Q6P - Mathematical Methods in Physical Sciences

Question

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

$\mathbf{cos}\mathbf{ }\mathbf{2}\mathit{z}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{z}\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{ }\mathit{z}$

Step-by-Step Solution

Answer

The equation $\cos 2z={\cos }^{2}z={\sin }^2 z$is verified using the equations (11.4), (12.2) and (12.3).

Step 1: Given information

The given equation is, $\cos 2z={\cos }^{2 }z-{\sin }^{2}z.$.

Step 2: 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.

Step 3: Use exponential form to expand the equation

The exponential form of the given equation is,

$\cos 2z=\frac{e^{2zi}+e^{-2zi}}{2}$                                                                                                 …. (1)

Multiply the equation (1) by $\frac{2}{2}$.

$$\begin{gathered} \cos 2z=\frac{e^{2zi}+e^{-2zi}}{2}\times \frac{2}{2} \\ \cos 2z=\frac{2e^{2zi}+2e^{-2zi}}{4} \end{gathered}$$

                                                                                                …. (2)

 

Add $2\left[e^{zi}{e}^{-zi}\right]-2\left[e^{zi}{e}^{-zi}\right]$ to square the numerator.

$$\begin{gathered} \cos 2z=\frac{2e^{2zi}+2e^{-2zi}+2\left[e^{zi}{e}^{-zi}\right]}{4}+\frac{e^{2zi}+e^{-2zi}+-2\left[e^{zi}{e}^{-zi}\right]}{4} \\ \cos 2z={\left(\frac{e^{zi}+e^{-zi}}{2}\right)}^2+{\left(\frac{e^{zi}+e^{-zi}}{2}\right)}^2 \\ \cos 2z={\cos }^{2}z-{\sin }^{2}z \end{gathered}$$

 

Hence the equation is verified.