Q 12P. - Mathematical Methods in Physical Sciences

Question

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

${\mathbf{cos}}^{\mathbf{4}}\mathbf{ }\mathbf{z}\mathbf{+}{\mathbf{sin}}^{\mathbf{4}}\mathbf{z}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{2}\mathbf{z}$

Step-by-Step Solution

Answer

The equation ${\cos }^4 \mathrm{z}+{\sin }^4\mathrm{z}=1-\frac{1}{2}{\sin }^{2}2\mathrm{z}$ is verified using the equations (11.4), (12.2) and (12.3).

Step 1: Given Information

Given equation is ${\cos }^4 \mathrm{z}+{\sin }^4\mathrm{z}=1-\frac{1}{2}{\sin }^{2}2\mathrm{z}$.

Step 2: Definition of Hyperbolic Function.

The term "Hyperbolic Function" refers to the relationship between a point on a hyperbola's distance from its origin and its coordinate axes, expressed as a function of an angle.

Step 3: Solve Left Hand Side(LHS) to prove the given equation.

Given the equation is ${\cos }^4 \mathrm{z}+{\sin }^4\mathrm{z}=1-\frac{1}{2}{\sin }^{2}2\mathrm{z}$.

Takeleft hand side of the given equation ${\cos }^4 \mathrm{z}+{\sin }^4\mathrm{z}$ and prove the right hand side.

 

Now, add and subtract $2{\cos }^2 \mathrm{z}+{\sin }^{2}z$ to the left hand side of the given equation.

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

 

Use property $co{s}^2\theta +si{n}^2\theta =1$ in the above step.

$$\begin{gathered} =1-2{\cos }^2\left(z\right){\sin }^2\left(z\right) \\ =1-\frac{2}{2}\times {\cos }^2\left(z\right){\sin }^2\left(z\right) \\ =1-\frac{1}{2}\times 4{\cos }^2\left(z\right){\sin }^2\left(z\right) \\ =1-\frac{1}{2}{\left[2\cos \left(z\right)\sin \left(z\right)\right]}^2 \end{gathered}$$

Use property $2 \cos \theta \sin \theta =\sin 2 \theta$ in the above step.

$$\begin{gathered} =1-\frac{1}{2}{\left[\sin 2z\right]}^2 \\ =1-\frac{1}{2}{\sin }^2\left(2z\right) \end{gathered}$$

The result is equal to right hand side. Hence, the equation is verified.