Q28P - Mathematical Methods in Physical Sciences

Question

As in Problem 27, find the formulas for $\mathbf{(}\mathit{f}\mathit{o}\mathit{r}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{3}\mathit{\theta }\mathbf{,}\mathbf{ }\mathbf{cos}\mathbf{3}\mathit{\theta }\mathbf{)}$.

Step-by-Step Solution

Answer

The formula is $\cos 3\theta ={\cos }^3\theta -3 \cos \theta {\sin }^2 \theta , \sin 3\theta =3 \sin \theta .{\cos }^2\theta -{\sin }^2\theta$, 

Step 1: Given Information

To find the formulas for $\cos 3\theta and \sin 3\theta$ and .

Step 2: Definition of the complex number

Complex numbers possess real numbers and imaginary numbers; a complex  can be written in the form of: 

$\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$  

 

Here x and y are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{\mathbf{-}\mathbf{1}}$.

Step 3: Finding an expression for s i n   3 θ   a n d   c o s   3 θ and

Exponential form for z ;

$$\begin{gathered} \mathrm{U}={\mathrm{e}}^{{\left(\mathrm{i\theta }\right)}^3} \\ ={\mathrm{e}}^{\left(3\mathrm{i\theta }\right)} \\ =\cos 3\mathrm{\theta }+\mathrm{i} \sin 3\mathrm{\theta } \end{gathered}$$ 

 

Use the same principle to get,

$$\begin{gathered} U=e^{{\left(i\theta \right)}^3} \\ ={\left[\cos \theta + \sin \theta \right]}^3 \end{gathered}$$ 

                                                         ……. (1)

Simplifying using Newton Theorem to get the expansion of (1),

 $$\begin{gathered} {\left[\cos \theta +\sin \theta \right]}^3={\cos }^3\theta + 3 {\cos }^2 \theta \left[i \sin \theta \right]+3 {\cos }^2 \theta {\left[i \sin \theta \right]}^2+{\left[i \sin \theta \right]}^3 \\ ={\cos }^3\theta + 3 {\cos }^2 \theta \sin \theta i - 3 \cos \theta {\sin }^2\theta -{\sin }^3 \theta i \\ ={\cos }^3\theta - 3 \cos \theta {\sin }^2 \theta +\left[3 {\cos }^2\theta \sin \theta -{\sin }^3\theta \right]i \end{gathered}$$    ...(2)

Step 4: Comparing the equations

Compare equations (1) and (2) as:

$$\begin{gathered} \cos 3\theta ={\cos }^3\theta - 3 \cos \theta {\sin }^2 \theta \\ \sin 3\theta =3 \sin \theta . {\cos }^2 \theta - {\sin }^3\theta \end{gathered}$$ 

 

Hence the formula will be,

 $\cos 3\theta ={\cos }^3\theta -3 \cos \theta {\sin }^2 \theta , \sin 3\theta =3 \sin \theta .{\cos }^2\theta -{\sin }^2\theta$