos z=-\sin z, (c) \frac{d}{d z} \tan z=\sec ^2 z.
(a) We have w = sin z = sin(x + iy) = sin x cosh y + i cos x sinh y. Then
u = sin x cosh y, v = cos x sinh y
Now \partial u / \partial x=\cos x \cosh y=\partial v / \partial y and \partial v / \partial x=-\sin x \sinh y=-(\partial u / \partial y) so that the Cauchy-Riemann equations are satisfied. Hence, by Problem 3.5, the required derivative is equal to
\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=-i \frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}=\cos x \cosh y-i \sin x \sinh y=\cos (x+i y)=\cos zAnother Method
Since \sin z=\frac{e^{i z}-e^{-i z}}{2 i}, we have, using Problem 3.11(b),
\frac{d}{d z} \sin z=\frac{d}{d z}\left(\frac{e^{i z}-e^{-i z}}{2 i}\right)=\frac{1}{2 i} \frac{d}{d z} e^{i z}-\frac{1}{2 i} \frac{d}{d z} e^{-i z}=\frac{1}{2} e^{i z}+\frac{1}{2} e^{-i z}=\cos z(b)
\begin{aligned}\frac{d}{d z} \cos z &=\frac{d}{d z}\left(\frac{e^{i z}+e^{-i z}}{2}\right)=\frac{1}{2} \frac{d}{d z} e^{i z}+\frac{1}{2} \frac{d}{d z} e^{-i z} \\&=\frac{i}{2} e^{i z}-\frac{i}{2} e^{-i z}=-\frac{e^{i z}-e^{-i z}}{2 i}=-\sin z\end{aligned}
The first method of part (a) can also be used.
(c) By the quotient rule of Problem 3.10(c), we have