Question

Verify each of the following by using equations (11.4), (12.2), and (12.3).

$\frac{\mathbf{d}}{\mathbf{dz}}\mathbf{cos}\mathbf{}\mathbf{z}\mathbf{=}\mathbf{-}\mathbf{sin}\mathbf{}\mathbf{z}$

Short answer

The equation $\frac{d}{dz}\cos z=-\sin \mathrm{z}$is verified using the equations (11.4), (12.2) and (12.3).

Step-by-step solution

Given information

The given function is $\frac{d}{dz}\cos z=-\sin z$.

Definition of Hyperbolic Function.

A hyperbolic function is a representation of the relationship between a point's distances from the origin to the coordinate axes as a function of an angle.

 Use exponential form to expand the equation

The exponential form of the given equation is,

$$\begin{gathered} \cos z=\frac{e^{zi}+e^{-zi}}{2} \\ \left(1\right) \end{gathered}$$ …. (1)

Differentiate equation (1) with respect to z.

$\frac{d}{dz}\cos \left(z\right)=\frac{1}{2}\times \frac{d}{dz}\left(e^{zi}+e^{-zi}\right)$ …. (2)

Open derivative using algebra of derivatives.

$$\begin{gathered} \frac{d}{dz}\cos \left(z\right)=\frac{1}{2}\times \left(e^{zi}-e^{-zi}\right) \\ =\frac{1}{2}\times \left(e^{zi}-e^{-zi}\right) \end{gathered}$$

Multiply the equation by $\frac{i}{i}$.

$$\begin{gathered} \frac{d}{dz}\cos \left(z\right)=\frac{i}{2}\times \left(e^{zi}-e^{-zi}\right)\times \frac{i}{i} \\ \frac{d}{dz}\cos \left(z\right)=-\frac{\left(e^{zi}-e^{-zi}\right)}{2i} \\ \frac{d}{dz}\cos \left(z\right)=-\sin z \end{gathered}$$

Hence the equation is verified.