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

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

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 $\sqrt{\mathbf{-}\mathbf{1}}$.

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)

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,

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